Second-order nonhomogeneous diff-eq

  • #1
Hey guys. It's been a few years since I've taken diff-eq and I can't remember how to solve second-order problems like this one:

Find the general solution of
u'' + a^2*u = cos(bx)

I know that if it were homogeneous, I would solve for r^2 + a^2 = 0, and get u = ce^(rx). But for the life of me I can't remember what to do with that cosine on the right side of the equation. Can anyone help?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi BucketOfFish! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)

You need to find a particular solution (ie any solution to the equation), which you can then add to the general solution to the homogeneous equation.

In this case, try a combination of cos(bx) and sin(bx). :smile:

(if b = ±a, that doesn't work, and you'll need xcos(bx) and xsin(bx))
 
  • #3
Thanks a lot; that really helped!
 
  • #4
19
0
I bet you could also use the function:

u''+a2u' = eibx and split the real and imaginary portions with Euler's formula.

I find e easier to work with than sin and cos.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
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hi evad1089! :smile:
I find e easier to work with than sin and cos.

even when cos is already on the RHS? :wink:
u'' + a^2*u = cos(bx)
 

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