Cauchy-Euler Differential Equation

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Homework Statement

Consider the equation x²y''-2xy'+2y=x with bountary conditions y(1)=0, y(2)=0.

I don't know how to solve this without treating it as a Cauchy-Euler equation but I'm struggling because the equation equals x.

The Attempt at a Solution

By treating this as a Cauchy-Euler equation x²y''-2xy'+2y=0 and using y=x^r, I get that r=2,1.
I can't find a particular integral for the equation though and I'm not even sure this is a Cauchy-Euler equation anymore.

 

[HallsofIvy]

Yes that is a 'Cauchy-Euler' equation and, yes, its characteristic equation is r(r-1)- 2r+ 2=(r- 2)(r- 1)= 0 so that x and $x^2$ are solutions. Since "x" is already a solution, try y= Ax ln(x) as a solution to the entire equation.

That works (and Cauchy-Euler equations are especially easy) because the change of variable $x= e^t$ reduces that equation to the equation with constant coefficients $y''- 3y'+ 2y= e^t$ (the primes now indicate differentiation with respect to t).