Solved using series method

Ordinary point and regular singular point are supported. irregular singular point support will be added in the future. Expansion around point other than zero is also supported, including initial conditions. All three cases of regular point are supported, these are when the roots on indicial equation are repeated, or diﬀer by an integer, or diﬀer by non integer. case of Complex roots of indicial equation is also supported. Only second order and ﬁrst order series solution is supported. Higher order ode support will be added in the future.

4.3.2.5.1 Ordinary point using Taylor series method ode internal name "second_order_taylor_series_method_ordinary_point"

Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives

\begin{align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\end{align*}

\begin{align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end{align}

And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as

\begin{align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6}\end{align}

\begin{equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7}\end{equation}

4.3.2.5.2 Ordinary point using power series method ode internal name "second_order_power_series_method_ordinary_point"

Expansion point is an ordinary point. Using standard power series. For an ordinary point, and for inhomogeneous. ode, always generate the full solution directly from the summation. Do not split the problem into \(y_{h},y_{p}\,\). To be able to do this, we have to express the RHS as Taylor series (expand it around the same expansion point). If the RHS is already a polynomial in \(x\) then there is nothing to do as it is already in Taylor series form. Examples below show how to do this. When the RHS is not zero, do not attempt to ﬁnd recurrence relation as the RHS will get in the way, If the RHS is zero, then ﬁnd recurrence relation.

In this method, we let Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n}\) and replace this in the above ode and solve for \(a_{n}\) using recurrence relation. Examples below show how these methods work.

\begin{align*} y^{\prime \prime }+xy^{\prime }+y & =2x+x^{2}+x^{4}\\ y^{\prime \prime } & =-xy^{\prime }-y+2x+x^{2}+x^{4}\\ y^{\prime \prime } & =f\left ( x,y,y^{\prime }\right ) \end{align*}

\begin{align*} F_{0} & =f\left ( x,y,y^{\prime }\right ) \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0}\end{align*}

\begin{align*} F_{1} & =\frac {\partial \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) }{\partial x}+\frac {\partial \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) }{\partial y}y^{\prime }+\frac {\partial \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) }{\partial y^{\prime }}y^{\prime \prime }\\ & =\left ( 4x^{3}+2x-y^{\prime }+2\right ) -y^{\prime }-xy^{\prime \prime }\\ & =2x-2y^{\prime }-xy^{\prime \prime }+4x^{3}+2 \end{align*}

\begin{align*} F_{2} & =\frac {d}{dx}\left ( F_{n-1}\right ) \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) +\\ & +\left ( \frac {\partial }{\partial y}\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) \right ) y^{\prime }\\ & +\left ( \frac {\partial }{\partial y^{\prime }}\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) \right ) y^{\prime \prime }\\ & =\left ( y-4x+2xy^{\prime }+9x^{2}-5x^{4}+2\right ) +xy^{\prime }+\left ( -2+x^{2}\right ) y^{\prime \prime }\\ & =y-4x-2y^{\prime \prime }+3xy^{\prime }+x^{2}y^{\prime \prime }+9x^{2}-5x^{4}+2 \end{align*}

\begin{align*} F_{2} & =y-4x-2\left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) +3xy^{\prime }+x^{2}\left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) +9x^{2}-5x^{4}+2\\ & =3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2 \end{align*}

\begin{align*} F_{3} & =\frac {d}{dx}\left ( F_{2}\right ) \\ & =\frac {\partial }{\partial x}F_{2}+\left ( \frac {\partial F_{2}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{2}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}\left ( 3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2\right ) \\ & +\left ( \frac {\partial }{\partial y}\left ( 3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2\right ) \right ) y^{\prime }\\ & +\left ( \frac {\partial }{\partial y^{\prime }}\left ( 3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2\right ) \right ) y^{\prime \prime }\\ & =14x+5y^{\prime }-3x^{2}y^{\prime }-2xy+6x^{2}-24x^{3}+6x^{5}-8+\left ( 3-x^{2}\right ) y^{\prime }+\left ( 5x-x^{3}\right ) y^{\prime \prime }\end{align*}

\begin{align*} F_{3} & =14x+5y^{\prime }-3x^{2}y^{\prime }-2xy+6x^{2}-24x^{3}+6x^{5}-8+\left ( 3-x^{2}\right ) y^{\prime }+\left ( 5x-x^{3}\right ) \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) \\ & =14x+8y^{\prime }+x^{3}y-9x^{2}y^{\prime }+x^{4}y^{\prime }-7xy+16x^{2}-19x^{3}-2x^{4}+10x^{5}-x^{7}-8 \end{align*}

And so on. Evaluating each of the above at \(x=0,y=y_{0},y^{\prime }=y_{0}^{\prime }\) gives

\begin{align*} F_{0} & =\left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) _{x=0,y_{0},y_{0}^{\prime }}=-y_{0}\\ F_{1} & =\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) _{x=0,y_{0},y_{0}^{\prime }}=\left ( -2y_{0}^{\prime }+2\right ) \\ F_{2} & =3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2=3y_{0}+2\\ F_{3} & =14x+8y^{\prime }+x^{3}y-9x^{2}y^{\prime }+x^{4}y^{\prime }-7xy+16x^{2}-19x^{3}-2x^{4}+10x^{5}-x^{7}-8=8y_{0}^{\prime }-8 \end{align*}

\begin{align*} y\left ( x\right ) & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}F_{0}+\frac {x^{3}}{6}F_{1}+\frac {x^{4}}{24}F_{2}+\frac {x^{5}}{5!}F_{3}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left ( -y_{0}\right ) +\frac {x^{3}}{6}\left ( -2y_{0}^{\prime }+2\right ) +\frac {x^{4}}{24}\left ( 3y_{0}+2\right ) +\frac {x^{5}}{5!}\left ( 8y_{0}^{\prime }-8\right ) +\cdots \\ & =y_{0}\left ( 1-\frac {x^{2}}{2}+\frac {1}{8}x^{4}+\cdots \right ) +y_{0}^{\prime }\left ( x-\frac {x^{3}}{3}+\frac {1}{15}x^{4}\cdots \right ) +\left ( \frac {1}{3}x^{3}+\frac {1}{12}x^{4}-\frac {1}{15}x^{4}\right ) \\ & =c_{1}\left ( 1-\frac {x^{2}}{2}+\frac {1}{8}x^{4}+\cdots \right ) +c_{2}\left ( x-\frac {x^{3}}{3}+\frac {1}{15}x^{4}\cdots \right ) +\left ( \frac {1}{3}x^{3}+\frac {1}{12}x^{4}-\frac {1}{15}x^{4}\right ) \end{align*}

Comparing the homogenous ode to \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0\) shows that \(p\left ( x\right ) =x,q\left ( x\right ) =1\). These are deﬁned everywhere. Let the expansion point be \(x_{0}=0\). This is ordinary point since \(p\left ( x\right ) ,q\left ( x\right ) \) are deﬁned at \(x_{0}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n}\). Hence \(y^{\prime }=\sum _{n=0}^{\infty }na_{n}x^{n-1}=\sum _{n=1}^{\infty }na_{n}x^{n-1}\) and \(y^{\prime \prime }=\sum _{n=1}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}=\sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}\). The homogenous ode becomes

\begin{align} \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+x\sum _{n=1}^{\infty }na_{n}x^{n-1}+\sum _{n=0}^{\infty }a_{n}x^{n} & =2x+x^{2}+x^{4}\nonumber \\ \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+\sum _{n=1}^{\infty }na_{n}x^{n}+\sum _{n=0}^{\infty }a_{n}x^{n} & =2x+x^{2}+x^{4}\nonumber \end{align}

\begin{align*} 2a_{2}+a_{0} & =0\\ a_{2} & =-\frac {1}{2}a_{0}\end{align*}

\begin{align*} 6a_{3}+2a_{1} & =2\\ a_{3} & =\frac {2-2a_{1}}{6}\\ & =\frac {1}{3}-\frac {1}{3}a_{1}\end{align*}

\begin{align*} 12a_{4}+3a_{2} & =1\\ a_{4} & =\frac {1-3a_{2}}{12}\\ & =\frac {1-3\left ( -\frac {1}{2}a_{0}\right ) }{12}\\ & =\frac {1}{8}a_{0}+\frac {1}{12}\end{align*}

\begin{align*} 20a_{5}+4a_{3} & =0\\ a_{5} & =\frac {-4a_{3}}{20}\\ & =\frac {-4\left ( \frac {1}{3}-\frac {1}{3}a_{1}\right ) }{20}\\ & =\frac {1}{15}a_{1}-\frac {1}{15}\end{align*}

\begin{align*} 30a_{6}+5a_{4} & =1\\ a_{6} & =\frac {1-5a_{4}}{30}\\ & =\frac {1-5\left ( \frac {1}{8}a_{0}+\frac {1}{12}\right ) }{30}\\ & =\frac {7}{360}-\frac {1}{48}a_{0}\end{align*}

\begin{align*} \left ( n\right ) \left ( n-1\right ) a_{n}+\left ( n-2\right ) a_{n-2}+a_{n-2} & =0\\ a_{n} & =-\frac {n-1}{n\left ( n-1\right ) }a_{n-2}\end{align*}

\begin{align*} a_{7} & =-\frac {6}{42}a_{5}\\ & =-\frac {6}{42}\left ( \frac {1}{15}a_{1}-\frac {1}{15}\right ) \\ & =\frac {1}{105}-\frac {1}{105}a_{1}\end{align*}

\begin{align*} a_{8} & =-\frac {7}{\left ( 8\right ) \left ( 7\right ) }a_{6}\\ & =-\frac {7}{\left ( 8\right ) \left ( 7\right ) }\left ( \frac {7}{360}-\frac {1}{48}a_{0}\right ) \\ & =\frac {1}{384}a_{0}-\frac {7}{2880}\end{align*}

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \\ & =a_{0}+a_{1}x-\frac {1}{2}a_{0}x^{2}+\left ( \frac {1}{3}-\frac {1}{3}a_{1}\right ) x^{3}+\left ( \frac {1}{8}a_{0}+\frac {1}{12}\right ) x^{4}+\left ( \frac {1}{15}a_{1}-\frac {1}{15}\right ) x^{5}+\left ( \frac {7}{360}-\frac {1}{48}a_{0}\right ) x^{6}+\left ( \frac {1}{105}-\frac {1}{105}a_{1}\right ) x^{7}+\cdots \\ & =a_{0}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{8}x^{4}-\frac {1}{48}x^{6}+\cdots \right ) +a_{1}\left ( x-\frac {1}{3}x^{3}+\frac {1}{15}x^{5}-\frac {1}{105}x^{7}-\cdots \right ) +\left ( \frac {1}{3}x^{3}+\frac {1}{12}x^{4}-\frac {1}{15}x^{5}+\frac {7}{360}x^{6}+\frac {1}{105}x^{7}+\cdots \right ) \end{align*}

Which is the same answer given using the Taylor series method. We see that the Taylor series method is much simpler, but requires using the computer to calculate the derivatives as they become very complicated as more terms are needed.

Even though the expansion point is ordinary, we can also solve this using Frobenius series as follows. Comparing the ode \(y^{\prime \prime }+xy^{\prime }+y=0\) to

Hence \(p\left ( x\right ) =x,q\left ( x\right ) =1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) & =0\\ r & =1,0 \end{align*}

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\end{align*}

\begin{align*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=2}^{\infty }\left ( n+r-2\right ) a_{n-2}x^{n+r-2}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r-2}=0 \tag {1}\end{equation}

\begin{equation} y^{\prime \prime }+xy^{\prime }+y=r\left ( r-1\right ) a_{0}x^{r} \tag {2}\end{equation}

For \(r=1\) we see that \(a_{1}=0\). But for \(r=0\) then the above gives \(0b_{1}=0\). This means \(b_{1}\) can be any value and we choose \(b_{1}=0\) in this case.

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-2\right ) a_{n-2}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-\left ( n+r-2\right ) a_{n-2}-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) }=\frac {-\left ( n+r-1\right ) a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) } \tag {3}\end{align}

Now we ﬁnd \(y_{1}\) which is associated with \(r=1\). From (3) and for \(r=1\) it becomes

\begin{equation} a_{n}=-\frac {n}{\left ( n+1\right ) n}a_{n-2}=-\frac {1}{n+1}a_{n-2} \tag {4}\end{equation}

\begin{align*} y_{1} & =\sum a_{n}x^{n+r_{1}}\\ & =x\sum a_{n}x^{n}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+\cdots \right ) \\ & =x\left ( 1-\frac {1}{2}x^{2}+\frac {1}{10}x^{4}-\cdots \right ) \\ & =x-\frac {1}{3}x^{3}+\frac {1}{15}x^{5}-\cdots \end{align*}

Now we ﬁnd \(y_{2}\) associated with \(r=0\). From (3) this becomes (using \(b\) instead of \(a\)) and \(r=0\)

\begin{align} b_{n} & =\frac {-\left ( n+r-1\right ) b_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) }\nonumber \\ & =\frac {-\left ( n-1\right ) b_{n-2}}{\left ( n\right ) \left ( n-1\right ) }\nonumber \\ & =-\frac {b_{n-2}}{n} \tag {5}\end{align}

From above, we found that \(b_{1}=0\). Now we use (5) to ﬁnd all \(b_{n}\) for \(n\geq 2\). For \(n=2\)

\begin{align*} y_{2} & =\sum b_{n}x^{n+r_{2}}\\ & =\sum b_{n}x^{n}\\ & =\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\left ( 1-\frac {1}{2}x^{2}+\frac {1}{8}x^{4}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x-\frac {1}{3}x^{3}+\frac {1}{15}x^{5}-\cdots \right ) +c_{2}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{8}x^{4}+\cdots \right ) \end{align*}

We see this is the same \(y_{h}\) obtained using standard power series. This shows that we can also use Frobenius series to solve for ordinary point. The roots will always be \(r_{1}=1,r_{2}=0\). But this requires more work than using standard power series. The main advantage of using Frobenius series for ordinary point comes in when the RHS has no series expansion at \(x=0\). For example, if the RHS in this ode was say \(\sqrt {x}\) then we must use Frobenius to be able to solve it as standard power series will fail, since\(\sqrt {x}\) has no series representation at \(x=0\). Examples below shows how to do this.

\begin{align*} y^{\prime \prime } & =-x^{5}\left ( y^{\prime }+y\right ) \\ & =-x^{5}y-x^{5}y^{\prime }\\ y^{\prime \prime } & =f\left ( x,y,y^{\prime }\right ) \end{align*}

\begin{align*} F_{1} & =\frac {\partial \left ( -x^{5}y-x^{5}y^{\prime }\right ) }{\partial x}+\frac {\partial \left ( -x^{5}y-x^{5}y^{\prime }\right ) }{\partial y}y^{\prime }+\frac {\partial \left ( -x^{5}y-x^{5}y^{\prime }\right ) }{\partial y^{\prime }}y^{\prime \prime }\\ & =\left ( -5x^{4}y-5x^{4}y^{\prime }\right ) -x^{5}y^{\prime }-x^{5}y^{\prime \prime }\end{align*}

\begin{align*} F_{1} & =\left ( -5x^{4}y-5x^{4}y^{\prime }\right ) -x^{5}y^{\prime }-x^{5}\left ( -x^{5}y-x^{5}y^{\prime }\right ) \\ & =x^{10}y-5x^{4}y-5x^{4}y^{\prime }-x^{5}y^{\prime }+x^{10}y^{\prime }\end{align*}

\begin{align*} F_{2} & =\frac {d}{dx}\left ( F_{n-1}\right ) \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}\left ( x^{10}y-5x^{4}y-5x^{4}y^{\prime }-x^{5}y^{\prime }+x^{10}y^{\prime }\right ) +\\ & +\left ( \frac {\partial }{\partial y}\left ( x^{10}y-5x^{4}y-5x^{4}y^{\prime }-x^{5}y^{\prime }+x^{10}y^{\prime }\right ) \right ) y^{\prime }\\ & +\left ( \frac {\partial }{\partial y^{\prime }}\left ( x^{10}y-5x^{4}y-5x^{4}y^{\prime }-x^{5}y^{\prime }+x^{10}y^{\prime }\right ) \right ) y^{\prime \prime }\\ & =\left ( 10x^{9}y-20x^{3}y-20x^{3}y^{\prime }-5x^{4}y^{\prime }+10x^{9}y^{\prime }\right ) +x^{4}\left ( x^{6}-5\right ) y^{\prime }+\left ( -5x^{4}-x^{5}+x^{10}\right ) y^{\prime \prime }\end{align*}

\begin{align*} F_{2} & =\left ( 10x^{9}y-20x^{3}y-20x^{3}y^{\prime }-5x^{4}y^{\prime }+10x^{9}y^{\prime }\right ) +x^{4}\left ( x^{6}-5\right ) y^{\prime }+\left ( -5x^{4}-x^{5}+x^{10}\right ) \left ( -x^{5}\left ( y^{\prime }+y\right ) \right ) \\ & =-x^{3}\left ( 20y+20y^{\prime }+10xy^{\prime }-15x^{6}y-x^{7}y+x^{12}y-15x^{6}y^{\prime }-2x^{7}y^{\prime }+x^{12}y^{\prime }\right ) \end{align*}

\begin{align*} F_{3} & =\frac {d}{dx}\left ( F_{2}\right ) \\ & =\frac {\partial }{\partial x}F_{2}+\left ( \frac {\partial F_{2}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{2}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}\left ( -x^{3}\left ( 20y+20y^{\prime }+10xy^{\prime }-15x^{6}y-x^{7}y+x^{12}y-15x^{6}y^{\prime }-2x^{7}y^{\prime }+x^{12}y^{\prime }\right ) \right ) \\ & +\left ( \frac {\partial }{\partial y}\left ( -x^{3}\left ( 20y+20y^{\prime }+10xy^{\prime }-15x^{6}y-x^{7}y+x^{12}y-15x^{6}y^{\prime }-2x^{7}y^{\prime }+x^{12}y^{\prime }\right ) \right ) \right ) y^{\prime }\\ & +\left ( \frac {\partial }{\partial y^{\prime }}\left ( -x^{3}\left ( 20y+20y^{\prime }+10xy^{\prime }-15x^{6}y-x^{7}y+x^{12}y-15x^{6}y^{\prime }-2x^{7}y^{\prime }+x^{12}y^{\prime }\right ) \right ) \right ) y^{\prime \prime }\\ & =-5x^{2}\left ( 12y+12y^{\prime }+8xy^{\prime }-27x^{6}y-2x^{7}y+3x^{12}y-27x^{6}y^{\prime }-4x^{7}y^{\prime }+3x^{12}y^{\prime }\right ) +x^{3}\left ( -x^{12}+x^{7}+15x^{6}-20\right ) y^{\prime }+\left ( -20x^{3}-10\allowbreak x^{4}+15x^{9}+2x^{10}-x^{15}\right ) y^{\prime \prime }\end{align*}

\begin{align*} F_{3} & =-5x^{2}\left ( 12y+12y^{\prime }+8xy^{\prime }-27x^{6}y-2x^{7}y+3x^{12}y-27x^{6}y^{\prime }-4x^{7}y^{\prime }+3x^{12}y^{\prime }\right ) +x^{3}\left ( -x^{12}+x^{7}+15x^{6}-20\right ) y^{\prime }+\left ( -20x^{3}-10\allowbreak x^{4}+15x^{9}+2x^{10}-x^{15}\right ) \left ( -x^{5}\left ( y^{\prime }+y\right ) \right ) \\ & =-x^{2}\left ( 60y+60y^{\prime }+60xy^{\prime }-155x^{6}y-20x^{7}y+30x^{12}y+2x^{13}y-x^{18}y-155x^{6}y^{\prime }-45x^{7}y^{\prime }-x^{8}y^{\prime }+30x^{12}y^{\prime }+3x^{13}y^{\prime }-x^{18}y^{\prime }\right ) \end{align*}

And so on. Since the derivatives become very complicated, the result was done on the computer which results in (Evaluating each of the above at \(x=0,y=y_{0},y^{\prime }=y_{0}^{\prime }\))

\begin{align*} F_{0} & =0\\ F_{1} & =0\\ F_{2} & =0\\ F_{3} & =0\\ F_{4} & =0\\ F_{5} & =-120y_{0}^{\prime }-120y_{0}\\ F_{6} & =-720y_{0}^{\prime }\\ F_{7} & =0\\ F_{8} & =0\\ F_{9} & =0\\ F_{10} & =0\\ F_{11} & =6652800y_{0}^{\prime }+6652800y_{0}\\ F_{12} & =79833600y_{0}^{\prime }+11404800y_{0}\\ F_{13} & =111196800y_{0}^{\prime }\\ F_{14} & =0\\ & \vdots \end{align*}

\begin{align*} y\left ( x\right ) & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{7}}{7!}\left ( -120y_{0}^{\prime }-120y_{0}\right ) -\frac {x^{8}}{8!}\left ( 720y_{0}^{\prime }\right ) +\frac {x^{13}}{13!}\left ( 6652800y_{0}^{\prime }+6652800y_{0}\right ) \\ & +\frac {x^{14}}{14!}\left ( 79833600y_{0}^{\prime }+11404800y_{0}\right ) +\frac {x^{15}}{15!}\left ( 111196800y_{0}^{\prime }\right ) +\cdots \\ & =y_{0}\left ( 1-\frac {120}{7!}x^{7}+\frac {6652800}{13!}x^{13}+\frac {11404800}{14!}x^{14}-\cdots \right ) +y_{0}^{\prime }\left ( x-\frac {120}{7!}x^{7}-\frac {720}{8!}x^{8}+\frac {6652800}{13!}x^{13}+\frac {79833600}{14!}x^{14}+\frac {111196800}{15!}x^{15}+\cdots \right ) \\ & =y_{0}\left ( 1-\frac {1}{42}x^{7}+\frac {1}{936}x^{13}+\frac {1}{7644}x^{14}+\cdots \right ) +y_{0}^{\prime }\left ( x-\frac {1}{42}x^{7}-\frac {1}{56}x^{8}+\frac {1}{936}x^{13}+\frac {1}{1092}x^{14}+\frac {1}{11\,760}x^{15}+\cdots \right ) \end{align*}

Expansion around \(x=0\). This is ordinary point. Since RHS is zero, we will ﬁnd recurrence relation.

Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n}\). Hence \(y^{\prime }=\sum _{n=0}^{\infty }na_{n}x^{n-1}=\sum _{n=1}^{\infty }na_{n}x^{n-1}\) and \(y^{\prime \prime }=\sum _{n=1}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}=\sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}\). The ode becomes

\begin{align} \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+\sum _{n=1}^{\infty }na_{n}x^{n-1}+\sum _{n=0}^{\infty }a_{n}x^{n} & =0\nonumber \\ \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-7}+\sum _{n=1}^{\infty }na_{n}x^{n-1}+\sum _{n=0}^{\infty }a_{n}x^{n} & =0\nonumber \end{align}

\begin{equation} \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-7}+\sum _{n=7}^{\infty }\left ( n-6\right ) a_{n-6}x^{n-7}+\sum _{n=7}^{\infty }a_{n-7}x^{n-7}=0 \tag {1}\end{equation}

For \(n=2,3,4,5,6\) it generates \(a_{2}=0,a_{3}=0,a_{4}=0,a_{5}=0,a_{6}=0\) since there is only one term in each one of these and the RHS is zero.

\begin{align} \left ( n\right ) \left ( n-1\right ) a_{n}+\left ( n-6\right ) a_{n-6}+a_{n-7} & =0\tag {2}\\ a_{n} & =-\frac {\left ( n-6\right ) a_{n-6}+a_{n-7}}{\left ( n+2\right ) \left ( n+1\right ) }\nonumber \end{align}

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{7}x^{7}+a_{13}x^{13}+\cdots \end{align*}

\begin{align*} y & =a_{0}+a_{1}x+\left ( -\frac {1}{42}a_{0}-\frac {1}{42}a_{1}\right ) x^{7}+\left ( \frac {1}{936}a_{0}+\frac {1}{936}a_{1}\right ) x^{13}+\cdots \\ & =a_{0}\left ( 1-\frac {1}{42}x^{7}+\frac {1}{936}x^{13}+\cdots \right ) +a_{1}\left ( x-\frac {1}{42}x^{7}+\frac {1}{936}x^{13}+\cdots \right ) \end{align*}

Expansion around \(x=0\). This is ordinary point. Since RHS is not zero, do not ﬁnd recurrence relation. Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n}\). Hence \(y^{\prime }=\sum _{n=0}^{\infty }na_{n}x^{n-1}=\sum _{n=1}^{\infty }na_{n}x^{n-1}\) and \(y^{\prime \prime }=\sum _{n=1}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}=\sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}\). The ode becomes

\begin{align} \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+x^{2}\sum _{n=1}^{\infty }na_{n}x^{n-1}+x^{2}\sum _{n=0}^{\infty }a_{n}x^{n} & =x^{2}\sin x\nonumber \\ \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+\sum _{n=1}^{\infty }na_{n}x^{n+1}+\sum _{n=0}^{\infty }a_{n}x^{n+2} & =x^{2}\sin x\nonumber \end{align}

To be able to continue, we have to expand \(\sin x\) as Taylor series around \(x\). The above becomes

\begin{align*} \sum _{n=0}^{\infty }\left ( n+2\right ) \left ( n+1\right ) a_{n+2}x^{n}+\sum _{n=2}^{\infty }\left ( n-1\right ) a_{n-1}x^{n}+\sum _{n=2}^{\infty }a_{n-2}x^{n} & =x^{2}\left ( x-\frac {1}{6}x^{3}+\frac {1}{120}x^{5}-\frac {1}{5040}x^{7}+\cdots \right ) \\ \sum _{n=0}^{\infty }\left ( n+2\right ) \left ( n+1\right ) a_{n+2}x^{n}+\sum _{n=2}^{\infty }\left ( n-1\right ) a_{n-1}x^{n}+\sum _{n=2}^{\infty }a_{n-2}x^{n} & =x^{3}-\frac {1}{6}x^{5}+\frac {1}{120}x^{7}-\frac {1}{5040}x^{9}+\cdots \end{align*}

\begin{align*} 2a_{2} & =0\\ a_{2} & =0 \end{align*}

\begin{align*} \left ( 3\right ) \left ( 2\right ) a_{3} & =0\\ a_{3} & =0 \end{align*}

\begin{align*} \left ( 2+2\right ) \left ( 2+1\right ) a_{4}+\left ( 2-1\right ) a_{1}+a_{0} & =0\\ 12a_{4}+a_{1}+a_{0} & =0\\ a_{4} & =\frac {-a_{1}-a_{0}}{12}\end{align*}

\begin{align*} 20a_{5}+2a_{2}+a_{1} & =1\\ a_{5} & =\frac {1-a_{1}}{20}\end{align*}

\begin{align*} 30a_{6}+3a_{3}+a_{2} & =0\\ a_{6} & =0 \end{align*}

\begin{align*} 42a_{7}+4a_{4}+a_{3} & =-\frac {1}{6}\\ a_{7} & =\frac {-\frac {1}{6}-4a_{4}}{42}=\frac {-\frac {1}{6}-4\left ( \frac {-a_{1}-a_{0}}{12}\right ) }{42}=\frac {1}{126}a_{0}+\frac {1}{126}a_{1}-\frac {1}{252}\end{align*}

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \\ & =a_{0}+a_{1}x+\left ( \frac {-a_{1}-a_{0}}{12}\right ) x^{4}+\left ( \frac {1-a_{1}}{20}\right ) x^{5}+\left ( \frac {1}{126}a_{0}+\frac {1}{126}a_{1}-\frac {1}{252}\right ) x^{7}+\cdots \\ & =a_{0}\left ( 1-\frac {1}{12}x^{4}+\frac {1}{126}x^{7}+\cdots \right ) +a_{1}\left ( x-\frac {1}{12}x^{4}-\frac {1}{20}x^{5}+\frac {1}{126}x^{7}+\cdots \right ) +\left ( \frac {1}{20}x^{5}-\frac {1}{252}x^{7}+\cdots \right ) \end{align*}

4.3.2.5.3 Regular singular point using Frobenius series method. expansion point is regular singular point. Four sub methods depending on type of roots of the indicial equations.

Roots of indicial equation are complex ode internal name "second_order_series_method_regular_singular_point_complex_roots"

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

Where \(r_{1},r_{2}\) are roots of the indicial equation. \(a_{0},b_{0}\) are set to \(1\) as arbitrary.

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {1}{x^{2}}\). There is one singular point at \(x_{0}=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}1=1\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r+1 & =0\\ r^{2}+1 & =0\\ r & =\pm i \end{align*}

Hence \(r_{1}=i,r_{2}=-i\). Expansion around \(x=0\). This is regular singular point. Let

\begin{align*} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

\begin{equation} \left ( r\left ( r-1\right ) +r+1\right ) a_{0}x^{r}=0 \tag {1}\end{equation}

Since \(a_{0}\neq 0\), then \(\left ( r\left ( r-1\right ) +r+1\right ) =0\) or \(\allowbreak r^{2}+1=0\). Therefore \(r=\pm i\) as was found above. The homogenous ode therefore satisﬁes

Since when \(r=\pm i\), the RHS is zero. For \(n\geq 1\) the recurrence relation is

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n} & =0\nonumber \\ \left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) +1\right ) a_{n} & =0\nonumber \\ \left ( n^{2}+2nr+r^{2}+1\right ) a_{n} & =0 \tag {2}\end{align}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+i}\\ & =x^{i}\left ( a_{0}+a_{1}x+\cdots \right ) \\ & =a_{0}x^{i}\\ & =x^{i}\end{align*}

For \(r=-i\). For \(n=1\) and using \(b\) instead of \(a\), we obtain (also using \(b_{0}=1\))

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n-i}\\ & =x^{-i}\left ( b_{0}+b_{1}x+\cdots \right ) \\ & =b_{0}x^{-i}\\ & =x^{-i}\end{align*}

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}x^{i}+c_{2}x^{-i}\end{align*}

Relabel \(r=m,a_{0}=c_{0}\) to avoid confusion with terms used above, then we balance RHS, hence

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

Using the recurrence relation (2) found above, but now using the values found \(m=0\) and \(c_{0}=1\), then (2) becomes

\begin{align*} \left ( n^{2}+2nm+m^{2}+1\right ) c_{n} & =0\\ \left ( n^{2}+1\right ) c_{n} & =0 \end{align*}

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =c_{0}\\ & =1 \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}x^{i}+c_{2}x^{-i}+1 \end{align*}

Roots of indicial equation diﬀer by non integer ode internal name "second_order_series_method_regular_singular_point_diﬀerence_not_integer"

If one of the roots is an integer, and the ode is inhomogeneous. ode, then we do not need to split the solution into \(y_{h},y_{p}\) and can use the integer root to ﬁnd \(y_{p}\) directly. If both roots are non-integer, we have to split the problem into \(y_{h},y_{p}\). This is because it will not be possible to match powers on \(x\) from the left side to the right side. Because the RHS will be polynomial in \(x\), but the LHS will not be polynomial in \(x\) because of the non integer powers on \(x.\)In this case the solution is

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

And \(r_{1},r_{2}\) are roots of the indicial equation. \(a_{0},b_{0}\) are set to \(1\) as arbitrary.

Hence \(p\left ( x\right ) =\frac {3}{2x},q\left ( x\right ) =\frac {-1}{2x}\). There is one singular point at \(x=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3}{2}=\frac {3}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}-\frac {x}{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {3}{2}r+0 & =0\\ r\left ( 2r+1\right ) & =0\\ r & =0,-\frac {1}{2}\end{align*}

Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\), hence

\begin{align*} 2x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-x\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r+1} & =0 \end{align*}

\begin{align*} 2\left ( r\right ) \left ( r-1\right ) a_{0}x^{r}+3\left ( r\right ) a_{0}x^{r} & =0\\ \left ( r\left ( 2r+1\right ) \right ) a_{0}x^{r} & =0 \end{align*}

Since \(a_{0}\neq 0\) then \(r\left ( 2r+1\right ) =0\) and \(r=0,r=-\frac {1}{2}\) as was found above. Therefore the homogenous ode satisﬁes

Where the RHS will be zero when \(r=0\) or \(r=-\frac {1}{2}\). For \(n\geq 1\) the recurrence relation is

\begin{align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+3\left ( n+r\right ) a_{n}-a_{n-1} & =0\nonumber \\ a_{n} & =\frac {a_{n-1}}{2\left ( n+r\right ) \left ( n+r-1\right ) +3\left ( n+r\right ) }\nonumber \\ & =\frac {a_{n-1}}{2n^{2}+4nr+n+2r^{2}+r} \tag {1}\end{align}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r}=\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \end{align*}

And for \(r=-\frac {1}{2}\) the recurrence relation (2) becomes, and using \(b\) instead of \(a\)

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\\ & =\frac {1}{\sqrt {x}}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\frac {1}{\sqrt {x}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) \end{align*}

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) \end{align*}

To ﬁnd \(y_{p}\), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used for \(y_{h}\). Then the above becomes

The RHS is \(x^{2}+2x\). We balance each term at a time, this ﬁnds a particular solution for each term on the RHS, then these particular solutions are added at the end. For the input \(2x\) the balance equation is

Therefore \(\left ( m\left ( 2m+1\right ) \right ) c_{0}=2\), or \(c_{0}\left ( 1\left ( 2+1\right ) \right ) =2\) or \(3c_{0}=2\) or

The recurrence relation now becomes (using \(m\) for \(r\) and \(c_{0}\) for \(a_{0}\))

\begin{align*} y_{p_{1}} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x\left ( \frac {2}{3}+\frac {1}{15}x+\frac {1}{315}x^{2}+\frac {1}{11\,340}x^{3}+\cdots \right ) \\ & =\left ( \frac {2}{3}x+\frac {1}{15}x^{2}+\frac {1}{315}x^{3}+\frac {1}{11\,340}x^{4}+\cdots \right ) \end{align*}

\begin{align*} \left ( 2\left ( 4+1\right ) \right ) c_{0} & =1\\ c_{0} & =\frac {1}{10}\end{align*}

\begin{align*} y_{p_{2}} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{10}+\frac {1}{210}x+\frac {1}{7560}x^{2}+\frac {1}{415\,800}x^{3}+\cdots \right ) \\ & =\left ( \frac {1}{10}x^{2}+\frac {1}{210}x^{3}+\frac {1}{7560}x^{4}+\frac {1}{415\,800}x^{5}+\cdots \right ) \end{align*}

The particular solution is the sum of all the particular solutions found above, which is

\begin{align*} y_{p} & =y_{p_{1}}+y_{p_{2}}\\ & =\left ( \frac {2}{3}x+\frac {1}{15}x^{2}+\frac {1}{315}x^{3}+\frac {1}{11\,340}x^{4}+\cdots \right ) +\left ( \frac {1}{10}x^{2}+\frac {1}{210}x^{3}+\frac {1}{7560}x^{4}+\frac {1}{415\,800}x^{5}+\cdots \right ) \\ & =\frac {2}{3}x+\left ( \frac {1}{15}+\frac {1}{10}\right ) x^{2}+\left ( \frac {1}{315}+\frac {1}{210}\right ) x^{3}+\left ( \frac {1}{11\,340}+\frac {1}{7560}\right ) x^{4}+\cdots \\ & =\frac {2}{3}x+\frac {1}{6}x^{2}+\frac {1}{126}x^{3}+\frac {1}{4536}x^{4}+\cdots \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1+\frac {1}{3}x+\frac {1}{30}x^{2}+\frac {1}{630}x^{3}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1+x+\frac {1}{6}x^{2}+\frac {1}{90}x^{3}+\cdots \right ) +\frac {2}{3}x+\frac {1}{6}x^{2}+\frac {1}{126}x^{3}+\frac {1}{4536}x^{4}+\cdots \end{align*}

Hence \(p\left ( x\right ) =\frac {\left ( x+1\right ) }{2x},q\left ( x\right ) =\frac {3}{2x}\). There is one singular point at \(x=0\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {\left ( x+1\right ) }{2}=\frac {1}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3x}{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {1}{2}r+0 & =0\\ r\left ( 2r-1\right ) & =0\\ r & =0,\frac {1}{2}\end{align*}

Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align*} 2x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\left ( x+1\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+3\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3a_{n}x^{n+r} & =0 \end{align*}

\begin{align*} \left ( 2\left ( r\right ) \left ( r-1\right ) a_{0}+ra_{0}\right ) x^{r-1} & =0\\ \left ( 2r\left ( r-1\right ) +r\right ) a_{0} & =0 \end{align*}

Since \(a_{0}\neq 0\) then the ﬁrst term above will vanish only when \(2r\left ( r-1\right ) +r=0\) or \(r\left ( 2r-1\right ) =0\). Hence \(r=0,r=\frac {1}{2}\) as was found above. For \(n\geq 1\)

\begin{align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-1\right ) a_{n-1}+\left ( n+r\right ) a_{n}+3a_{n-1} & =0\nonumber \\ a_{n} & =-\frac {n+r+2}{\left ( n+r\right ) \left ( 2r+2n-1\right ) }a_{n-1} \tag {1}\end{align}

\begin{equation} 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=r\left ( 2r-1\right ) a_{0}x^{r-1} \tag {2}\end{equation}

The RHS above will be zero when \(r=0\) or \(r=\frac {1}{2}\). When \(r=0\) the recurrence relation (1) becomes

And when \(r=\frac {1}{2}\) the recurrence relation is (using \(b\) in place of \(a\) and letting \(b_{0}=1\) also)

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \right ) +c_{2}\left ( \sqrt {x}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) \right ) \end{align*}

Now we ﬁnd \(y_{p}\). From (2), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used

Therefore we need to balance \(m\left ( 2m-1\right ) c_{0}x^{m-1}=5\) since the RHS is \(5\). This implies \(m-1=0\) or \(m=1\). Therefore \(m\left ( 2m-1\right ) c_{0}=5\) or \(\left ( 2-1\right ) c_{0}=5\) which gives \(c_{0}=5\). Hence

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

To ﬁnd \(c_{n}\), the same recurrence relation (1) is used by with \(r\) replaced by \(m\) and \(a\) replaced by \(c\). This gives

\begin{align*} y_{p} & =x\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x\left ( c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \right ) \\ & =x\left ( 5-\frac {10}{3}x+\frac {10}{9}x^{2}-\frac {5}{21}x^{3}+\cdots \right ) \\ & =\left ( 5x-\frac {10}{3}x^{2}+\frac {10}{9}x^{3}-\frac {5}{21}x^{4}+\cdots \right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \right ) +\sqrt {x}c_{2}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) +\left ( 5x-\frac {10}{3}x^{2}+\frac {10}{9}x^{3}-\frac {5}{21}x^{4}+\cdots \right ) \end{align*}

This is the same problem as above but diﬀerent RHS. As shown above, we obtained that the diﬀerential equation satisﬁes

To ﬁnd \(y_{p}\), and using \(m\) in place of \(r\) and \(c\) in place of \(a\) so not to confuse terms with the \(y_{h}\) terms, then the above becomes

The RHS above will be zero when \(m=0\) or \(m=\frac {1}{2}\). We now need to balance the RHS against given RHS which is \(x\). Hence

To balance this we need \(m-1=1\) or \(m=2\). Hence \(2\left ( 4-1\right ) c_{0}=1\) or \(c_{0}=\frac {1}{6}\). Using the recurrence relation we found above, which is for \(n\geq 1\) (again, calling \(r\) as \(m\) so not to confuse \(y_{h}\) terms with \(y_{p}\) terms), we obtain

\begin{align*} c_{1} & =-\frac {1+4}{\left ( 1+2\right ) \left ( 4+2-1\right ) }c_{0}\\ & =-\frac {1}{3}c_{0}\\ & =-\frac {1}{3}\left ( \frac {1}{6}\right ) =-\frac {1}{18}\end{align*}

\begin{align*} c_{2} & =-\frac {6}{\left ( 2+2\right ) \left ( 4+4-1\right ) }c_{1}\\ & =-\frac {3}{14}c_{1}=-\frac {3}{14}\left ( -\frac {1}{18}\right ) =\frac {1}{84}\end{align*}

\begin{align*} c_{3} & =-\frac {3+4}{\left ( 3+2\right ) \left ( 4+6-1\right ) }c_{2}\\ & =-\frac {7}{45}c_{2}=-\frac {7}{45}\left ( \frac {1}{84}\right ) =-\frac {1}{540}\end{align*}

\begin{align*} c_{4} & =-\frac {4+4}{\left ( 4+2\right ) \left ( 4+8-1\right ) }c_{3}\\ & =-\frac {4}{33}c_{3}=-\frac {4}{33}\left ( -\frac {1}{540}\right ) =\frac {1}{4455}\end{align*}

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+r}\\ & =x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{6}-\frac {1}{18}x+\frac {1}{84}x^{2}-\frac {1}{540}x^{3}+\frac {1}{4455}x^{4}+\cdots \right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \right ) +c_{2}\left ( \sqrt {x}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) \right ) +x^{2}\left ( \frac {1}{6}-\frac {1}{18}x+\frac {1}{84}x^{2}-\frac {1}{540}x^{3}+\frac {1}{4455}x^{4}+\cdots \cdots \right ) \end{align*}

Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. Comparing the ode to

Hence \(p\left ( x\right ) =\frac {x+1}{x^{2}},q\left ( x\right ) =\frac {1}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x+1}{x}\) which is not deﬁned. Hence not possible to solve this using series solution.

Hence \(p\left ( x\right ) =\frac {-x}{2x^{2}}=-\frac {1}{2x},q\left ( x\right ) =\frac {\left ( 1-x^{2}\right ) }{2x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {-1}{2}=\frac {-1}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {\left ( 1-x^{2}\right ) }{2}=\frac {1}{2}\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) -\frac {1}{2}r+\frac {1}{2} & =0\\ r^{2}-\frac {3}{2}r+\frac {1}{2} & =0\\ r & =1,\frac {1}{2}\end{align*}

Therefore \(r_{1}=0,r_{2}=-\frac {1}{2}\). Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\), hence

\begin{align*} 2x^{2}y^{\prime \prime }-xy^{\prime }+\left ( 1-x^{2}\right ) y & =0\\ 2x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\left ( 1-x^{2}\right ) \sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}-\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r+2} & =0 \end{align*}

\begin{align*} \left ( 2\left ( n+r\right ) \left ( n+r-1\right ) a_{0}-\left ( n+r\right ) a_{0}+a_{0}\right ) x^{r} & =0\\ \left ( 2r\left ( r-1\right ) -r+1\right ) a_{0}x^{r} & =0\\ \left ( 2r^{2}-3r+1\right ) a_{0}x^{r} & =0 \end{align*}

Since \(a_{0}\neq 0\) then \(2r^{2}-3r+1=0\), hence \(r=1,r=\frac {1}{2}\) as was found above. Therefore the homogenous ode satisﬁes

\begin{align*} 2\left ( 1+r\right ) \left ( 1+r-1\right ) a_{1}-\left ( 1+r\right ) a_{1}+a_{1} & =0\\ \left ( 2\left ( 1+r\right ) \left ( 1+r-1\right ) -\left ( 1+r\right ) +1\right ) a_{1} & =0\\ r\left ( 2r+1\right ) a_{1} & =0 \end{align*}

\begin{align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}-\left ( n+r\right ) a_{n}+a_{n}-a_{n-2} & =0\nonumber \\ a_{n} & =\frac {a_{n-2}}{2\left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) +1}\nonumber \\ & =\frac {a_{n-2}}{2\left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) +1} \tag {1}\end{align}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r}=x\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+\cdots \right ) \\ & =x\left ( 1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+\cdots \right ) \end{align*}

And for \(r=\frac {1}{2}\) the recurrence relation (1) becomes, and using \(b\) instead of \(a\)

\begin{align*} b_{n} & ==\frac {b_{n-2}}{2\left ( n+r\right ) \left ( n+r-1\right ) -\left ( n+r\right ) +1}\\ & =\frac {b_{n-2}}{2\left ( n+\frac {1}{2}\right ) \left ( n+\frac {1}{2}-1\right ) -\left ( n+\frac {1}{2}\right ) +1}\\ & =\frac {b_{n-2}}{n\left ( 2n-1\right ) }\end{align*}

Notice also that \(b_{1}=0\) just like \(a_{1}=0\) from above. Now, for \(n=2\) and using \(b_{0}=1\)

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\\ & =\sqrt {x}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\sqrt {x}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \end{align*}

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x\left ( 1+\frac {x^{2}}{10}+\frac {x^{4}}{360}+\cdots \right ) \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \\ & =c_{1}\left ( x+\frac {x^{3}}{10}+\frac {x^{5}}{360}+\cdots \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) \end{align*}

To ﬁnd \(y_{p}\), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used for \(y_{h}\). Then the above becomes

Therefore \(\left ( 2m^{2}-3m+1\right ) c_{0}=1\), or \(\left ( 8-6+1\right ) c_{0}=1\) or

The recurrence relation (1) from above now becomes (using \(m\) for \(r\) and \(c_{0}\) for \(a_{0}\))

\begin{align*} c_{n} & =\frac {c_{n-2}}{2\left ( n+2\right ) \left ( n+1\right ) -\left ( n+2\right ) +1}\\ & =\frac {c_{n-2}}{2n^{2}+5n+3}\end{align*}

For \(n=1\) we use \(c_{1}=0\) the same as was found for \(a_{1},b_{1}\). For \(n\geq 2\) the above is used. Hence for \(n=2\)

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}=x^{2}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{2}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{2}\left ( \frac {1}{3}+\frac {1}{63}x^{2}+\frac {1}{3465}x^{4}+\cdots \right ) \\ & =\frac {1}{3}x^{2}+\frac {1}{63}x^{4}+\frac {1}{3465}x^{6}+\cdots \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( x+\frac {x^{3}}{10}+\frac {x^{5}}{360}+\cdots \right ) +c_{2}\sqrt {x}\left ( 1+\frac {1}{6}x^{2}+\frac {1}{168}x^{4}+\cdots \right ) +\left ( \frac {1}{3}x^{2}+\frac {1}{63}x^{4}+\frac {1}{3465}x^{6}+\cdots \right ) \end{align*}

Alternative way to ﬁnd \(y_{p}\) is the the following. Let \(y_{p}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \) then \(y_{p}^{\prime }=c_{1}+2c_{2}x+3c_{3}x^{2}+\cdots \) and \(y_{p}^{\prime \prime }=2c_{2}+6c_{3}x+\cdots \). Hence the ode becomes

\begin{align*} 2x^{2}\left ( 2c_{2}+6c_{3}x+\cdots \right ) -x\left ( c_{1}+2c_{2}x+3c_{3}x^{2}+\cdots \right ) +\left ( 1-x^{2}\right ) \left ( c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\cdots \right ) & =x^{2}\\ c_{0}+x\left ( -c_{1}+c_{1}\right ) +x^{2}\left ( 4c_{2}-2c_{2}+c_{2}-c_{0}\right ) +x^{3}\left ( \cdots \right ) & =x^{2}\end{align*}

Hence \(c_{0}=0,4c_{2}-2c_{2}+c_{2}-c_{0}=1\) or \(3c_{2}-c_{0}=1\) or \(c_{2}=\frac {1}{3}\). We need to keep adding more equations and solving them simultaneously. This method is not as easy to use as the method used above, which uses the balance equation to ﬁnd to \(y_{p}\). Also this method could fail, since in practice we should not use undetermined coeﬃcients method (which is what this does) on an ode with variable coeﬃcients. So I will not use this any more.

Hence \(p\left ( x\right ) =\frac {1}{2x},q\left ( x\right ) =\frac {1}{2x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {1}{2}=\frac {1}{2}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x}{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {1}{2}r & =0\\ r\left ( 2r-1\right ) & =0\\ r & =0,\frac {1}{2}\end{align*}

Therefore \(r_{1}=0,r_{2}=\frac {1}{2}\). Expansion around \(x=x_{0}=0\). This is regular singular point. Hence Frobenius is needed. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\), hence

\begin{align*} xy^{\prime \prime }+y^{\prime }+y & =0\\ 2x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

\begin{align*} 2\left ( r\right ) \left ( r-1\right ) a_{0}x^{r-1}+ra_{0}x^{r-1} & =0\\ \left ( 2r\left ( r-1\right ) +r\right ) a_{0}x^{r-1} & =0\\ \left ( r\left ( 2r-1\right ) \right ) a_{0}x^{r-1} & =0 \end{align*}

Since \(a_{0}\neq 0\) then \(r\left ( 2r-1\right ) =0\), hence \(r=0,r=\frac {1}{2}\) as was found above. Therefore the homogenous ode satisﬁes

Where the RHS will be zero when \(r=1,r=\frac {1}{2}\). For \(n\geq 1\) the recurrence relation is

\begin{align} 2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n} & =-a_{n-1}\nonumber \\ a_{n} & =\frac {-a_{n-1}}{2\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) }\nonumber \\ & =\frac {-a_{n-1}}{2n^{2}+4nr-n+2r^{2}-r} \tag {1}\end{align}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1-x+\frac {1}{6}x^{2}-\frac {1}{90}x^{3}+\cdots \end{align*}

To ﬁnd \(y_{2}\), using (1) but replacing \(a\) by \(b\) and using \(r=\frac {1}{2}\) and letting \(b_{0}=1\) and following the above process gives

\begin{align*} y_{2} & =\sqrt {x}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\sqrt {x}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\sqrt {x}\left ( 1-\frac {1}{3}x+\frac {1}{30}x^{2}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{1}\\ & =c_{1}\left ( 1-x+\frac {1}{6}x^{2}-\frac {1}{90}x^{3}+\cdots \right ) +c_{2}\left ( \sqrt {x}\left ( 1-\frac {1}{3}x+\frac {1}{30}x^{2}+\cdots \right ) \right ) \end{align*}

Hence \(p\left ( x\right ) =\frac {3}{4x},q\left ( x\right ) =\frac {3}{4x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3}{4}=\frac {3}{4}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3x}{4}=0\). Hence \(x=0\) is regular singular point. The indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {3}{4}r+0 & =0\\ r\left ( r-1\right ) +\frac {3}{4}r & =0\\ r & =\frac {1}{4},0 \end{align*}

Frobenius is now used. Roots diﬀer by non integer. First we ﬁnd \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\).

\begin{align*} 4x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+3\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3a_{n}x^{n+r} & =0 \end{align*}

\begin{align*} 4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+3\left ( n+r\right ) a_{n}x^{n+r-1} & =0\\ 4r\left ( r-1\right ) a_{0}+3ra_{0} & =0\\ \left ( 4r\left ( r-1\right ) +3r\right ) a_{0} & =0 \end{align*}

Since \(a_{0}\neq 0\) then \(4r\left ( r-1\right ) +3r=0\), hence \(r=0,r=\frac {1}{4}\) as was found above. Therefore the homogenous ode satisﬁes

Hence the balance equation is that we will use to ﬁnd the particular solution is

We will get back to the above after ﬁnding \(y_{h}\). Going over the same steps as before, we ﬁnd the recurrence relation

\begin{align*} y_{h} & =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \\ & =c_{1}x^{\frac {1}{4}}\left ( 1-\frac {3}{5}x+\frac {1}{10}x^{2}-\frac {1}{130}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {3}{14}x^{2}-\frac {3}{154}x^{3}+\cdots \right ) \end{align*}

Hence \(m-1=\frac {1}{2}\) or \(m=\frac {3}{2}\). And \(\left ( 4m\left ( m-1\right ) +3m\right ) c_{0}=1\), hence \(\left ( 4\left ( \frac {3}{2}\right ) \left ( \frac {3}{2}-1\right ) +3\left ( \frac {3}{2}\right ) \right ) c_{0}=1\), which gives \(c_{0}=\frac {2}{15}\). Therefore

\begin{align*} y_{p} & =x^{m}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {3}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {3}{2}}\left ( \frac {2}{15}+c_{1}x+c_{2}x^{2}+\cdots \right ) \end{align*}

We now just need to determined \(c_{n}\) for \(n>0\). For this we use the same recurrence relation as found above. We can use \(a_{n}\) or \(b_{n}\) as they are the same, but change \(a_{n}\) to \(c_{n}\) and \(r\) to \(c\) (so not to confuse notations). This gives

\begin{align*} c_{1} & =-\frac {3c_{0}}{4+8\left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-1-\frac {3}{2}}\\ & =-\frac {3\left ( \frac {2}{15}\right ) }{4+8\left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-1-\frac {3}{2}}\\ & =-\frac {4}{225}\end{align*}

\begin{align*} c_{1} & =-\frac {3c_{1}}{4\left ( 2\right ) ^{2}+8\left ( 2\right ) \left ( \frac {3}{2}\right ) +\left ( \frac {3}{2}\right ) ^{2}-2-\left ( \frac {3}{2}\right ) }\\ & =-\frac {3\left ( -\frac {4}{225}\right ) }{4\left ( 2\right ) ^{2}+8\left ( 2\right ) \left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-2-\left ( \frac {3}{2}\right ) }\\ & =\frac {8}{6825}\end{align*}

\begin{align*} y_{p} & =x^{\frac {3}{2}}\left ( \frac {2}{15}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {3}{2}}\left ( \frac {2}{15}-\frac {4}{225}x+\frac {8}{6825}x^{2}-\frac {16}{348075}x^{3}+\cdots \right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}x^{\frac {1}{4}}\left ( 1-\frac {3}{5}x+\frac {1}{10}x^{2}-\frac {1}{130}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {3}{14}x^{2}-\frac {3}{154}x^{3}+\cdots \right ) +x^{\frac {3}{2}}\left ( \frac {2}{15}-\frac {4}{225}x+\frac {8}{6825}x^{2}-\frac {16}{348075}x^{3}+\cdots \right ) \end{align*}

Roots of indicial equation diﬀer by integer. Good case ode internal name "second_order_series_method_regular_singular_point_diﬀerence_is_integer_good_case".

There are two sub cases that show up when roots diﬀer by integer. First sub case is when the second solution \(y_{2}\) is obtained similar to how \(y_{1}\) is obtained. i.e. using standard Frobenius series but with the second root. The second sub case is the harder one, this is when \(y_{2}\) fails to be obtained using the standard method due to \(b_{N}\) being undeﬁned where \(N\) is the diﬀerence between the roots. In this sub case we need to use a modiﬁed Frobenius series method where, which is explained more using examples below. Therefore for sub case one (called the good case) we have

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

For the second subcase (called the bad case) ﬁrst we will ﬁnd the bad root \(r\) of the indicial equation which causes the recurrence relation to become undeﬁned at some \(n\). Call it \(r_{bad}\), then we ﬁrst ﬁnd \(\overline {y}\) deﬁned as

Where \(a_{n}\) is found using the recurrence relation (but \(r\) is kept symbolic). \(y_{1}\) is then found from by evaluating it \(r=r_{bad}\)

And also setting \(a_{0}=1\). Note that some terms will vanish above but not all, since there will be cancellation of \(\left ( r-r_{bad}\right ) \) during the process. \(y_{2}\) is next found using

\begin{align*} y_{2} & =\left ( \frac {d}{dr}\overline {y}\right ) _{r=r_{bad}}\\ & =y_{1}\ln \left ( x\right ) +x^{r_{bad}}\sum _{n=0}^{\infty }\left ( \frac {d}{dr}\left ( \left ( r-r_{bad}\right ) a_{n}x^{n}\right ) \right ) _{r=r_{bad}}\end{align*}

Comparing the above to \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0\) shows that \(p\left ( x\right ) =\frac {3}{x\left ( 1-x\right ) },q\left ( x\right ) =\frac {2}{x\left ( x-1\right ) }\). Hence there are two singular points, one at \(x=0\) and one at \(x=1\). Let the expansion be around \(x=0\). This means the solution will deﬁne up to \(x=1\), which is the next nearest singular point.

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +3r & =0\\ r^{2}-r+3r & =0\\ r^{2}+2r & =0\\ r\left ( r+2\right ) & =0 \end{align*}

Therefore \(r=0,r=-2\). They diﬀer by an integer \(N=2\). Therefore two linearly independent solutions can be constructed using

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

Where \(C\) above can be zero depending on a condition given below. Now we will work out the solution for a general \(r\). Let

\begin{align*} \left ( x-x^{2}\right ) y^{\prime \prime }+3y^{\prime }+2y & =0\\ \left ( x-x^{2}\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+3\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+2\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }2a_{n}x^{n+r} & =0 \end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=1}^{\infty }\left ( n+r-1\right ) \left ( n+r-2\right ) a_{n-1}x^{n+r-1}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }2a_{n-1}x^{n+r-1}=0 \tag {1A}\end{equation}

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+3\left ( n+r\right ) a_{n}x^{n+r-1} & =0\nonumber \\ \left ( r\left ( r-1\right ) +3r\right ) a_{0}x^{r-1} & =\nonumber \\ \left ( r^{2}+2r\right ) a_{0}x^{r-1} & =0 \tag {1B}\end{align}

Since \(a_{0}\neq 0\), then \(r=0,r=-2\) as was found above. Hence \(N=2\) which is the diﬀerence between the two roots. The homogenous ode therefore satisﬁes

Since when \(r=0,r=-2\) the RHS is zero. The term on the right of the above is important as it will be used to determine the particular solution. The recurrence relation is when \(n\geq 1\) from (1A) and is given by

\begin{align} a_{n} & =\frac {-2+\left ( n+r-1\right ) \left ( n+r-2\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) +3\left ( n+r\right ) }a_{n-1}\nonumber \\ a_{n} & =\frac {n+r-3}{n+r+2}a_{n-1} \tag {1}\end{align}

\begin{equation} a_{n}=\frac {n-3}{n+2}a_{n-1} \tag {2}\end{equation}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}\\ & =1-\frac {2}{3}x+\frac {1}{6}x^{2}\end{align*}

Now we need to ﬁnd \(y_{2}\). We ﬁrst check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\). For this we calculate \(b_{N}=b_{2}\) using same recurrence relation (1) to see if it is deﬁned or not. If it is deﬁned, then we continue, else we have to use the modiﬁed Frobenius method. From (1) and using \(b\) instead of \(a\) and using \(r=r_{2}=-2\) gives

\begin{align*} b_{n} & =\frac {n+r-3}{n+r+2}b_{n-1}\\ & =\frac {n-2-3}{n-2+2}b_{n-1}\\ & =\frac {n-5}{n}b_{n-1}\end{align*}

Since \(b_{N}\) is deﬁned, we can continue and \(y_{2}\) is found using same recurrence relation. Hence this is subcase one. For \(n=3\)

\begin{align*} y_{2} & =\frac {1}{x^{2}}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\frac {1}{x^{2}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}\right ) \\ & =\frac {1}{x^{2}}\left ( 1-4x+6x^{2}-4x^{3}+x^{4}\right ) \end{align*}

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-\frac {2}{3}x+\frac {1}{6}x^{2}\right ) +c_{2}\left ( \frac {1}{x^{2}}\left ( 1-4x+6x^{2}-4x^{3}+x^{4}\right ) \right ) \end{align*}

Now we ﬁnd \(y_{p}\). From earlier we found in (1B) the balance equation which gives

Relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used in ﬁnding \(y_{h}\) the above becomes

Therefore we need to balance \(\left ( m^{2}+2m\right ) c_{0}x^{m-1}=3x^{2}\). This implies \(m-1=2\) or \(m=3\). Therefore \(\left ( m^{2}+2m\right ) c_{0}=3\) or \(\left ( 9+6\right ) c_{0}=3\) which gives \(c_{0}=\frac {3}{15}=\frac {1}{5}\). Hence

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x^{3}\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

To ﬁnd \(c_{n}\), the same recurrence relation given in (1) is used again but now \(r\) is replaced by \(m\) and \(a\) replaced by \(c\). This gives the recurrence relation to ﬁnd coeﬃcients of the particular solution as

\begin{align*} c_{n} & =\frac {n+3-3}{n+3+2}c_{n-1}\\ & =\frac {n}{n+5}c_{n-1}\end{align*}

\begin{align*} y_{p} & =x^{3}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{3}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{3}\left ( \frac {1}{5}+\frac {1}{30}x+\frac {1}{105}x^{2}+\cdots \right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-\frac {2}{3}x+\frac {1}{6}x^{2}\right ) +c_{2}\left ( \frac {1}{x^{2}}\left ( 1-4x+6x^{2}-4x^{3}+x^{4}\right ) \right ) +\left ( \frac {1}{5}x^{3}+\frac {1}{30}x^{4}+\frac {1}{105}x^{5}+\cdots \right ) \end{align*}

If we try to ﬁnd \(y_{p}\) by assuming \(y_{p}=\sum _{n=0}^{\infty }c_{n}x^{n}\) and substituting into the ode and try to match coeﬃcients, we can not always be successful. The above method using the balance equation always works and that is what I am using in my solver.

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {4x^{2}-1}{4x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {4x^{2}-1}{4}=-\frac {1}{4}\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r-\frac {1}{4} & =0\\ r^{2}-\frac {1}{4} & =0\\ r & =-\frac {1}{2},\frac {1}{2}\end{align*}

Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align} 4x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+4x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+4x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) \left ( n+r-1\right ) +4\left ( n+r\right ) -1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4n^{2}+8nr+4r^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }4a_{n}x^{n+r+2} & =0 \tag {1}\end{align}

\begin{equation} \sum _{n=0}^{\infty }\left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }4a_{n-2}x^{n+r}=0 \tag {2}\end{equation}

Since \(a_{0}\neq 0\), then \(r_{1}=\frac {1}{2},r_{2}=-\frac {1}{2}\) as was found above. The ode therefore satisﬁes

Since when \(r_{1}=\frac {1}{2},r_{2}=-\frac {1}{2}\) the RHS is zero. When \(n=1\) then (2) gives

\begin{equation} \left ( 4\left ( 1+r\right ) ^{2}-1\right ) a_{1}=0 \tag {3}\end{equation}

\begin{align} \left ( 4\left ( n+r\right ) ^{2}-1\right ) a_{n}+4a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-4}{4\left ( n+r\right ) ^{2}-1}a_{n-2} \tag {4}\end{align}

Since roots diﬀer by an integer \(N=1\) then there two linearly independent solutions can be constructed using

\begin{align*} y_{1} & =x^{r_{1}}\sum _{n=0}^{\infty }a_{n}x^{n}\\ y_{2} & =Cy_{1}\ln \left ( x\right ) +x^{r_{2}}\sum _{n=0}^{\infty }b_{n}x^{n}\end{align*}

\(C\) above can come out to be zero. We start by ﬁnding \(y_{1}\) (the one with the larger \(r\)).

\begin{align*} \left ( 4\left ( 1+\frac {1}{2}\right ) ^{2}-1\right ) a_{1} & =0\\ 8a_{1} & =0\\ a_{1} & =0 \end{align*}

\begin{align} a_{n} & =\frac {-4}{4\left ( n+\frac {1}{2}\right ) ^{2}-1}a_{n-2}\nonumber \\ & =-\frac {1}{n^{2}+n}a_{n-2} \tag {5}\end{align}

\begin{align*} a_{2} & =-\frac {1}{6}a_{0}\\ & =-\frac {1}{6}\end{align*}

\begin{align*} a_{3} & =-\frac {1}{12}a_{1}\\ & =0 \end{align*}

\begin{align*} a_{4} & =-\frac {1}{20}a_{2}\\ & =-\frac {1}{20}\left ( -\frac {1}{6}\right ) =\frac {1}{120}\end{align*}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+\frac {1}{2}}\\ & =x^{\frac {1}{2}}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =\sqrt {x}\left ( 1-\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) \end{align*}

Now we need to ﬁnd \(y_{2}\). We ﬁrst check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\). For this we calculate \(b_{N}=b_{1}\) using same recurrence relation (1) to see if it is deﬁned or not. If it is deﬁned, then we continue, else we have to use the modiﬁed Frobenius method. From (1) and using \(b\) instead of \(a\) and using \(r=r_{2}=-\frac {1}{2}\) gives

\begin{align*} \left ( 4\left ( 1-\frac {1}{2}\right ) ^{2}-1\right ) b_{1} & =0\\ 0b_{1} & =0 \end{align*}

Hence \(b_{1}\) is arbitrary. Let \(b_{1}=0\). Since \(b_{N}=b_{1}\) is deﬁned, we can continue and \(y_{2}\) is found using same recurrence relation. Hence this is subcase one. From (4) and using \(r=-\frac {1}{2}\) it becomes

\begin{align} b_{n} & =\frac {-4}{4\left ( n-\frac {1}{2}\right ) ^{2}-1}b_{n-2}\nonumber \\ & =-\frac {1}{n\left ( n-1\right ) }b_{n-2} \tag {6}\end{align}

\begin{align*} b_{2} & =-\frac {1}{2\left ( 2-1\right ) }b_{0}\\ & =-\frac {1}{2}\end{align*}

\begin{align*} b_{3} & =-\frac {1}{3\left ( 3-1\right ) }b_{1}\\ & =0 \end{align*}

\begin{align*} b_{4} & =-\frac {1}{4\left ( 4-1\right ) }b_{2}\\ & =--\frac {1}{12}\left ( -\frac {1}{2}\right ) =\frac {1}{24}\end{align*}

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n-\frac {1}{2}}\\ & =\frac {1}{\sqrt {x}}\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\frac {1}{\sqrt {x}}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\sqrt {x}\left ( 1-\frac {1}{6}x^{2}+\frac {1}{120}x^{4}+\cdots \right ) +c_{2}\frac {1}{\sqrt {x}}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \end{align*}

This ode is here because the RHS has no series expansion at \(x=0\). Comparing the ode to

Hence \(p\left ( x\right ) =1,q\left ( x\right ) =1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}x=0\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) & =0\\ r & =0,1 \end{align*}

Expansion around \(x=0\). This is regular singular point (due to the RHS not having series expansion). Hence Frobenius is needed. Let

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \tag {1}\end{equation}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) a_{n-1}x^{n+r-2}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r-2}=0 \tag {2}\end{equation}

Since \(a_{0}\neq 0\), then \(r_{1}=1,r_{2}=0\) as was found above. The ode therefore satisﬁes

\begin{align} \left ( 1+r\right ) \left ( r\right ) a_{1}+ra_{0} & =0\nonumber \\ a_{1} & =\frac {-a_{0}}{1+r} \tag {3}\end{align}

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-1\right ) a_{n-1}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-\left ( n+r-1\right ) a_{n-1}-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) } \tag {4}\end{align}

Since roots diﬀer by an integer \(N=1\) then there two linearly independent solutions can be constructed using

\begin{align*} y_{1} & =x^{r_{1}}\sum _{n=0}^{\infty }a_{n}x^{n}\\ y_{2} & =Cy_{1}\ln \left ( x\right ) +x^{r_{2}}\sum _{n=0}^{\infty }b_{n}x^{n}\end{align*}

\(C\) above can come out to be zero. We start by ﬁnding \(y_{1}\) (the one with the larger \(r\)).

\begin{align*} a_{1} & =\frac {-a_{0}}{2}\\ a_{1} & =\frac {-1}{2}\end{align*}

\begin{equation} a_{2}=\frac {-2a_{1}-a_{0}}{\left ( 2+1\right ) \left ( 2\right ) }=\frac {-2a_{1}-a_{0}}{6}=\frac {-2\left ( \frac {-1}{2}\right ) -1}{6}=0\nonumber \end{equation}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+1}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) \end{align*}

Now we need to ﬁnd \(y_{2}\). We ﬁrst check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\). For this we look at \(a_{1}=\frac {-a_{0}}{1+r}\) and see this is deﬁned for \(r=0\). Next we look at the recurrence relation \(a_{n}=\frac {-\left ( n+r-1\right ) a_{n-1}-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) }\) and see this is also deﬁned for \(r=1\). Hence \(C=0\) and we can ﬁnd \(y_{2}\) using same series expansion and using \(b_{0}=1\).

\begin{equation} b_{n}=\frac {-\left ( n-1\right ) b_{n-1}-b_{n-2}}{n\left ( n-1\right ) } \tag {5}\end{equation}

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+0}\\ & =\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \end{align*}

\begin{align*} y_{h} & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \right ) \end{align*}

Now we ﬁnd \(y_{p}\). From above \(y^{\prime \prime }+y^{\prime }+y=r\left ( r-1\right ) a_{0}x^{r-2}\), and relabeling \(r\) as \(m\) and \(a\) as \(c\) so not to confuse terms used

Therefore we need to balance \(m\left ( m-1\right ) c_{0}x^{m-2}=x^{\frac {1}{2}}\) since the RHS is \(\sqrt {x}\). This implies \(m-2=\frac {1}{2}\) or \(m=\frac {5}{2}\). Therefore \(m\left ( m-1\right ) c_{0}=1\) or \(\frac {5}{2}\left ( \frac {5}{2}-1\right ) c_{0}=1\), \(c_{0}=\frac {4}{15}\). Hence

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x^{\frac {5}{2}}\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

To ﬁnd \(c_{n}\), the same recurrence relation (4) is used by with \(r\) replaced by \(m\) and \(a\) replaced by \(c\). This gives

\begin{align} c_{n} & =\frac {-\left ( n+m-1\right ) c_{n-1}-c_{n-2}}{\left ( n+m\right ) \left ( n+m-1\right ) }\nonumber \\ & =\frac {-\left ( n+\frac {5}{2}-1\right ) c_{n-1}-c_{n-2}}{\left ( n+\frac {5}{2}\right ) \left ( n+\frac {5}{2}-1\right ) }\nonumber \\ & =-4\frac {\frac {3}{2}c_{n-1}+c_{n-2}+nc_{n-1}}{\left ( 2n+3\right ) \left ( 2n+5\right ) } \tag {6}\end{align}

The above is only for \(n\geq 2\). For \(n=1\), using \(a_{1}=\frac {-a_{0}}{1+r}\) and replacing \(a\) by \(c\) and \(r\) by \(m\) gives

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n+\frac {5}{2}}\\ & =x^{\frac {5}{2}}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {5}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {5}{2}}\left ( \frac {4}{15}-\frac {8}{105}x+\frac {32}{10\,395}x^{3}+\cdots \right ) \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \right ) +x^{\frac {5}{2}}\left ( \frac {4}{15}-\frac {8}{105}x+\frac {32}{10\,395}x^{3}+\cdots \right ) \end{align*}

Roots of indicial equation diﬀer by integer. Bad case ode internal name "second_order_series_method_regular_singular_point_diﬀerence_is_integer_bad_case".

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {x^{2}-4}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}-4=-4\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r-4 & =0\\ r^{2}-4 & =0\\ r & =2,-2 \end{align*}

Therefore \(r_{1}=2,r_{2}=-2\). Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\left ( x^{2}-4\right ) \sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-4\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+2}-\sum _{n=0}^{\infty }4a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+2} & =0\nonumber \end{align}

\begin{equation} \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r}=0 \tag {2}\end{equation}

\begin{align*} \left ( r\left ( r-1\right ) +r-4\right ) a_{0}x^{r} & =0\\ \left ( r^{2}-4\right ) a_{0}x^{r} & =0 \end{align*}

Since \(a_{0}\neq 0\), then \(r^{2}=4\) or \(r_{1}=2,r_{2}=-2\) as was found above. The ode therefore satisﬁes

Since when \(r_{1}=2\) or \(r_{2}=-2\) then the RHS is zero. When \(n=1\) then (2) gives

\begin{align*} \left ( \left ( 1+r\right ) r+\left ( 1+r\right ) -4\right ) a_{1} & =0\\ \left ( r^{2}+2r-3\right ) a_{1} & =0 \end{align*}

\begin{align} \left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) a_{n}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-a_{n-2}}{\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -4\right ) } \tag {4}\end{align}

We check ﬁrst if this is subcase one or two. To do this, we check if the recurrence relation is deﬁned for both roots for all \(n\geq 2\). The above for \(r=2\) gives

We see that it is deﬁned for all \(n\geq 2\). Now we check the other root \(r_{2}=-2\). (4) now becomes

We see that this is the diﬃcult root as at \(n=4\) it is not deﬁned as it gives \(1/0\) error. Hence

Therefore this is subcase two. For this case we do the following. We ﬁrst ﬁnd the solution using symbolic \(r\) using (4), and at the end replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}=\left ( r+2\right ) b_{0}\). From (4) and for \(n=2\)

\[ a_{4}=\frac {-a_{2}}{\left ( \left ( 4+r\right ) \left ( 3+r\right ) +\left ( 4+r\right ) -4\right ) }=-\frac {a_{2}}{\left ( r+6\right ) \left ( r+2\right ) }=-\frac {-\frac {1}{r}\frac {a_{0}}{r+4}}{\left ( r+6\right ) \left ( r+2\right ) }=\frac {1}{r}\frac {a_{0}}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }\]

\begin{align*} a_{6} & =\frac {-a_{4}}{\left ( \left ( 6+r\right ) \left ( 5+r\right ) +\left ( 6+r\right ) -4\right ) }=-\frac {a_{4}}{\left ( r+8\right ) \left ( r+4\right ) }\\ & =-\frac {\frac {1}{r}\frac {a_{0}}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }}{\left ( r+8\right ) \left ( r+4\right ) }\\ & =-\frac {1}{r}\frac {a_{0}}{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }\end{align*}

\begin{align*} \overline {y} & =x^{r}\left ( a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{r}\frac {1}{r+4}x^{2}+\frac {1}{r}\frac {1}{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {1}{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) \end{align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}\left ( r+2\right ) \) the above becomes

\begin{equation} \overline {y}=x^{r}b_{0}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) \tag {5}\end{equation}

\begin{align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=-2}\\ & =x^{-2}b_{0}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) _{r=-2}\\ & =x^{-2}b_{0}\left ( \frac {1}{r}\frac {1}{\left ( r+4\right ) \left ( r+6\right ) }x^{4}-\frac {1}{r}\frac {1}{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) }x^{6}+\cdots \right ) _{r=-2}\\ & =x^{-2}b_{0}\left ( -\frac {1}{16}x^{4}+\frac {1}{192}x^{6}-\cdots \right ) \end{align*}

\begin{align*} y_{1} & =\left ( -\frac {1}{16}x^{2}+\frac {1}{192}x^{4}-\cdots \right ) \\ & =-\frac {1}{16}\left ( x^{2}-\frac {1}{12}x^{4}-\cdots \right ) \end{align*}

We can removing the leading \(-\frac {1}{16}\) since it will be absorbed by the \(c_{1}\) constant. Hence

Notice the derivative is evaluated also at the bad root \(r=r_{bad}=-2\) same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives

\begin{align*} y_{2} & =\frac {d}{dr}\left ( x^{r}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) \right ) _{r=-2}\\ & =\overline {y}_{r=-2}\ln x+x^{r}\frac {d}{dr}\left ( \left ( r+2\right ) -\frac {1}{r}\frac {\left ( r+2\right ) }{r+4}x^{2}+\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{4}-\frac {1}{r}\frac {\left ( r+2\right ) }{\left ( r+8\right ) \left ( r+4\right ) \left ( r+4\right ) \left ( r+6\right ) \left ( r+2\right ) }x^{6}+\cdots \right ) _{r=-2}\end{align*}

\begin{align*} y_{2} & =y_{1}\ln x+x^{r}\left ( 1+\frac {\left ( r^{2}+4r+8\right ) }{r^{2}\left ( r+4\right ) ^{2}}x^{2}-\frac {1}{r^{2}}\frac {3r^{2}+20r+24}{\left ( r^{2}+10r+24\right ) ^{2}}x^{4}+\frac {\left ( 5r^{3}+68r^{2}+256r+192\right ) }{r^{2}\left ( r+4\right ) ^{3}\left ( r^{2}+14r+48\right ) ^{2}}x^{6}+\cdots \right ) _{r=-2}\\ & =y_{1}\ln x+x^{-2}\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {11}{2304}x^{6}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x^{2}-\frac {1}{12}x^{4}-\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( x^{2}-\frac {1}{12}x^{4}-\cdots \right ) +\left ( \frac {1}{4}+\frac {1}{x^{2}}+\frac {1}{64}x^{2}-\frac {11}{2304}x^{4}+\cdots \cdots \right ) \right ) \end{align*}

Hence \(p\left ( x\right ) =\frac {-3}{x},q\left ( x\right ) =1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\left ( -3\right ) =-3\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) -3r & =0\\ r^{2}-4r & =0\\ r\left ( r-4\right ) & =0\\ r & =0,4 \end{align*}

Therefore \(r_{1}=4,r_{2}=0\). Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align*} xy^{\prime \prime }-3y^{\prime }+xy & =0\\ x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-3\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+x\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r+1} & =0 \end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r-1}=0 \tag {2}\end{equation}

\begin{align*} r\left ( r-1\right ) a_{0}x^{r-1}-3ra_{0}x^{r-1} & =0\\ \left ( r\left ( r-4\right ) \right ) a_{0}x^{r-1} & =0 \end{align*}

Since \(a_{0}\neq 0\), then \(r\left ( r-4\right ) =0\) or \(r_{1}=0,r_{2}=4\) as was found above. The ode therefore satisﬁes

Since when \(r_{1}=4\) or \(r_{2}=0\) then the RHS is zero. When \(n=1\) then (2) gives

\begin{align*} \left ( 1+r\right ) \left ( r\right ) a_{1}-3\left ( 1+r\right ) a_{1} & =0\\ \left ( r^{2}-2r-3\right ) a_{1} & =0 \end{align*}

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}-3\left ( n+r\right ) a_{n}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) -3\left ( n+r\right ) } \tag {4}\end{align}

We check ﬁrst if this is subcase one or two. To do this, we check if the recurrence relation is deﬁned for both roots for all \(n\geq 2\). The above for \(r=4\) gives

Which is not deﬁned for \(n=4\). Hence this is subcase two, where \(y_{2}\) does not exist using standard method. Hence

For this case we do the following. We ﬁnd the solution using symbolic \(r\) and replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}\). From (4) and for \(n=2\)

\begin{align*} a_{6} & =\frac {-a_{4}}{\left ( 6+r\right ) \left ( 5+r\right ) -3\left ( 6+r\right ) }=\frac {-\frac {a_{0}}{r\left ( r+4\right ) \left ( r^{2}-4\right ) }}{r^{2}+8r+12}\\ & =\frac {-a_{0}}{\left ( r^{2}+8r+12\right ) r\left ( r+4\right ) \left ( r^{2}-4\right ) }\end{align*}

\begin{align*} \overline {y} & =x^{r}\left ( a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{r^{2}-4}x^{2}+\frac {1}{r\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) r\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) \end{align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{2}\right ) \) or \(b_{0}r\) since \(r_{2}=0\), the above becomes

\begin{equation} \overline {y}=x^{r}b_{0}\left ( r-\frac {r}{r^{2}-4}x^{2}+\frac {1}{\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) \left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) \tag {5}\end{equation}

\begin{align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=0}\\ & =b_{0}\left ( \frac {1}{\left ( 4\right ) \left ( -4\right ) }x^{4}-\frac {1}{\left ( 12\right ) \left ( 4\right ) \left ( -4\right ) }x^{6}+\cdots \right ) \\ & =b_{0}\left ( -\frac {1}{16}x^{4}+\frac {1}{192}x^{6}+\cdots \right ) \end{align*}

\begin{align*} y_{1} & =\left ( -\frac {1}{16}x^{4}+\frac {1}{192}x^{6}+\cdots \right ) \\ & =-\frac {1}{16}\left ( x^{4}-\frac {1}{12}x^{6}+\cdots \right ) \end{align*}

We can removing the leading \(-\frac {1}{16}\) since it will be absorbed by the \(c_{1}\) constant. Hence

\begin{align*} y_{1} & =c_{1}\left ( x^{4}-\frac {1}{12}x^{6}+\cdots \right ) \\ & =x^{4}c_{1}\left ( 1-\frac {1}{12}x^{2}+\cdots \right ) \end{align*}

Notice the derivative is evaluated also at root \(r=r_{bad}=0\) the same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives

\begin{align*} y_{2} & =\frac {d}{dr}\left ( x^{r}\left ( r-\frac {r}{r^{2}-4}x^{2}+\frac {1}{\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) \left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+x^{r}\frac {d}{dr}\left ( r-\frac {r}{r^{2}-4}x^{2}+\frac {1}{\left ( r+4\right ) \left ( r^{2}-4\right ) }x^{4}-\frac {1}{\left ( r^{2}+8r+12\right ) \left ( r+4\right ) \left ( r^{2}-4\right ) }x^{6}+\cdots \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+x^{0}\left ( 1+\frac {\left ( r^{2}+4\right ) }{\left ( r^{2}-4\right ) ^{2}}x^{2}-\frac {3r^{2}+8r-4}{\left ( r^{3}+4r^{2}-4r-16\right ) ^{2}}x^{4}+\frac {1}{\left ( r+2\right ) ^{3}}\frac {5r^{3}+38r^{2}+44r-88}{\left ( r^{3}+8r^{2}+4r-48\right ) ^{2}}x^{6}-\cdots \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {11}{2304}x^{6}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =x^{4}c_{1}\left ( 1-\frac {1}{12}x^{2}+\cdots \right ) \\ & +c_{2}\left ( \ln x\left ( x^{4}\left ( 1-\frac {1}{12}x^{2}+\cdots \right ) \right ) +\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {11}{2304}x^{6}+\cdots \right ) \right ) \end{align*}

Show that \(p\left ( x\right ) =\frac {\left ( x^{2}-2x\right ) }{x^{2}}=\frac {\left ( x-2\right ) }{x},q\left ( x\right ) =\frac {2}{x^{2}}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\left ( x-2\right ) =-2\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}2=2\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) -2r+2 & =0\\ r^{2}-3r+2 & =0\\ r & =2,1 \end{align*}

Therefore \(r_{1}=2,r_{2}=1\). Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align*} x^{2}y^{\prime \prime }+\left ( x^{2}-2x\right ) y^{\prime }+2y & =0\\ x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\left ( x^{2}-2x\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+2\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-2x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }2a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r+1}-\sum _{n=0}^{\infty }2\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }2a_{n}x^{n+r} & =0 \end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) a_{n-1}x^{n+r}-\sum _{n=0}^{\infty }2\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }2a_{n}x^{n+r}=0 \tag {2}\end{equation}

\begin{align*} r\left ( r-1\right ) a_{0}x^{r}-2ra_{0}x^{r}+2a_{0}x^{r} & =0\\ \left ( r\left ( r-1\right ) -2r+2\right ) a_{0}x^{r} & =0\\ \left ( r^{2}-3r+2\right ) a_{0}x^{r} & =0 \end{align*}

Since \(a_{0}\neq 0\), then \(r^{2}-3r+2=0,\)or \(r_{1}=2,r_{2}=1\) as was found above. The ode therefore satisﬁes

\begin{equation} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-1\right ) a_{n-1}-2\left ( n+r\right ) a_{n}+2a_{n}=0\nonumber \end{equation}

\begin{align} a_{n} & =-\frac {\left ( n+r-1\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) -2\left ( n+r\right ) +2}a_{n-1}\nonumber \\ & =-\frac {1}{n+r-2}a_{n-1} \tag {3}\end{align}

We check ﬁrst if this is subcase one or two. To do this, we check if the above recurrence relation is deﬁned for both roots for all \(n\geq 1\). The above for \(r=r_{1}=2\) gives

Which is not deﬁned for \(n=1\). Hence this is subcase two, where \(y_{2}\) does not exist using standard method. Hence

For this case we do the following. We ﬁnd the solution using symbolic \(r\) and replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}\). From (3) and for \(n=1\)

\begin{align*} \overline {y} & =x^{r}\left ( a_{0}+a_{1}x+a_{2}x^{2}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{r-1}x+\frac {1}{\left ( r\right ) \left ( r-1\right ) }x^{2}-\frac {1}{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) }x^{3}+\frac {1}{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \end{align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}\left ( r-1\right ) \) since \(r_{bad}=1\), the above becomes

\begin{align} \overline {y} & =x^{r}b_{0}\left ( \left ( r-1\right ) -\frac {\left ( r-1\right ) }{r-1}x+\frac {\left ( r-1\right ) }{\left ( r\right ) \left ( r-1\right ) }x^{2}-\frac {\left ( r-1\right ) }{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) }x^{3}+\frac {\left ( r-1\right ) }{\left ( r\right ) \left ( r-1\right ) \left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \nonumber \\ & =x^{r}b_{0}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \tag {5}\end{align}

\begin{align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=1}\\ & =xb_{0}\left ( -x+x^{2}-\frac {1}{2}x^{3}+\frac {1}{\left ( 1\right ) \left ( 2\right ) \left ( 3\right ) }x^{4}-\cdots \right ) \\ & =xb_{0}\left ( -x+x^{2}-\frac {1}{2}x^{3}+\frac {1}{6}x^{4}-\cdots \right ) \end{align*}

\begin{align*} y_{1} & =x\left ( -x+x^{2}-\frac {1}{2}x^{3}+\frac {1}{6}x^{4}-\cdots \right ) \\ & =-x^{2}+x^{3}-\frac {1}{2}x^{4}+\frac {1}{6}x^{5}-\cdots \end{align*}

Notice the derivative is evaluated also at root \(r=r_{bad}=1,\) the same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives

\begin{align*} y_{2} & =\frac {d}{dr}\left ( x^{r}b_{0}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) \right ) _{r=1}\\ & =\overline {y}_{r=1}\ln x+x^{r=1}\frac {d}{dr}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) _{r=1}\\ & =y_{1}\ln x+x\frac {d}{dr}\left ( \left ( r-1\right ) -x+\frac {1}{r}x^{2}-\frac {1}{r\left ( r+1\right ) }x^{3}+\frac {1}{r\left ( r+1\right ) \left ( r+2\right ) }x^{4}-\cdots \right ) _{r=1}\\ & =y_{1}\ln x+x\left ( 1-\frac {1}{r^{2}}x^{2}+\frac {1}{r^{2}}\frac {2r+1}{\left ( r+1\right ) ^{2}}x^{3}-\frac {1}{r^{2}}\frac {3r^{2}+6r+2}{\left ( r^{2}+3r+2\right ) ^{2}}x^{4}-\cdots \right ) _{r=1}\\ & =y_{1}\ln x+x\left ( 1-x^{2}+\frac {3}{4}x^{3}-\frac {11}{36}x^{4}-\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( -x^{2}+x^{3}-\frac {1}{2}x^{4}+\frac {1}{6}x^{5}-\cdots \right ) \\ & +c_{2}\left ( \ln x\left ( -x^{2}+x^{3}-\frac {1}{2}x^{4}+\frac {1}{6}x^{5}-\cdots \right ) +\left ( x-x^{3}+\frac {3}{4}x^{4}-\frac {11}{36}x^{5}-\cdots \right ) \right ) \end{align*}

Hence \(p\left ( x\right ) =\frac {x}{x-1},q\left ( x\right ) =\frac {1}{x\left ( x-1\right ) }\). There is a singular point at \(x=0\) and at \(x=1\). For \(x=0\), \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =0\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) & =0\\ r & =0,1 \end{align*}

For expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align*} \left ( x-1\right ) y^{\prime \prime }+xy^{\prime }+\frac {y}{x} & =0\\ \left ( x-1\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+x^{-1}\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r-1} & =0 \end{align*}

\begin{equation} \sum _{n=1}^{\infty }\left ( n+r-1\right ) \left ( n+r-2\right ) a_{n-1}x^{n+r-2}-\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=2}^{\infty }\left ( n+r-2\right ) a_{n-2}x^{n+r-2}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-2}=0 \tag {2}\end{equation}

\begin{align*} \left ( r\right ) \left ( r-1\right ) a_{0}-\left ( 1+r\right ) \left ( r\right ) a_{1}+a_{0} & =0\\ a_{1} & =\frac {a_{0}+\left ( r\right ) \left ( r-1\right ) a_{0}}{\left ( 1+r\right ) \left ( r\right ) }=\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }a_{0}\end{align*}

For \(r=0\) the above is not deﬁned. Therefore this falls into case two (diﬃcult case). Hence \(r_{bad}=0\). For \(r=1\) we see \(a_{1}\) is deﬁned.

For this case we do the following. We ﬁnd the solution using symbolic \(r\) and replace \(a_{0}\) by \(\left ( r-r_{bad}\right ) b_{0}=rb_{0}\). For \(n=1\)

\begin{equation} a_{n}=\frac {\left ( n+r-1\right ) \left ( n+r-2\right ) +1}{\left ( n+r\right ) \left ( n+r-1\right ) }a_{n-1}+\frac {\left ( n+r-2\right ) }{\left ( n+r\right ) \left ( n+r-1\right ) }a_{n-2} \tag {3}\end{equation}

\begin{align*} a_{2} & =\frac {\left ( 1+r\right ) \left ( r\right ) +1}{\left ( 2+r\right ) \left ( 1+r\right ) }a_{1}+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }a_{0}\\ & =\frac {r\left ( 1+r\right ) +1}{\left ( 2+r\right ) \left ( 1+r\right ) }\left ( \frac {1+r\left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }a_{0}\right ) +\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }a_{0}\\ & =\left ( \frac {r\left ( 1+r\right ) +1}{\left ( 2+r\right ) \left ( 1+r\right ) }\frac {1+r\left ( r-1\right ) }{r\left ( 1+r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) a_{0}\\ & =\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) a_{0}\end{align*}

\begin{align*} a_{3} & =\frac {\left ( 2+r\right ) \left ( 1+r\right ) +1}{\left ( 3+r\right ) \left ( 2+r\right ) }a_{2}+\frac {\left ( 1+r\right ) }{\left ( 3+r\right ) \left ( 2+r\right ) }a_{1}\\ & =\frac {\left ( 2+r\right ) \left ( 1+r\right ) +1}{\left ( 3+r\right ) \left ( 2+r\right ) }\left ( \left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) a_{0}\right ) +\frac {\left ( 1+r\right ) }{\left ( 3+r\right ) \left ( 2+r\right ) }\left ( \frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }a_{0}\right ) \\ & =\left [ \frac {\left ( 2+r\right ) \left ( 1+r\right ) +1}{\left ( 3+r\right ) \left ( 2+r\right ) }\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) +\frac {\left ( 1+r\right ) }{\left ( 3+r\right ) \left ( 2+r\right ) }\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }\right ] a_{0}\end{align*}

\begin{align*} \overline {y} & =x^{r}\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \end{align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}r\) since \(r_{bad}=0\), the above becomes

\begin{align} \overline {y} & =x^{r}b_{0}\left ( r+r\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) \left ( r\right ) }x+r\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) \left ( r\right ) }+\frac {r}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \nonumber \\ & =x^{r}b_{0}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \tag {5}\end{align}

\begin{align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=0}\\ & =x^{0}b_{0}\left ( x+\left ( \frac {1}{\left ( 2\right ) \left ( 1\right ) \left ( 1\right ) }\right ) x^{2}+\left [ \frac {\left ( 2\right ) \left ( 1\right ) +1}{\left ( 3\right ) \left ( 2\right ) }\left ( \frac {\left ( 1\right ) \left ( 1\right ) }{\left ( 2\right ) \left ( 1\right ) \left ( 1\right ) }\right ) +\frac {\left ( 1\right ) }{\left ( 3\right ) \left ( 2\right ) }\right ] x^{3}\cdots \right ) \\ & =b_{0}\left ( x+\frac {1}{2}x^{2}+\frac {5}{12}x^{3}+\cdots \right ) \end{align*}

Notice the derivative is evaluated also at root \(r=r_{bad}=0,\) the same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives

\begin{align*} y_{2} & =\frac {d}{dr}\left ( x^{r}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) \right ) _{r=0}\\ & =\overline {y}_{r=0}\ln x+x^{r=0}\frac {d}{dr}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) _{r=0}\end{align*}

\[ y_{2}=y_{1}\ln x+\frac {d}{dr}\left ( r+\frac {1+\left ( r\right ) \left ( r-1\right ) }{\left ( 1+r\right ) }x+\left ( \frac {\left ( r\left ( 1+r\right ) +1\right ) \left ( 1+r\left ( r-1\right ) \right ) }{\left ( 2+r\right ) \left ( 1+r\right ) \left ( 1+r\right ) }+\frac {r^{2}}{\left ( 2+r\right ) \left ( 1+r\right ) }\right ) x^{2}+\cdots \right ) _{r=0}\]

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x+\frac {1}{2}x^{2}+\frac {5}{12}x^{3}+\cdots \right ) \\ & +c_{2}\left ( \ln x\left ( x+\frac {1}{2}x^{2}+\frac {5}{12}x^{3}+\cdots \right ) +\left ( 1-2x-\frac {5}{4}x^{2}+\cdots \right ) \right ) \end{align*}

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r-1 & =0\\ r^{2}-1 & =0\\ r & =1,-1 \end{align*}

Therefore \(r_{1}=1,r_{2}=-1\). Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\left ( x^{2}-1\right ) \sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+2}-\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \end{align}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r}-\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \tag {2}\end{equation}

\begin{align*} r\left ( r-1\right ) a_{0}x^{r}+ra_{0}x^{r}-a_{0} & =0\\ \left ( r\left ( r-1\right ) +r-1\right ) a_{0}x^{r} & =0\\ \left ( r^{2}-1\right ) a_{0}x^{r} & =0 \end{align*}

Since \(a_{0}\neq 0\), then \(r^{2}=1\) or \(r_{1}=,r_{2}=-1\) as was found above. The ode therefore satisﬁes

\begin{equation} x^{2}y^{\prime \prime }+xy^{\prime }+\left ( x^{2}-1\right ) y=\left ( r^{2}-1\right ) a_{0}x^{r} \tag {2A}\end{equation}

\begin{align*} \left ( 1+r\right ) \left ( r\right ) a_{1}+\left ( 1+r\right ) a_{1}-a_{1} & =0\\ \left ( \left ( 1+r\right ) \left ( r\right ) +\left ( 1+r\right ) -1\right ) a_{1} & =0\\ \left ( r\left ( r+2\right ) \right ) a_{1} & =0 \end{align*}

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n-2}-a_{n} & =0\nonumber \\ a_{n} & =\frac {-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) -1} \tag {4}\end{align}

We check ﬁrst if this is subcase one or two. To do this, we check if the recurrence relation is deﬁned for both roots for all \(n\geq 2\). The above for \(r=1\) gives

We see that it is deﬁned for all \(n\geq 2\). Now we check the other root \(r_{2}=-1\). (4) now becomes

We see that this is the diﬃcult root as at \(n=2\) it is not deﬁned as it gives \(1/0\) error. Hence

\[ a_{4}=\frac {-a_{2}}{\left ( \left ( 4+r\right ) \left ( 3+r\right ) +\left ( 4+r\right ) -1\right ) }=-\frac {a_{2}}{\left ( r+5\right ) \left ( r+3\right ) }=-\frac {\frac {-a_{0}}{\left ( r+1\right ) \left ( r+3\right ) }}{\left ( r+5\right ) \left ( r+3\right ) }=\frac {a_{0}}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }\]

\begin{align*} a_{6} & =\frac {-a_{4}}{\left ( \left ( 6+r\right ) \left ( 5+r\right ) +\left ( 6+r\right ) -1\right ) }=-\frac {a_{4}}{\left ( r+7\right ) \left ( r+5\right ) }=-\frac {\frac {a_{0}}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }}{\left ( r+7\right ) \left ( r+5\right ) }\\ & =-\frac {a_{0}}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }\end{align*}

\begin{align*} \overline {y} & =x^{r}\left ( a_{0}+a_{2}x^{2}+a_{4}x^{4}+\cdots \right ) \\ & =x^{r}a_{0}\left ( 1-\frac {1}{\left ( r+1\right ) \left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \end{align*}

Replacing \(a_{0}\) by \(b_{0}\left ( r-r_{bad}\right ) \) or \(b_{0}\left ( r+1\right ) \) the above becomes

\begin{align} \overline {y} & =x^{r}b_{0}\left ( \left ( r+1\right ) -\frac {\left ( r+1\right ) }{\left ( r+1\right ) \left ( r+3\right ) }x^{2}+\frac {\left ( r+1\right ) }{\left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{4}-\frac {\left ( r+1\right ) }{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+1\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \nonumber \\ & =x^{r}b_{0}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \tag {5}\end{align}

\begin{align*} y_{1} & =\overline {y}_{r=r_{bad}}\\ & =\overline {y}_{r=-1}\\ & =x^{-1}b_{0}\left ( -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) _{r=-1}\\ & =x^{-1}b_{0}\left ( -\frac {1}{\left ( -1+3\right ) }x^{2}+\frac {1}{\left ( -1+5\right ) \left ( -1+3\right ) \left ( -1+3\right ) }x^{4}-\frac {1}{\left ( -1+7\right ) \left ( -1+5\right ) \left ( -1+5\right ) \left ( -1+3\right ) \left ( -1+3\right ) }x^{6}+\cdots \right ) \\ & =x^{-1}b_{0}\left ( -\frac {1}{2}x^{2}+\frac {1}{16}x^{4}-\frac {1}{384}x^{6}+\cdots \right ) \end{align*}

\begin{align*} y_{1} & =\frac {1}{x}\left ( -\frac {1}{2}x^{2}+\frac {1}{16}x^{4}-\frac {1}{384}x^{6}+\cdots \right ) \\ & =\left ( -\frac {1}{2}x+\frac {1}{16}x^{3}-\frac {1}{384}x^{5}+\cdots \right ) \\ & =-\frac {1}{2}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) \end{align*}

We can remove the leading \(-\frac {1}{2}\) since it will be absorbed by the \(c_{1}\) constant. Hence

Notice the derivative is evaluated also at the bad root \(r=r_{bad}=-2\) same as for \(y_{1}\). Hence, and using \(b_{0}=1\) and using (5) the above gives

\begin{align*} y_{2} & =\frac {d}{dr}\left ( x^{r}b_{0}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) \right ) _{r=-1}\\ & =\overline {y}_{r=-1}\ln x+x^{r}\frac {d}{dr}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) _{r=-1}\end{align*}

\begin{align*} y_{2} & =y_{1}\ln x+x^{-1}\frac {d}{dr}\left ( \left ( r+1\right ) -\frac {1}{\left ( r+3\right ) }x^{2}+\frac {1}{\left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{4}-\frac {1}{\left ( r+7\right ) \left ( r+5\right ) \left ( r+5\right ) \left ( r+3\right ) \left ( r+3\right ) }x^{6}+\cdots \right ) _{r=-1}\\ & =y_{1}\ln x+x^{-1}\left ( 1+\frac {1}{\left ( r+3\right ) ^{2}}x^{2}-\frac {3r+13}{\left ( r+3\right ) ^{3}\left ( r+5\right ) ^{2}}x^{4}+\frac {1}{\left ( r+7\right ) ^{2}}\frac {5r^{2}+52r+127}{\left ( r^{2}+8r+15\right ) ^{3}}x^{6}+\cdots \right ) _{r=-1}\\ & =y_{1}\ln x+x^{-1}\left ( 1+\frac {1}{4}x^{2}-\frac {5}{64}x^{4}+\frac {5}{1152}x^{6}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\left ( \frac {1}{x}+\frac {1}{4}x-\frac {5}{64}x^{3}+\frac {5}{1152}x^{5}+\cdots \right ) \right ) \end{align*}

This is same example as above but with non zero in the RHS. So we can use the solution for \(y_{h}\) obtained above, but need to ﬁnd \(y_{p}\) here and add these to obtain the general solution. From above we found that

\begin{align*} y_{h} & =c_{1}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\left ( \frac {1}{x}+\frac {1}{4}x-\frac {5}{64}x^{3}+\frac {5}{1152}x^{5}+\cdots \right ) \right ) \end{align*}

And from (2A) in the above example we also found the balance equation, which is always the starting point to ﬁnding \(y_{p}\), which is

Therefore, and as we did all the time, relabel \(r\) as \(m\) and \(a\) as \(c\) so not to confuse notations. Therefore we have

Now we ﬁnd \(y_{p}\) using the same recursive relation found when ﬁnding \(y_{h}\) terms but using \(r=m=0\) now and using \(a_{0}=c_{0}=-1\) (instead of \(a_{0}=1\,\) as is always done when ﬁnding \(y_{h}\)). Also let \(c_{1}=0\) as that is the same as \(a_{1}\). Now we get to the recurrence relation (4) in last example which is

Using \(c\) in place of \(a\) and using \(m\) in place \(r\) it becomes for \(n\geq 2\)

\begin{align*} y_{p} & =x^{m}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =c_{0}+c_{2}x^{2}+c_{4}x^{4}+\cdots \\ & =-1+\frac {1}{3}x^{2}-\frac {1}{45}x^{4}+\frac {1}{1575}x^{6}+\cdots \end{align*}

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\\ & c_{2}\left ( \ln \left ( x\right ) \left ( x-\frac {1}{8}x^{3}+\frac {1}{192}x^{5}+\cdots \right ) +\left ( \frac {1}{x}+\frac {1}{4}x-\frac {5}{64}x^{3}+\frac {5}{1152}x^{5}+\cdots \right ) \right ) \\ & +\left ( -1+\frac {1}{3}x^{2}-\frac {1}{45}x^{4}+\frac {1}{1575}x^{6}+\cdots \right ) \end{align*}

This is same example as above but with \(\frac {1}{x}\) instead of \(1\) in the RHS to show that there will not be a series solution in this. From (2A) in the above example we found the balance equation, which is always the starting point to ﬁnding \(y_{p}\), which is

Therefore, and as we did all the time, relabel \(r\) as \(m\) and \(a\) as \(c\) so not to confuse notations. Therefore we have

\begin{align*} \left ( \left ( -1\right ) ^{2}-1\right ) c_{0} & =1\\ 0c_{0} & =1 \end{align*}

Therefore no solution exists. This is why there is no series solution for this ode. If we try to solve this using Maple, will will get no answer and the above explains why.

Roots of indicial equation are repeated ode internal name "second_order_series_method_regular_singular_point_repeated_root".

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ y_{2} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r_{2}}\end{align*}

\(r_{1},r_{2}\) are roots of the indicial equation. \(a_{0},b_{0}\) are set to \(1\) as arbitrary. The coeﬃcients \(b_{n}\) are not found from the recurrence relation but found using using \(b_{n}=\frac {d}{dr}a_{n}\left ( r\right ) \) after ﬁnding \(a_{n}\) ﬁrst, and the result evaluated at root \(r_{2}\). (notice that \(r=r_{1}=r_{2}\) in this case). Notice there is no \(C\) term in from of the \(\ln \) in this case as when root diﬀer by an integer and the sum on \(b_{n}\) starts at \(1\) since \(b_{0}\) is always zero due to \(\frac {d}{dr}a_{0}\left ( r\right ) =0\) always as \(a_{0}=1\) by default.

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =\frac {1}{x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r & =0\\ r^{2} & =0\\ r & =0,0 \end{align*}

Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align*} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+x\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+1} & =0 \end{align*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r}=0 \tag {1}\end{equation}

\begin{align*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r} & =0\\ \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n} & =0\\ \left ( r\right ) \left ( r-1\right ) a_{0}+ra_{0} & =0\\ a_{0}\left ( \left ( r^{2}-r\right ) +r\right ) & =0\\ a_{0}r^{2} & =0 \end{align*}

Hence \(r_{1}=0,r_{2}=0\). Since the roots are repeated then two linearly independent solutions can be constructed using

\begin{align*} y_{1} & =x^{r_{1}}\sum _{n=0}^{\infty }a_{n}x^{n}=\sum _{n=0}^{\infty }a_{n}x^{n}\\ y_{2} & =y_{1}\ln \left ( x\right ) +x^{r_{2}}\sum _{n=1}^{\infty }b_{n}x^{n}=y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n}\end{align*}

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n-1} & =0\nonumber \\ a_{n} & =-\frac {a_{n-1}}{\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) }\nonumber \\ & =-\frac {a_{n-1}}{\left ( n+r\right ) ^{2}} \tag {1}\end{align}

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \\ & =1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\cdots \end{align*}

In the case of duplicate roots, \(b_{n}\) is found using \(b_{n}=\frac {d}{dr}a_{n}\left ( r\right ) \). And this is evaluated at \(r=r_{0}=0\) in this case since \(r_{0}=0\) here. So we need to ﬁnd \(a_{n}\left ( r\right ) \). This is done from (1). For \(n=1\)

\begin{align*} b_{1} & =\frac {d}{dr}\left ( a_{1}\left ( r\right ) \right ) \\ b_{1} & =\frac {d}{dr}\left ( -\frac {a_{0}}{\left ( 1+r\right ) ^{2}}\right ) =\frac {d}{dr}\left ( -\frac {1}{\left ( 1+r\right ) ^{2}}\right ) =\frac {2}{\left ( r+1\right ) ^{3}}\end{align*}

\begin{align*} b_{2} & =\frac {d}{dr}\left ( a_{2}\left ( r\right ) \right ) \\ b_{2} & =\frac {d}{dr}\left ( -\frac {a_{1}}{\left ( 2+r\right ) ^{2}}\right ) =\frac {d}{dr}\left ( -\frac {-\frac {1}{\left ( 1+r\right ) ^{2}}}{\left ( 2+r\right ) ^{2}}\right ) =\frac {d}{dr}\left ( \frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}\right ) =-2\frac {2r+3}{\left ( r^{2}+3r+2\right ) ^{3}}\end{align*}

And so on. Just remember when replacing the \(a_{n}\) in the above, is to use the original \(a_{n}\left ( r\right ) \) as function of \(r\) and not the actual \(a_{n}\) values from above. It has to be function of \(r\) ﬁrst before taking derivatives, Hence

\begin{align*} y_{2} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n}\\ & =y_{1}\ln \left ( x\right ) +b_{1}x+b_{2}x^{2}+b_{3}x^{3}+\cdots \\ & =y_{1}\ln \left ( x\right ) +2x-\frac {3}{4}x^{2}+\cdots \\ & =y_{1}\ln \left ( x\right ) +\left ( 2x-\frac {3}{4}x^{2}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\cdots \right ) +c_{2}\left ( y_{1}\ln \left ( x\right ) +\left ( 2x-\frac {3}{4}x^{2}+\cdots \right ) \right ) \end{align*}

The homogenous ode was solved up, so we just need to ﬁnd \(y_{p}\). To ﬁnd \(y_{p}\), and using \(m\) in place of \(r\) and \(c\) in place of \(a\) so not to confuse terms with the \(y_{h}\) terms, then from the above problem, we found the indicial equation. Hence the balance equation is

To balance this we need \(m=0\). Hence \(0c_{0}=1\) which is not possible. Hence no particular solution exists. No solution in series exists.

This is the same ode as above but with diﬀerent RHS. So we will go directly to ﬁnding \(y_{p}\). From above we found that the balance equation is

Which implies \(m=-1\) and therefore \(m^{2}c_{0}=1\) or \(c_{0}=1\). Using the recurrence equation (1) in the above problem using using \(c_{n}\) in place of \(a_{n}\) and \(m\) in place or \(r\) gives

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =\frac {1}{x}\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

Which is not deﬁned. Hence no \(y_{p}\) exist. There is no solution in terms of series solution.

This is the same ode as above, where we found \(y_{h}\) but with diﬀerent RHS. So we will go directly to ﬁnding \(y_{p}\). From above we found that the balance equation is

Which implies \(m=1\) and therefore \(m^{2}c_{0}=1\) or \(c_{0}=1\). Using the recurrence equation (1) in the above problem and using \(c_{n}\) in place of \(a_{n}\) and \(m\) in place or \(r\) gives

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+m}\\ & =x\sum _{n=0}^{\infty }c_{n}x^{n}\end{align*}

\begin{align*} y_{p} & =x\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x\left ( 1-\frac {1}{4}x+\frac {1}{36}x^{2}-\frac {1}{576}x^{3}+\cdots \right ) \\ & =\left ( x-\frac {1}{4}x^{2}+\frac {1}{36}x^{3}-\frac {1}{576}x^{4}+\cdots \right ) \end{align*}

Using \(y_{h}\) found in the above problem since that does not change, then the general solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}\left ( 1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\frac {1}{576}x^{4}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( 1-x+\frac {1}{4}x^{2}-\frac {1}{36}x^{3}+\frac {1}{576}x^{4}+\cdots \right ) +\left ( 2x-\frac {3}{4}x^{2}+\frac {14}{108}x^{3}+\cdots \right ) \right ) \\ & +\left ( x-\frac {1}{4}x^{2}+\frac {1}{36}x^{3}-\frac {1}{576}x^{4}+\cdots \right ) \end{align*}

Hence \(p\left ( x\right ) =\frac {1}{x},q\left ( x\right ) =-1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}1=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r & =0\\ r^{2} & =0\\ r & =0,0 \end{align*}

Therefore \(r_{1}=0,r_{2}=0\). Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

\begin{align*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-x\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }a_{n}x^{n+r+1} & =0 \end{align*}

\begin{align} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }a_{n-2}x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( \left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) \right ) a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }a_{n-2}x^{n+r-1} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) ^{2}a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }a_{n-2}x^{n+r-1} & =0 \tag {1}\end{align}

Hence \(r_{1}=0,r_{2}=0\) as found earlier. Since the roots are repeated then two linearly independent solutions can be constructed using

\begin{align*} \left ( 1+r\right ) \left ( r\right ) a_{1}+\left ( 1+r\right ) a_{1} & =0\\ \left ( r+1\right ) ^{2}a_{1} & =0 \end{align*}

Hence \(a_{1}=0\). The recurrence relation is obtained for \(n\geq 2\). From (1)

\begin{align} n+r\left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}-a_{n-2} & =0\nonumber \\ a_{n} & =\frac {a_{n-2}}{\left ( n+r\right ) ^{2}} \tag {1}\end{align}

Since we need to diﬀerentiate \(y_{1}\) to obtain \(y_{2}\) and the diﬀerentiation is w.r.t \(r\), we will carry the calculations with \(r\) in place and at the end replace \(r\) by its value (which happened to be \(zero\) in this example). We do this only in the case of repeated roots.

For \(n=5\), we will ﬁnd \(a_{5}=0\) (for all odd \(n\) this is the case). For \(n=6\)

And so on. We see that \(n^{th}\) term is \(a_{n}=\Pi _{j=1}^{k}\frac {1}{\left ( 2j+r\right ) ^{2}}\). Now we can substitute the \(r=0\) value into the above to obtain

\begin{align*} a_{2} & =\frac {1}{4}\\ a_{4} & =\frac {1}{64}\\ a_{6} & =\frac {1}{2304}\end{align*}

\begin{align*} y_{1} & =\sum a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \end{align*}

To ﬁnd \(y_{2}\) we use \(b_{n}=\frac {d}{dr}a_{n}\) and evaluate this at \(r=r_{2}\) which in this case is zero. Hence

\begin{align*} b_{2} & =\frac {d}{dr}a_{2}=\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}}\right ) =\left ( -\frac {2}{\left ( r+2\right ) ^{3}}\right ) _{r=0}=-\frac {2}{8}=-\frac {1}{4}\\ b_{4} & =\frac {d}{dr}a_{4}=\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}}\right ) =\left ( -4\frac {r+3}{\left ( r^{2}+6r+8\right ) ^{3}}\right ) _{r=0}=\left ( -4\frac {3}{\left ( 8\right ) ^{3}}\right ) =-\frac {3}{128}\\ b_{6} & =\frac {d}{dr}a_{6}\\ & =\frac {d}{dr}\left ( \frac {1}{\left ( 2+r\right ) ^{2}\left ( 4+r\right ) ^{2}\left ( 6+r\right ) ^{2}}\right ) \\ & =\left ( -2\frac {3r^{2}+24r+44}{\left ( r^{3}+12r^{2}+44r+48\right ) ^{3}}\right ) _{r=0}\\ & =-2\frac {44}{\left ( 48\right ) ^{3}}\\ & =-\frac {11}{13\,824}\end{align*}

\begin{align*} y_{1} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r_{2}}\\ & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n}\\ & =y_{1}\ln \left ( x\right ) +\left ( b_{2}x^{2}+b_{4}x^{4}+b_{6}x^{6}+\cdots \right ) \\ & =y_{1}\ln \left ( x\right ) +\left ( -\frac {1}{4}x^{2}-\frac {3}{128}x^{4}+-\frac {11}{13\,824}x^{6}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \right ) \\ & +c_{2}\left ( \ln \left ( x\right ) \left ( 1+\frac {1}{4}x^{2}+\frac {1}{64}x^{4}+\frac {1}{2304}x^{6}+\cdots \right ) +\left ( -\frac {1}{4}x^{2}-\frac {3}{128}x^{4}+-\frac {11}{13\,824}x^{6}+\cdots \right ) \right ) \end{align*}

Hence \(p\left ( x\right ) =\frac {1}{\sin \left ( x\right ) },q\left ( x\right ) =\frac {1}{\sin x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x}{x-\frac {x^{2}}{3!}+\frac {x^{5}}{5!}-\cdots }=\frac {1}{1-\frac {x}{3!}+\frac {x^{4}}{5!}-}=1\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {x^{2}}{x-\frac {x^{2}}{3!}+\frac {x^{5}}{5!}-\cdots }=\frac {x}{1-\frac {x^{2}}{3!}+\frac {x^{5}}{5!}-\cdots }=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +r & =0\\ r^{2} & =0\\ r & =0,0 \end{align*}

Therefore \(r_{1}=0,r_{2}=0\). Expansion around \(x=0\). This is regular singular point. Hence Frobenius is needed. Let

Using \(O\left ( x^{7}\right ) \) terms as the Order of the series (if more terms are needed we will use more terms from the \(\sin x\) series). This means we have to now only expand up to \(n=7\) as that is the order used for the series of \(\sin x\). The above becomes

\begin{multline*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-\frac {x^{3}}{3!}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\\ +\frac {x^{5}}{5!}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \end{multline*}

\begin{multline*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\frac {1}{6}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+1}\\ +\sum _{n=0}^{\infty }\frac {1}{120}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+3}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r}=0 \end{multline*}

\begin{multline*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{n-2}x^{n+r-1}\\ +\sum _{n=4}^{\infty }\frac {1}{120}\left ( n+r-4\right ) \left ( n+r-5\right ) a_{n-4}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1}=0 \end{multline*}

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) ^{2}a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {\left ( n+r-2\right ) \left ( n+r-3\right ) }{6}a_{n-2}x^{n+r-1}+\sum _{n=4}^{\infty }\frac {\left ( n+r-4\right ) \left ( n+r-5\right ) }{120}a_{n-4}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1}=0 \tag {1}\end{equation}

Hence \(r_{1}=0,r_{2}=0\) as found earlier. Since the roots are repeated then two linearly independent solutions can be constructed using

\begin{align*} \left ( 1+r\right ) ^{2}a_{1}+a_{0} & =0\\ a_{1} & =-\frac {a_{0}}{\left ( 1+r\right ) ^{2}}\\ & =-\frac {1}{\left ( 1+r\right ) ^{2}}\end{align*}

\begin{align*} \left ( 2+r\right ) ^{2}a_{2}-\frac {\left ( r\right ) \left ( r-1\right ) }{6}a_{0}+a_{1} & =0\\ \left ( 2+r\right ) ^{2}a_{2} & =-a_{1}+\frac {\left ( r\right ) \left ( r-1\right ) }{6}a_{0}\\ a_{2} & =\frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}\end{align*}

\begin{align*} \left ( 3+r\right ) ^{2}a_{3}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6}a_{1}+a_{2} & =0\\ a_{3} & =-\frac {a_{2}}{\left ( 3+r\right ) ^{2}}+\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}}a_{1}\\ & =-\frac {\frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}}{\left ( 3+r\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}}\frac {1}{\left ( 1+r\right ) ^{2}}\\ & =-\frac {\left ( r^{4}+r^{3}-r^{2}-r+6\right ) }{6\left ( r+3\right ) ^{2}\left ( r^{2}+3r+2\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}\left ( 1+r\right ) ^{2}}\end{align*}

\begin{equation} a_{n}=-\frac {a_{n-1}}{\left ( n+r\right ) ^{2}}+\frac {\left ( n+r-2\right ) \left ( n+r-3\right ) }{6\left ( n+r\right ) ^{2}}a_{n-2}-\frac {\left ( n+r-4\right ) \left ( n+r-5\right ) }{120\left ( n+r\right ) ^{2}}a_{n-4} \tag {2}\end{equation}

\(\allowbreak \)Since we need to diﬀerentiate \(y_{1}\) to obtain \(y_{2}\) and the diﬀerentiation is w.r.t \(r\), we will carry the calculations with \(r\) in place and at the end replace \(r\) by its value (which happened to be \(zero\) in this example). We do this only in the case of repeated roots.

\begin{align*} a_{4} & =-\frac {a_{3}}{\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}a_{2}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}a_{0}\\ & =-\frac {-\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) ^{2}}}{\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}a_{2}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}a_{0}\\ & =\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) ^{2}\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}\end{align*}

And so on. Now we replace \(r=0\) to ﬁnd \(y_{1}\). Just remember not to use anything over \(n=5\) since we cut oﬀ the series for \(\sin \left ( x\right ) \) at \(x^{5}\).

\begin{align*} a_{1} & =-\frac {1}{\left ( 1+r\right ) ^{2}}=-1\\ a_{2} & =\frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}=\frac {1}{4}\\ a_{3} & =-\frac {\left ( r^{4}+r^{3}-r^{2}-r+6\right ) }{6\left ( r+3\right ) ^{2}\left ( r^{2}+3r+2\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}\left ( 1+r\right ) ^{2}}=-\frac {1}{\left ( 2\right ) ^{2}\left ( 3\right ) ^{2}}=-\frac {1}{36}\\ a_{4} & =\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}\left ( r+3\right ) ^{2}\left ( 4+r\right ) ^{2}}+\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{6\left ( 4+r\right ) ^{2}}\frac {1}{\left ( r+1\right ) ^{2}\left ( r+2\right ) ^{2}}-\frac {\left ( r\right ) \left ( -1+r\right ) }{120\left ( 4+r\right ) ^{2}}\\ & =\frac {1}{\left ( 2\right ) ^{2}\left ( 3\right ) ^{2}\left ( 4\right ) ^{2}}+\frac {\left ( 2\right ) }{6\left ( 4\right ) ^{2}}\frac {1}{\left ( 2\right ) ^{2}}\\ & =\frac {1}{144}\end{align*}

\begin{align*} a_{5} & =-\frac {a_{4}}{\left ( 5+r\right ) ^{2}}+\frac {\left ( 3+r\right ) \left ( 2+r\right ) }{6\left ( 5+r\right ) ^{2}}a_{3}-\frac {\left ( 1+r\right ) \left ( r\right ) }{120\left ( 5+r\right ) ^{2}}a_{1}\\ & =-\frac {\frac {1}{144}}{5^{2}}+\frac {\left ( 3\right ) \left ( 2\right ) }{6\left ( 5\right ) ^{2}}\left ( -\frac {1}{36}\right ) \\ & =-\frac {1}{720}\end{align*}

\begin{align*} a_{6} & =-\frac {a_{5}}{\left ( 6+r\right ) ^{2}}+\frac {\left ( 4+r\right ) \left ( 3+r\right ) }{6\left ( 6+r\right ) ^{2}}a_{4}-\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{120\left ( 6+r\right ) ^{2}}a_{2}\\ & =-\frac {-\frac {1}{720}}{6^{2}}+\frac {\left ( 4\right ) \left ( 3\right ) }{6\left ( 6\right ) ^{2}}\frac {1}{144}-\frac {\left ( 2\right ) }{120\left ( 6\right ) ^{2}}\frac {1}{4}\\ & =\frac {1}{3240}\end{align*}

\begin{align*} a_{7} & =-\frac {a_{6}}{\left ( 7+r\right ) ^{2}}+\frac {\left ( 5+r\right ) \left ( 4+r\right ) }{6\left ( 7+r\right ) ^{2}}a_{5}-\frac {\left ( 3+r\right ) \left ( 2+r\right ) }{120\left ( 7+r\right ) ^{2}}a_{3}\\ & =-\frac {\frac {1}{3240}}{\left ( 7\right ) ^{2}}+\frac {\left ( 5\right ) \left ( 4\right ) }{6\left ( 7\right ) ^{2}}\left ( -\frac {1}{720}\right ) -\frac {\left ( 3\right ) \left ( 2\right ) }{120\left ( 7\right ) ^{2}}\left ( -\frac {1}{36}\right ) \\ & =-\frac {23}{317\,520}\end{align*}

\begin{align*} a_{8} & =-\frac {a_{7}}{\left ( 8+r\right ) ^{2}}+\frac {\left ( 6+r\right ) \left ( 5+r\right ) }{6\left ( 8+r\right ) ^{2}}a_{6}-\frac {\left ( 4+r\right ) \left ( 3+r\right ) }{120\left ( 8+r\right ) ^{2}}a_{4}\\ & =-\frac {\left ( -\frac {23}{317\,520}\right ) }{\left ( 8\right ) ^{2}}+\frac {\left ( 6\right ) \left ( 5\right ) }{6\left ( 8\right ) ^{2}}\left ( \frac {1}{3240}\right ) -\frac {\left ( 4\right ) \left ( 3\right ) }{120\left ( 8\right ) ^{2}}\left ( \frac {1}{144}\right ) \\ & =\frac {13}{903\,168}\end{align*}

Which is now the wrong value. It should be \(\frac {1}{62720}\). So using 3 terms from \(\sin x\) we obtain up to \(a_{7}\) correct terms. Hence

\begin{align*} y_{1} & =\sum a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+\cdots \\ & =1-\frac {1}{2}x+\frac {1}{4}x^{2}+\frac {1}{36}x^{3}+\frac {1}{144}x^{4}-\frac {1}{720}x^{5}+\frac {1}{3240}x^{6}-\frac {23}{317\,520}x^{7}+\cdots \end{align*}

What would have happened if we expanded \(\sin \left ( x\right ) \) only for two terms? Lets ﬁnd out. The ode becomes

\begin{align*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}-\frac {x^{3}}{3!}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=0}^{\infty }\frac {1}{6}\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r+1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0 \end{align*}

\begin{align*} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{n-2}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) ^{2}a_{n}x^{n+r-1}-\sum _{n=2}^{\infty }\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{n-2}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1} & =0 \end{align*}

\begin{align*} \left ( 1+r^{2}\right ) a_{1}+a_{0} & =0\\ a_{1} & =-\frac {a_{0}}{\left ( 1+r^{2}\right ) }=-\frac {1}{\left ( 1+r^{2}\right ) }\end{align*}

\begin{align} \left ( n+r\right ) ^{2}a_{n}-\frac {1}{6}\left ( n+r-2\right ) \left ( n+r-3\right ) a_{n-2}+a_{n-1} & =0\nonumber \\ a_{n} & =-\frac {a_{n-1}}{\left ( n+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( n+r-2\right ) \left ( n+r-3\right ) }{\left ( n+r\right ) ^{2}}a_{n-2} \tag {2A}\end{align}

\begin{align*} a_{2} & =-\frac {a_{1}}{\left ( 2+r\right ) ^{2}}+\frac {1}{6}\frac {r\left ( -1+r\right ) }{\left ( 2+r\right ) ^{2}}a_{0}\\ & =-\frac {-\frac {1}{\left ( 1+r^{2}\right ) }}{\left ( 2+r\right ) ^{2}}+\frac {1}{6}\frac {r\left ( -1+r\right ) }{\left ( 2+r\right ) ^{2}}\end{align*}

\begin{align*} a_{3} & =-\frac {a_{2}}{\left ( 3+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 1+r\right ) \left ( r\right ) }{\left ( 3+r\right ) ^{2}}a_{1}\\ & =-\frac {\frac {1}{4}}{\left ( 3+r\right ) ^{2}}-\frac {1}{6}\frac {\left ( 1+r\right ) \left ( r\right ) }{\left ( 3+r\right ) ^{2}}\end{align*}

\begin{align*} a_{4} & =-\frac {a_{3}}{\left ( 4+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 2+r\right ) \left ( 1+r\right ) }{\left ( 4+r\right ) ^{2}}a_{2}\\ & =-\frac {a_{3}}{\left ( 4\right ) ^{2}}+\frac {1}{6}\frac {\left ( 2\right ) \left ( 1\right ) }{\left ( 4\right ) ^{2}}a_{2}\\ & =-\frac {\left ( -\frac {1}{36}\right ) }{\left ( 4\right ) ^{2}}+\frac {1}{6}\frac {\left ( 2\right ) \left ( 1\right ) }{\left ( 4\right ) ^{2}}\left ( \frac {1}{4}\right ) \\ & =\frac {1}{144}\end{align*}

\begin{align*} a_{5} & =-\frac {a_{4}}{\left ( 5+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 3+r\right ) \left ( 2+r\right ) }{\left ( 5+r\right ) ^{2}}a_{3}\\ & =-\frac {\frac {1}{144}}{\left ( 5\right ) ^{2}}+\frac {1}{6}\frac {\left ( 3\right ) \left ( 2\right ) }{\left ( 5\right ) ^{2}}\left ( -\frac {1}{36}\right ) =-\frac {1}{720}\end{align*}

\begin{align*} a_{6} & =-\frac {a_{5}}{\left ( 6+r\right ) ^{2}}+\frac {1}{6}\frac {\left ( 6+r-2\right ) \left ( 6+r-3\right ) }{\left ( 6+r\right ) ^{2}}a_{4}\\ & =-\frac {\left ( -\frac {1}{720}\right ) }{\left ( 6\right ) ^{2}}+\frac {1}{6}\frac {\left ( 4\right ) \left ( 3\right ) }{6^{2}}\frac {1}{144}\\ & =\frac {11}{25\,920}\end{align*}

Which is the wrong value. We see that using two terms only from the \(\sin \left ( x\right ) \) gave up correct \(a_{n}\) values up to \(a_{5}\). What if we used only one term? Lets ﬁnd out.

\begin{align*} \sin \left ( x\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \left ( x+\cdots \right ) \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) ^{2}a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }a_{n-1}x^{n+r-1} & =0 \end{align*}

\begin{align*} \left ( n+r\right ) ^{2}a_{n}+a_{n-1} & =0\\ a_{n} & =-\frac {a_{n-1}}{\left ( n+r\right ) ^{2}}\end{align*}

\begin{align*} a_{1} & =-\frac {a_{0}}{\left ( 1+r\right ) ^{2}}\\ & =-\frac {1}{\left ( 1+r\right ) ^{2}}\end{align*}

We see that this is the wrong value. So when using one term only we obtain correct \(a_{n}\) up to \(a_{3}\). What do we learn from all the above? It is that if we expand \(f\left ( x\right ) \) up to \(O\left ( x^{n}\right ) \) order, then we can only determine correct terms up to \(a_{n}\) and no more. In the above when we used \(\sin \left ( x\right ) =x-\frac {x^{3}}{6}+\frac {x^{5}}{120}+O\left ( x^{7}\right ) \) then we obtained correct terms up to \(a_{7}\). And when we used \(\sin \left ( x\right ) =x-\frac {x^{3}}{6}+O\left ( x^{5}\right ) \) then we obtained correct terms up to \(a_{5}\) and when we used \(\sin \left ( x\right ) =x+O\left ( x^{3}\right ) \) then we obtained correct terms up to \(a_{3}\). So we should keep this in mind from now on,.

To ﬁnd \(y_{2}\) we use \(b_{n}=\frac {d}{dr}a_{n}\) and evaluate this at \(r=r_{2}\) which in this case is zero. Hence

\begin{align*} b_{1} & =\frac {d}{dr}a_{1}=\frac {d}{dr}\left ( -\frac {1}{\left ( 1+r\right ) ^{2}}\right ) _{r=0}=\frac {2}{\left ( r+1\right ) ^{3}}=2\\ b_{2} & =\frac {d}{dr}a_{2}=\frac {d}{dr}\left ( \frac {1}{\left ( 1+r\right ) ^{2}\left ( 2+r\right ) ^{2}}+\frac {\left ( r\right ) \left ( r-1\right ) }{6\left ( 2+r\right ) ^{2}}\right ) =\left ( \frac {5r^{4}+13r^{3}+9r^{2}-25r-38}{6\left ( r^{2}+3r+2\right ) ^{3}}\right ) _{r=0}=\frac {-38}{6\left ( 2\right ) ^{3}}=-\frac {19}{24}\\ b_{3} & =\frac {d}{dr}a_{3}\\ & =\frac {d}{dr}\left ( -\frac {\left ( r^{4}+r^{3}-r^{2}-r+6\right ) }{6\left ( r+3\right ) ^{2}\left ( r^{2}+3r+2\right ) ^{2}}-\frac {\left ( 1+r\right ) \left ( r\right ) }{6\left ( 3+r\right ) ^{2}\left ( 1+r\right ) ^{2}}\right ) \\ & =\left ( \frac {\left ( 4r^{6}+18r^{5}+20r^{4}-15r^{3}-18r^{2}+93r+114\right ) }{6\left ( r^{3}+6r^{2}+11r+6\right ) ^{3}}\right ) _{r=0}\\ & =\frac {114}{6\left ( 6\right ) ^{3}}\\ & =\frac {19}{216}\end{align*}

\begin{align*} y_{1} & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r_{2}}\\ & =y_{1}\ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n}\\ & =y_{1}\ln \left ( x\right ) +\left ( 2x-\frac {19}{24}x^{2}+\frac {19}{216}x^{3}+\cdots \right ) \end{align*}

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( 1-\frac {1}{2}x+\frac {1}{4}x^{2}+\frac {1}{36}x^{3}+\frac {1}{144}x^{4}+\cdots \right ) \\ & +c_{2}\left ( \left ( 1-\frac {1}{2}x+\frac {1}{4}x^{2}+\frac {1}{36}x^{3}+\frac {1}{144}x^{4}+\cdots \right ) \ln \left ( x\right ) +\left ( 2x-\frac {19}{24}x^{2}+\frac {19}{216}x^{3}+\cdots \right ) \right ) \end{align*}

4.3.2.5.4 irregular singular point ode internal name "second_order_series_method_irregular_singular_point"

4.3.2.5 Solved using series method