General solution of a differential equation 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?
 A: 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.
A: Hint: there's a solution of the form $y=ax$. 
A: 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 $y_c$. 
A: 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.
