# Math help!November 11, 2011 1:47 AM   Subscribe

Homework help!? Solving a second-order linear ODE.

(tl;dr: please explain how to get this.)

The DE in question:

y'' + y = 1 with the conditions: y(2) = x(2), y'(2) = x'(2)

where x is the solution to the homogenous equation (y'' + y = 0) with the initial conditions x(0) = 1, x'(0) = 0

I am told to "use the Method of Undetermined Coefficients to solve this problem."

So, I went along the same way I did the 5 questions before it, finding the complementary solution (Acos(t) + Bsin(t)) and the particular solution (+1). But wait.. if the particular solution is a constant, that means y'(t) = x'(t) for all t, which means I can't solve for both A and B... Uh oh. Eventually I gave up and plugged the entire thing into wolfram|alpha -- huh?! How on earth did they get that?

Right now I am starting to think there is something I am fundamentally not understanding about the question... This is going to bother me enough to keep me up until I figure it out, and it really is getting quite late (or early in the morning, rather). Please help me get some sleep?
posted by btfreek to Education (3 answers total)

Best answer: I think your problem might be that you're assuming A and B for the complementary solution are the same for y and x?
You clearly know that y' = -A sin(t) + B cos(t)
From your initial conditions on x you can infer that x=cos(t), right? But that doesn't preclude y from having a sin term, and then you just have the conditions at t=2 to constrain A and B for it.
So your first condition on y tells you that A cos(2) + B sin(2) + 1 = cos(2), and your second that -A sin(2) + B cos(2) = -sin(2) .

You should then just be able to jiggle around to get A and B - you just have a pair of simultaneous equations in A and B.

Wolfram Alpha has just made things a bit confusing by changing the A cos(t) + B sin(t) to a different but equivalent expression.
posted by edd at 4:46 AM on November 11, 2011

Best answer: Unless I'm missing something (it's possible, it's early), forget your initial conditions for a second, and find the general solution to the corresponding homogeneous equation, y'' + y = 0. Call this solution yc.

Now you need only a particular solution to y'' + y = 1. Clearly, y=1 does the job.

By superposition, every solution to y'' + y = 1 should be of the form yc + 1. Now deal with your initial conditions to find what the coefficients in yc need to be.

Again, if I'm off the mark, apologies.
posted by King Bee at 5:23 AM on November 11, 2011

Response by poster: Got it! Turns out I had everything right except I just kept bungling up the algebra last night, ugh. Thanks!
posted by btfreek at 10:47 AM on November 11, 2011

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