2.2.2.0 Example 7 \(x^{2}y^{\prime \prime }+xy^{\prime }+xy=x\)

\[ x^{2}y^{\prime \prime }+xy^{\prime }+xy=x \]

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

\[ r^{2}c_{0}x^{r}=x \]

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

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

For \(r=1\)

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

Hence

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

Now to find few \(c_{n}\) terms. For \(n=1\)

\[ c_{1}=-\frac {c_{0}}{\left ( 2\right ) ^{2}}=-\frac {1}{4}\]

For \(n=2\)

\[ c_{2}=-\frac {c_{1}}{\left ( 2+1\right ) ^{2}}=\frac {\frac {1}{4}}{9}=\frac {1}{36}\]

For \(n=3\)

\[ c_{3}=-\frac {c_{2}}{\left ( 3+1\right ) ^{2}}=-\frac {\frac {1}{36}}{16}=-\frac {1}{576}\]

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}+\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*}