Reduction of order problem

Question

So I have a very simple question here that looks fairly easy.

Given that $y_1(x) = e^x$ is one solution of the equation $(x - 1)y''-xy'+y=0$, find a fundamental set of solutions.

I am certain this is a reduction of order problem but when I simplify everything by substituting $y_2(x)=u(x)e^x$ and the corresponding derivatives into the differential equation, while I do get the $u(x)$ to cancel, I can't solve the resulting linear ODE since I end up integrating an exponential of degree greater than 1.

I don't think I made a mistake as I checked several times but if someone could tell me how I would approach this otherwise, that would be much appreciated. Thanks.

Answer 1

Substitution $y(x)=e^xz(x)$ will be noted $y=e^xz$ for convenience.

We get $y'=e^x(z'+z)$ and $y''=e^x(z''+2z'+z)$.

So, the transformed equation is (after division by $e^x$):

$$(x-1)(z''+2z'+z)-x(z'+z)+z=0$$

which simplifies to :

$$(x-1)z''+(x-2)z'=0$$

So we get a first order linear ODE in terms of $z'$.

This leads to $z'=K(x-1)e^{-x}$ and a last integration by parts should do the job.