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

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

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

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

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

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

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

\[ m\left ( 2m-1\right ) c_{0}x^{m-1}=x \]

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

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

But now using \(m=2\)

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

Hence for \(n=1\)

\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*}

for \(n=2\)

\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*}

For \(n=3\)

\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*}

For \(n=4\)

\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*}

And so on. Hence

\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*}

Hence the solution is (\(y_{h}\) was found in the earlier problem)

\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*}