4.3.2.0 Ordinary point using power series method

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.

\[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \] 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.

Example 1 Solved using Taylor series method.\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*}

Hence \[ y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+1\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\] Where\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*}

Hence \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*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes \[ F_{1}=2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2 \] And\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*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes \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*}

And\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*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes \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*}

Hence\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*}

Solved using power series method.\[ y^{\prime \prime }+xy^{\prime }+y=2x+x^{2}+x^{4}\] 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}

Adjust all sums to lowest power on \(x\) gives\[ \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+\sum _{n=3}^{\infty }\left ( n-2\right ) a_{n-2}x^{n-2}+\sum _{n=2}^{\infty }a_{n-2}x^{n-2}=2x+x^{2}+x^{4}\] \(n=2\) gives \(x^{0}\) on the LHS with no match on the RHS. Hence\begin {align*} 2a_{2}+a_{0} & =0\\ a_{2} & =-\frac {1}{2}a_{0} \end {align*}

\(n=3\) gives \(x^{1}\) on the LHS with one match on the RHS. Hence\begin {align*} 6a_{3}+2a_{1} & =2\\ a_{3} & =\frac {2-2a_{1}}{6}\\ & =\frac {1}{3}-\frac {1}{3}a_{1} \end {align*}

\(n=4\) gives \(x^{2}\) on the LHS with one match on the RHS. Hence \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*}

\(n=5\) gives \(x^{3}\) on the LHS with no match on the RHS. Hence\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*}

\(n=6\) gives \(x^{4}\) on the LHS with one match on the RHS. Hence \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*}

And for \(n\geq 7\) we have recurrence relation\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*}

Hence for \(n=7\)\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*}

For \(n=8\)\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*}

And so on. Hence\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 \[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \] 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*}

Hence \(r_{1}=1,r_{2}=0\). All ordinary points will have the same roots. Let\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*}

The ode becomes\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*}

Reindex to lowest powers gives\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} For \(n=0\)\[ r\left ( r-1\right ) a_{0}x^{r-2}=0 \] The homogenous ode therefore satisﬁes\begin {equation} y^{\prime \prime }+xy^{\prime }+y=r\left ( r-1\right ) a_{0}x^{r} \tag {2} \end {equation} For \(n=1\), Eq (1) gives\[ \left ( 1+r\right ) \left ( r\right ) a_{1}=0 \] 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.

For \(n\geq 2\) we obtain the recurrence relation\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} For \(n=2\) and using \(a_{0}=1\)\[ a_{2}=-\frac {1}{3}a_{0}=-\frac {1}{3}\] For \(n=3\)\[ a_{3}=-\frac {1}{4}a_{1}=0 \] All odd \(a_{n}\) will be zero. For \(n=4\)\[ a_{4}=-\frac {1}{5}a_{2}=-\frac {1}{5}\left ( -\frac {1}{3}\right ) =\frac {1}{15}\] And so on. Hence\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 \(n=2\)\[ b_{2}=-\frac {b_{0}}{2}=-\frac {1}{2}\] For \(n=3\)\[ b_{3}=-\frac {b_{1}}{3}=0 \] For \(n=4\)\[ b_{4}=-\frac {b_{2}}{4}=\frac {1}{8}\] And so on. Hence \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*}

Hence the solution \(y_{h}\) is \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.

Example 2 \[ \frac {1}{x^{5}}y^{\prime \prime }+y^{\prime }+y=0 \] Solved using Taylor series method.\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*}

Hence \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*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes \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*}

And\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*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes \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*}

And\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*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes \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*}

And so on. Hence\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*}

Solved using power series method

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\[ x^{-5}y^{\prime \prime }+y^{\prime }+y=0 \] Hence\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}

Reindex so all powers start at lowest powers \(n-7\)\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.

For \(n\geq 7\) we have the recurrence relation\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}

Hence for \(n=7\)\[ a_{7}=-\frac {a_{1}+a_{0}}{42}\] For \(n=8\)\[ a_{8}=-\frac {2a_{2}+a_{1}}{\left ( 6+2\right ) \left ( 6+1\right ) }=\frac {-a_{1}}{56}\] For \(n=9\)\[ a_{9}=-\frac {\left ( 7-4\right ) a_{3}+a_{2}}{\left ( 7+2\right ) \left ( 7+1\right ) }=0 \] For \(n=10\)\[ a_{10}=-\frac {\left ( 8-4\right ) a_{4}+a_{3}}{\left ( 8+2\right ) \left ( 8+1\right ) }=0 \] For \(n=11\)\[ a_{11}=-\frac {\left ( 9-4\right ) a_{5}+a_{4}}{\left ( 9+2\right ) \left ( 9+1\right ) }=0 \] For \(n=12\)\[ a_{12}=-\frac {\left ( n-4\right ) a_{6}+a_{5}}{\left ( n+2\right ) \left ( n+1\right ) }=0 \] For \(n=13\)\[ a_{13}=-\frac {\left ( 11-4\right ) a_{7}+a_{6}}{\left ( 11+2\right ) \left ( 11+1\right ) }=-\frac {\left ( 11-4\right ) a_{7}}{\left ( 11+2\right ) \left ( 11+1\right ) }=-\frac {7}{156}a_{7}=-\frac {7}{156}\left ( -\frac {a_{1}+a_{0}}{42}\right ) =\frac {1}{936}a_{0}+\frac {1}{936}a_{1}\] And so on. Hence\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*}

Notice that all terms \(a_{n}=0\) for \(n=2\cdots 6\). The above becomes\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*}

Example 3 \[ \frac {1}{x^{2}}y^{\prime \prime }+y^{\prime }+y=\sin x \] 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\[ y^{\prime \prime }+x^{2}y^{\prime }+x^{2}y=x^{2}\sin x \] Hence\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}

Reindex so all powers to start from \(n\). This results in\[ \sum _{n=0}^{\infty }\left ( n+2\right ) \left ( n+1\right ) a_{n+2}x^{n}+\sum _{n=2}^{\infty }\left ( n-1\right ) a_{n-1}x^{n}+\sum _{n=2}^{\infty }a_{n-2}x^{n}=x^{2}\sin x \] 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*}

For \(n=0\)\begin {align*} 2a_{2} & =0\\ a_{2} & =0 \end {align*}

For \(n=1\)\begin {align*} \left ( 3\right ) \left ( 2\right ) a_{3} & =0\\ a_{3} & =0 \end {align*}

For \(n=2\)\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*}

For \(n=3\) (now we pick one term from the RHS which match on \(x^{3}\))\begin {align*} 20a_{5}+2a_{2}+a_{1} & =1\\ a_{5} & =\frac {1-a_{1}}{20} \end {align*}

For \(n=4\)\begin {align*} 30a_{6}+3a_{3}+a_{2} & =0\\ a_{6} & =0 \end {align*}

For \(n=5\)\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*}

And so on. Hence\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*}

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