Variation of Parameters Nonhomogeneous Differential Equation

[BarackObama]

Homework Statement

4y'' + y = cosx

Solve using variation of parameters

Homework Equations

The Attempt at a Solution

from a) -> yc(x) = c1cos(x/2) + c2sin(x/2)
let y1 = cos(x/2) , y2 = sin(x/2)
y1y2' - y2y1' = 1/2cos^x/2 + 1/2sin^x/2 = 1/2

u1' = ?

How do I find this?

 

[BarackObama]

Can someone help me out with variation of parameters? It's urgent!

 

[PAllen]

Your problem statement seems incomplete. Are you looking for a general expression for y with enough parameters to fit any initial value problem? Normally, I think of variation of parameters as applying to a particular initial value problem where you vary the parameters to match the initial conditions. Assuming you want some general expression, you have correctly found two linearly independent solutions of the homogenous equation that can be used in any linear combination with a particular solution of the inhomogenous equation. So it seems all you need is particular solution of the inhomogeneous equation. There are messy mechanical procedures you can use, or you can make a guess. In this case, making about the most obvious guess can lead you to a particular solution. Then you just have linear combination of your sin(x/2), cos(x/2) plus the particular solution.

[EDIT] It's been too long, I'm mixing up terminology above. Variation of parameters is one of the general, messy, ways of finding a particular solution. It is overkill for this example, since the most obvious guess as to form, with solving for coefficient, works, as I said above. However, if you must use variation of parameters, then you are looking for two unknown functions, u1 and u2, which you combine with y1=cos(x/2), y2=sin(x/2) to get a particular solution:

u1 y1 + u2 y2

Then you want to obtain u1 and u2 by solving:

u1' y1 + u2' y2 = 0

and

u1' y1' + u2' y2' = cos(x)/4

for u1' and u2'; then you integregrate each to get u1 and u2. Then you have the particular solution.