Let $J(x)$ and $N(x)$ be two linearly independent solutions of the equation
$$x^2y''+xy'+(-1+x^2)y=f(x)$$
It is known that the general solution of the equation is given by the formula
$$y=-J(x)\int N(x) xe^{-x^2} dx +N(x)\int J(x) xe^{-x^2} dx$$
Then
A) $ f(x)=x^2e^{-x^2} $
B) $ f(x)=xe^{-x^2} $
C) $ f(x)=e^{-x^2} $
D) $ f(x)=x^{-1}e^{-x^2} $
E) None of the above
I am not sure how to find the Wronskian Determinant of $J(x)$ and $N(x)$ because from what I think the solution would be in the form of a power series. If $W(x)$ is found then I can substitute in $\frac{f(x)}{W(x)}= xe^{-x^2}$, but I don't think that's how you suppose to do it.
Can you suggest any way to tackle this problem? Thank you.
Just take the derivatives and insert into the given formula, \begin{align} y'(x)&=-J'(x)\int N(x)g(x)dx+N'(x)\int J(x)g(x)dx\\ y''(x)&=(N'J-NJ')g(x)-J''(x)\int N(x)g(x)dx+N''(x)\int J(x)g(x)dx\\ f(x)=x^2y''+xy'+(x^2-1)y&=x^2(N'J-NJ')g(x) \end{align} and the Wronskian term $W=N'J-NJ'$ has the derivative $$ x^2W'=x^2(N''J-NJ'')=-x(N'J-NJ')-(x^2-1)(NJ-NJ)\implies W=\frac{C}{x} $$ and assuming $W(1)=1$, one has $C=1$.