Hello Mike,
We are given the nontrivial solution:
$$f(x)=xe^{-x}$$
So, let's set:
$$y(x)=v(x)f(x)$$ and substitute for the derivatives of $y(x)$ into the given ODE. Thus, computing the needed derivatives, we find:
$$y'(x)=v(x)f'(x)+v'(x)f(x)$$
$$y''(x)=v''(x)f(x)+2v'(x)f'(x)+v(x)f''(x)$$
Now we need to compute the derivatives of $f(x)$:
$$f'(x)=-xe^{-x}+e^{-x}=e^{-x}(1-x)$$
$$f''(x)=-e^{-x}+(x-1)e^{-x}=e^{-x}(x-2)$$
And so we find:
$$y'(x)=v(x)e^{-x}(1-x)+v'(x)xe^{-x}=e^{-x}\left(v(x)(1-x)+xv'(x) \right)$$
$$y''(x)=v''(x)xe^{-x}+2v'(x)e^{-x}(1-x)+v(x)e^{-x}(x-2)=e^{-x}\left(xv''(x)+2v'(x)(1-x)+v(x)(x-2) \right)$$
Substituting into the original ODE, we find:
$$e^{-x}\left(xv''(x)+2v'(x)(1-x)+v(x)(x-2) \right)+2\left(e^{-x}\left(v(x)(1-x)+xv'(x) \right) \right)+v(x)xe^{-x}=0$$
Since $$e^{-x}\ne0$$ we may divide through by this factor to obtain:
$$xv''(x)+2v'(x)(1-x)+v(x)(x-2)+2v(x)(1-x)+2xv'(x)+v(x)x=0$$
Distribute:
$$xv''(x)+2v'(x)-2xv'(x)+xv(x)-2v(x)+2v(x)-2xv(x)+2xv'(x)+xv(x)=0$$
Combine like terms:
$$xv''(x)+2v'(x)=0$$
Multiply through by $x$:
$$x^2v''(x)+2xv'(x)=0$$
Now the left side is the differentiation of a product:
$$\frac{d}{dx}\left(x^2v'(x) \right)=0$$
Integrate with respect to $x$ to obtain:
$$\int\,d\left(x^2v'(x) \right)=\int\,dx$$
$$x^2v'(x)=c_1$$
$$v'(x)=c_1x^{-2}$$
Integrate again with respect to $x$:
$$\int\,dv=c_1\int x^{-2}\,dx$$
$$v(x)=-c_1x^{-1}+c_2$$
For simplicity, let $c_1=-1$ and $c_2=0$ and so:
$$v(x)=\frac{1}{x}$$
And hence, we find our second linearly independent solution is:
$$y(x)=v(x)f(x)=\frac{1}{x}xe^{-x}=e^{-x}$$
Shown as required.
