Cauchy-Euler Differential Equation

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The discussion centers on solving the Cauchy-Euler differential equation x²y'' - 2xy' + 2y = x with boundary conditions y(1) = 0 and y(2) = 0. The equation is identified as a Cauchy-Euler type, leading to characteristic roots r = 2 and r = 1, with solutions x and x². A particular integral is sought, and it is suggested to try y = Ax ln(x) due to "x" being a solution. The transformation x = e^t simplifies the equation to one with constant coefficients, making it easier to solve. This approach effectively addresses the original equation's complexity.
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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=xr, 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.
 
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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).
 
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