# 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?

(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?

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 y

Now you need only

By superposition, every solution to y'' + y = 1 should be of the form y

Again, if I'm off the mark, apologies.

posted by King Bee at 5:23 AM on November 11, 2011

_{c}.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 y

_{c}+ 1.*Now*deal with your initial conditions to find what the coefficients in y_{c}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

posted by btfreek at 10:47 AM on November 11, 2011

This thread is closed to new comments.

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