# Thread: y'' + 2/x y' = cy

1. ## y'' + 2/x y' = cy

Hello!

I have this second order differential equation which is causing me some trouble:

y'' + 2/x y' = c y

where c is a constant. I've got no clear idea of where to start solving it, the 2/x term is the main source of my head ache here.

2. ## Re: y'' + 2/x y' = cy

First start by writing it clearly. Do you mean y''+ 2/(xy')= cy or y''+ (2/x)y'= cy? Since you refer to "2/x", I guess you mean the latter. That is a linear second order differential equation with variable coefficients which has a regular singularity at x= 0. If you multiply through by x you get xy''+ 2y'- cy= 0. The simplest thing to do is to look for a power series of the form $\displaystyle \sum_{i= 0}a_n(x- b)^n$ where b is NOT 0. Or look for a series solution of the form $\displaystyle \sum_{n= 0}^\infty a_nx^{n- c}$ where c is a positive number.

3. ## Re: y'' + 2/x y' = cy

Thank you! Yes, my apologies, I meant the latter. So xy''+ 2y'- cxy= 0 would be the result from the multiplication, if I guess that there is an x missing from the last term, and if so does it change the power series approach?

And a little follow-up question: Is it always possible to find a power series-solution to a differential equation?

4. ## Re: y'' + 2/x y' = cy

No need for power series in this case.
More simply, let y=z/x which leads to a second order ODE easy to solve for z(x)

5. ## Re: y'' + 2/x y' = cy

My power series did not really work out well so I'll get back to that later with more questions, but if I could solve it as an ODE then it would make my day but I got stuck on that one as well.

I tried calculating y' and y'' for y = z(x)/x but this gives pretty heavy expressions to substitute and becomes even tougher, so I suppose I'm working from the completely wrong direction.

6. ## Re: y'' + 2/x y' = cy

the substitution $y=z/x$ is the correct one and will lead you to a linear 2nd order diff eq w/constant coefficients which is easily solved.

Keep plugging away.

7. ## Re: y'' + 2/x y' = cy

Thank you guys, I think I got quite close right now. Wolfram alpha gets a slightly different answer than I get, but I'll try to put it all here tomorrow to see if someone agrees/disagrees with me.

And btw: Is it possible to get a solution through a power series for this one (and really any differential equation, or are there restrictions)? I'd, ultimately, at some point have to practice on doing those and this would be a good time if it's possible.