Solving an Equation using Power Series

Date: 05/18/99 at 10:25:03
From: Heinz
Subject: Diff. Equation

Hello Dr Math,

I'm stuck with this problem:

Solve the differential equation (t^4)*x'' + x = 0 for x.

How could one go about solving an equation of this sort?
I would greatly appreciate your help.

Sincerely,
Heinz

Date: 05/18/99 at 12:58:08
From: Doctor Nick
Subject: Re: Diff. Equation

Hi Heinz -

The way to solve an equation like this is with power series. Assume 
the solution has a power series in t around some point (for instance, 
0). Differentiating the power series twice, multiplying by t^4, and 
using the differential equation will give you a recurrence relation 
for the coefficients in the power series; that is, you'll get an 
expression that allows you to give the n-th coefficient as a function 
of earlier coefficients.

Take a look at just about any book on differential equations. There 
should be a chapter on power series solutions. It's quite a powerful 
method.

Feel free to write back if you need more help.

- Doctor Nick, The Math Forum
  http://mathforum.org/dr.math/   

Date: 05/18/99 at 13:27:50
From: Doctor Rob
Subject: Re: Diff. Equation

Thanks for writing to Ask Dr. Math!

A good approach is to regard x as a function of v = 1/t.  Then

   x' = dx/dt
      = (dx/dv)*(dv/dt)
      = (-1/t^2)*(dx/dv)
   x" = dx'/dt
      = (1/t^4)*(d^2x/dv^2) + (2/t^3)*(dx/dv)

so the equation for x in terms of v becomes

   d^2x/dv^2 + (2/v)*(dx/dv) + x = 0

Now write x = y/v, which will have the effect of clearing
fractions, so

   dx/dv = (v*dy/dv - y)/v^2
         = 1/v*dy/dv - y/v^2
   d^x*dv^2 = (1/v)*(dy^2/dv^2) - (2/v^2)*(dy/dv) + 2*y/v^3

and the equation is

   (1/v)*(d^2y/dv^2) + y/v = 0
   d^2y/dv^2 + y = 0

(v is not zero because it equals 1/t, and t is not infinite.) Now this 
is a familiar equation. The solutions have the form

   y = A*sin(v) + B*cos(v)

where A and B are arbitrary constants.  Then

   x = y/v = [A*sin(v) + B*cos(v)]/v
   x = t*[A*sin(1/t) + B*cos(1/t)]

which is the general solution.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/