What is the general solution to the 2nd order linear ODE xy''+2y'+4xy=0?

[sydneyfranke]

Homework Statement

Find general solution to:

xy''+2y'+4xy=0

Homework Equations

Frobenius Method or Bessel's Equation

The Attempt at a Solution

I know how to get the roots for this problem (which are r1 = 0 & r2 = -1). But not I don't know what to do with these roots. I know that this is considered "Case 3", but I seriously don't know what to do from here.

Furthermore, I have attempted to solve for y1(x) and came up with the answer:

y1(x) = a0(x^-1)cos(2x) + ((a1)/2)(x^-1)sin(2x)

If this is right, then I'm not sure what the next step is. Do I do reduction of order? I hope not because that would take FOREVER.

Wolfram Alpha has the general solution as

y= (c1(exp(-2ix)))/x - (ic2(exp(2ix)))/4x

I'm totally lost. Any help would be appreciated.

 

[haruspex]

y1(x) = a0(x^-1)cos(2x) + ((a1)/2)(x^-1)sin(2x)

Did you check that satisfies the original equation? (It doesn't look right to me. I get exponentials, not trig.)

 

[sydneyfranke]

No I didn't, but exponentials match the wolfram alpha solution. Can you maybe explain how you got the exponentials?

 

[haruspex]

The way it works on these fora is that you post your working and others try to spot where you went wrong.