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 \(y_{p}\).