1.1.3.3 Example 3. \(xy^{\prime }+y=1\)

\begin{equation} xy^{\prime }+y=1 \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.

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

Balance gives \(r=0\). Hence

\begin{align*} \left ( r+1\right ) c_{0} & =1\\ c_{0} & =1 \end{align*}

Since all sum terms in the summation equation of the first example have same starting index, then we know that all \(c_{n}=0\) for \(n>0\). Therefore

\begin{align*} y_{p} & =c_{0}x^{r}\\ & =1 \end{align*}

Therefore

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