# Homework Help: Second-Order PDE Help

1. Aug 14, 2011

### Scootertaj

1. Integrate (by calculus): u''(x) = -4u(x), 0 < x < pi

2. The attempt at a solution
I'm not really sure where to start on this one is my problem. I can see that it won't be a e^2x problem because of the negative, which leads me to believe that it will deal with the positive/negative relationship involved when you differentiate cos.
The answer is u(x) = c1sin(2x) + c2cos(2x) which makes sense since u'(x) = 2cos(2x) - 2sin(2x) and u''(x) = -4sin(2x) - 4cos(2x) = -4(sin(2x) + cos(2x)) = -4u(x)
But, how do I go about showing my work? How am I supposed to know it's c1sin(2x) + c2cos(2x) in the first place?

2. Aug 14, 2011

### rock.freak667

Rewrite your equation as

u''+ 4u = 0

and since this is a 2nd order ODE with constant coefficients, you can use a trial solution of u = erx to get auxiliary equation which will be for the form

ar2+br+c = 0 where you can solve for r.

For more info: here

3. Aug 14, 2011

### dynamicsolo

Have you discussed in your course yet how to interpret complex roots from the characteristic equation to a second-order DE? You will find the "Euler Identity" of use:

e^(ikx) = cos(kx) + i sin(kx) .

4. Aug 14, 2011

### vela

Staff Emeritus
You could say you made an educated guess as to what the solution is. Guessing the answer is a legitimate way to solve a differential equation. It's a second-order equation so you need two independent solutions. As you noted, the negative sign suggests the solutions will be sines and cosines. Then it's just a matter of figuring out what you need to do to get the factor of 4 out front.

5. Aug 15, 2011

### Scootertaj

Thank you guys for all your help!
I haven't actually started this class yet, I'm just preparing for it since I know it's going to be quite a *****. It's a graduate level PDE class, and I didn't learn much in my ODE class because of a terrible teacher (even though I got an A).
I was wondering if guessing the answer is sufficient enough, assuming I prove why it works.
When you say "you need two independent solutions" do you mean having c1 and c2?
Would just c1sin(2x) work? After all, if u = sin(2x), then u' = 2cos(2x) and u'' = -4sin(2x), which works.
Or does that not work since there need to be 2 constant c's because it's second order?

I haven't learned that (or anything yet), but how would I go about using that as an answer?
Would I just use e^(i2x)?

Interesting, I'll look into that more thank you! That makes sense, I need to study it more though.

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