Note: I changed $y$ to $y'$. Solve the second-order ODE $y''+\frac{1}{x}y'=a,\;\;a$ constant.
For me to use Variation of parameters, I need to find two linearly independent solutions $y_1$ and $y_2$, where the particular solution, $y_p$ is given by
\begin{align} y_p(x)=-y_1(x)\int\dfrac{r(x)y_2(x)}{W(y_1,y_2)}dx+y_2(x)\int\dfrac{r(x)y_1(x)}{W(y_1,y_2)}dx \end{align}
and $W,\;r(x)$ are Wronskian and the right-hand side, respectively. Here, $r(x)=a.$
So, I need to look for the solution to $y''+\frac{1}{x}y'=0$ to get $y_1$ and $y_2.$ If I get them, I'm done. Is there any way of getting these two?
Best Answer
This is solvable.$$xy''+y'=0\\(xy')'=ax\\xy'=C_1\\y=\int\frac {C_1}x\,dx\\y=C_1\ln x+C_2$$ So your two independent solutions can be $1$ and $\ln x$.