2.5.2.2 Example 2 \(2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=5\)

\[ 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=5 \]

Comparing the ode to

\[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \]

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*}

Therefore \(r_{1}=0,r_{2}=\frac {1}{2}\).  

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

\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*}

Re indexing to lowest powers on \(x\) gives

\[ \sum _{n=0}^{\infty }2\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) a_{n-1}x^{n+r-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }3a_{n-1}x^{n+r-1}=0 \]

For \(n=0\)

\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}

Therefore the differential equation satisfies

\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

\[ a_{n}=-\frac {n+2}{\left ( n\right ) \left ( 2n-1\right ) }a_{n-1}\]

Which gives (for \(a_{0}=1\)) (working out few terms using the above)

\[ y_{1}=1-3x+2x^{2}-\frac {2}{3}x^{3}+\cdots \]

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

\[ b_{n}=-\frac {n+\frac {5}{2}}{\left ( n+\frac {1}{2}\right ) \left ( 1+2n-1\right ) }b_{n-1}\]

Which gives (working out few terms)

\[ y_{2}=\sqrt {x}\left ( 1-\frac {7x}{6}+21\frac {x^{2}}{40}+\cdots \right ) \]

Hence the solution is

\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

\[ 2xy^{\prime \prime }+\left ( x+1\right ) y^{\prime }+3y=m\left ( 2m-1\right ) c_{0}x^{m-1}\]

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

\[ c_{n}=-\frac {n+m+2}{\left ( n+m\right ) \left ( 2m+2n-1\right ) }c_{n-1}\]

For \(m=1\) the above becomes

\[ c_{n}=-\frac {n+3}{\left ( n+1\right ) \left ( 1+2n\right ) }c_{n-1}\]

For \(n=1\)

\[ c_{1}=-\frac {1+3}{\left ( 1+1\right ) \left ( 1+2\right ) }c_{0}=-\frac {2}{3}c_{0}=-\frac {2}{3}\left ( 5\right ) =-\frac {10}{3}\]

For \(n=2\)

\[ c_{2}=-\frac {2+3}{\left ( 2+1\right ) \left ( 1+4\right ) }c_{1}=-\frac {1}{3}c_{1}=-\frac {1}{3}\left ( -\frac {10}{3}\right ) =\frac {10}{9}\]

For \(n=3\)

\[ c_{1}=-\frac {3+3}{\left ( 3+1\right ) \left ( 1+6\right ) }c_{2}=-\frac {3}{14}\left ( \frac {10}{9}\right ) =-\frac {2}{3}\left ( 5\right ) =-\frac {5}{21}\]

And so on. Hence

\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*}

Hence the final solution

\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*}