1.3.10 Example 10. \(xy^{\prime }+y=\frac {1}{x^{3}}\)

\begin{equation} xy^{\prime }+y=x^{-3} \tag {1}\end{equation}

This is same as example 1, but with nonzero on RHS. The solution is \(y=y_{h}+y_{p}\). Where \(y_{h}\) was found above as

\[ y_{h}=\frac {a_{0}}{x}\]

To find \(y_{p}\) we will use the balance equation, EQ (*) found in the first example when finding \(y_{h}\). We just need to rename \(a_{0}\) to \(c_{0}\) and add the \(x\) on the right side of the balance equation.

\[ \left ( r+1\right ) c_{0}x^{r}=x^{-3}\]

Hence \(r=-3\), therefore \(\left ( r+1\right ) c_{0}=1\) or \(c_{0}=-\frac {1}{2}\). Hence the particular solution is \(y_{p}=-\frac {1}{2}x^{-3}\). The solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =\frac {a_{0}}{x}-\frac {1}{2x^{3}}\end{align*}