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what is the general solution of this diff. equation $$x^2y''-4xy'+6y=x$$

Tried calling $y=xv$ but didnt work. ($x^2v''-2xv'+v=1$) what can I try else?

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up vote 3 down vote accepted

Hint: You can use the substitution $$ x=e^t. $$ You have $$ \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dt}e^{-t}. $$ In the same spirit you should find $$ \frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\ldots $$ After this substitution you will get a linear equation with constant coefficients.

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Hint: there's a solution of the form $y=ax$.

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Generally an ode like $$ax^2y''+bxy'+cy=g(x)$$ is called Cauchy-Euler equation. What you should do is the solve the homogenous equation $$x^2y''-4xy'+6y=0$$ firstly and then find the particular solution for non-homogenous equation $$x^2y''-4xy'+6y=x$$ by other proper method like variation of parameter secondly. For the first step you can use $y=x^m$ and find the proper $m$'s for finding the general solution $y_c$.

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Very helpful suggestions and "sketch! +1 – amWhy Apr 15 '13 at 0:23

Emmett, in book Elementary Differential Equations and Boundary Value Problems - Boyce and DiPrima, Chapter 3 deals with exactly this type of equation you're working. To be more precise, on page 185, exercise 28 shows a technique for solving problems like yours.

$$y'' + p(t)y' + q(t)y = g(t)$$

I even did some time ago with a similar method presented in this book. I could now follow the steps here and give solution. Then I would win a few points, and you the answer, but math.stackexchange loses it. Then look in chapter 3 and see what you can do. So tell us later. Hugs.

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Sorry my english, I'm trying to learn. – Manoel Dec 20 '12 at 6:04

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