1.3.6 Example 6. \(xy^{\prime }+y=x+x^{3}+2x^{4}\)

\begin{equation} xy^{\prime }+y=x+x^{3}+2x^{4} \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.

\[ \left ( r+1\right ) c_{0}x^{r}=x+x^{3}+2x^{4}\]

For each term on the right side, there is different balance equation. Hence we have

\begin{align} \left ( r+1\right ) c_{0}x^{r} & =x\tag {2}\\ \left ( r+1\right ) c_{0}x^{r} & =x^{3}\tag {3}\\ \left ( r+1\right ) c_{0}x^{r} & =2x^{4} \tag {4}\end{align}

Each one equation above, gives different \(y_{p}\) then at the end we will add them all. Starting with (2)

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

Hence \(r=1\), therefore \(\left ( r+1\right ) c_{0}=1\) or \(c_{0}=\frac {1}{2}\). Hence the first particular solution is \(y_{p_{1}}=\frac {1}{2}x\). Now we will look at (3)

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

Hence \(r=3\), therefore \(\left ( r+1\right ) c_{0}=1\) or \(4c_{0}=1\) or \(c_{0}=\frac {1}{4}\). Hence the second particular solution is \(y_{p_{2}}=\frac {1}{4}x^{3}\). Now we will look at (3)

\[ \left ( r+1\right ) c_{0}x^{r}=2x^{4}\]

Hence \(r=4\), therefore \(\left ( r+1\right ) c_{0}=1\) or \(5c_{0}=2\) or \(c_{0}=\frac {2}{5}\). Hence the third particular solution is \(y_{p_{3}}=\frac {2}{5}x^{4}\). Hence adding all the above gives

\[ y_{p}=\frac {1}{2}x+\frac {1}{4}x^{3}+\frac {2}{5}x^{4}\]

We found \(y_{p}\). The solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =\frac {a_{0}}{x}+\frac {1}{2}x+\frac {1}{4}x^{3}+\frac {2}{5}x^{4}\end{align*}

Notice here we also used that fact that since all sum term in the summation equation in the first example have same starting index, then we know that all \(c_{n}=0\) for \(n>0\) for each \(y_{p}\) we found above, so we only needed to find \(c_{0}\) only for each \(y_{p}\).