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