1.1.3.8 Example 8. \(xy^{\prime }+y=\frac {1}{x^{2}}\)
\begin{equation} xy^{\prime }+y=x^{-2} \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.
\[ \left ( r+1\right ) c_{0}x^{r}=\frac {1}{x^{2}}\]
Hence
\(r=-2\), therefore
\(\left ( r+1\right ) c_{0}=1\) or
\(-c_{0}=1\) or
\(c_{0}=-1\). Hence the first particular solution is
\(y_{p}=-x^{-2}\). Hence the
solution is
\begin{align*} y & =y_{h}+y_{p}\\ & =\frac {a_{0}}{x}-\frac {1}{x^{2}}\end{align*}