Using Wronskian to find particular solution of $y''+\frac{1}{x}y'-\frac{m^2}{x^2}y=\frac{1}{1-x}$

Question

The differential equation is $$\frac{d^2y(x)}{dx^2}+\frac{1}{x}\frac{dy(x)}{dx}-\frac{m^2}{x^2}y(x)=\frac{1}{1-x}$$ I'm given $y_1(x)=x^m$.

So, using $P(x) = \frac{1}{x}$ I found $W(x)=W_0e^{-\int{dxP(x)}}=\frac{W_0}{x}$.

I then used $W(x)$ and $y_1(x)$ to find $y_2(x)=y_1(x)\int{}dx\frac{W(X)}{y_1^2(x)}=-\frac{W_0}{m}+W_0cx^m$.

Does this all follow so far?

Now I want to find the particular solution $y_p(x)=-y_1(x)\int{dx\frac{y_2(x)d(x)}{a(x)W(x)}}+y_2(x)\int{dx\frac{y_1(x)d(x)}{a(x)W(x)}}$, where $d(x)=\frac{1}{1-x}$ and $a(x)=1$.

When I try to do this, it's extremely messy. I figured I messed up in one of the above steps. Can anyone confirm that or give me a clue as to how to solve this?

Answer 1

Hint:

$y_1=x^m$ then $$y_2=y_1\int\frac{e^{-\int p\,dx}}{y_1^2}=x^m\int\frac{e^{-\int\frac1x\,dx}}{x^{2m}}=x^m\frac{x^{-2m}}{-2m}=\frac{1}{-2mx^m}$$ the general solution is $$y_p=C_1x^m+C_2\frac{1}{-2mx^m}$$