I've already taken my Diff EQ class, but I can't remember a lot of the material that we covered. I just have a couple questions that I'm stuck on; I'd appreciate it if someone could get me started in the right direction, not necessarily just give me the answer.

B )
x'(t) - sin(t)x(t) = 2t*e^(-cos(t)) x(0) = 1

For this one, I solved the homogeneous equation, and got x(t) = c1*e^(-cost), but when I tried to find the value of the coefficient, I got 0, and I feel like that's wrong.

C)
y''(x) + 8y(x) = 0 y(0) = y(1) = 0
D)
y''(x) - 8y(x) = 0 y(0) = y(1) = 0

For both of these, I again solved the homogeneous equations, and again, when solving for the coefficients using the initial conditions, got 0 for both.

E)
y''(t) - 4y(t) = 2e^(2t)

At first, I tried solving this using the method of undetermined coefficients, but I couldn't solve for A in the particular solution.

This post was edited by SexualNinja on Aug 27 2013 09:15pm

a) i dont think you are applying the chain rule correctly. remember e^(f(t)) = f'(t)e^(f(t))

b ) im guessing its the same error. be careful in your application of the chain rule.

c) do you mean the A in the corresponding quadratic equation?

For the first two, I'll take a second look.
For the last one, I mean A as in when I'm solving for the undetermined coefficient.
So like yp(t) = Ae^(2t), yp''(t) = 4e^(2t). But when you plug those into the DiffEQ, you get Ae^(2t) - 4[ Ae^(2t) ] = 2e^(2t), which isn't true.

are you sure the equation is correct? because looking at

y''(t) - 4y(t) = 2e^(2t)

i can easily see that the solution has to be of the form y(t) = Ae^(2t), as you said, but taking 2 derivatives of that yields 4Ae^(2t), which cancels.

Positive, that's why I'm thinking there's another way to do it without the method of undetermined coefficients

did you try yp(t) = Ate^(2t)?

edit: i did this and got a = 1/2

This post was edited by khemist on Aug 27 2013 09:47pm

Aaaand that's right. Thank you so much.

As far as the first two, I think I'm doing them right. If we start with B,

x'(t) - sin(t)x(t) = 2t*e^(-cos(t)) x(0) = 1

I started by setting the Diff EQ equal to 0 and solving the homogeneous equation. That was a separable first order equation, and I got x(t) = Ce^(-cost), but I'm not sure what to do after that.

well you solved for x(t).

and you are given x(0) = 1.

so x(0) = c1*e^(-cos(0)) = 1

c1*e^(1) = 1
c1 = 1/e

the second i dont have time to finish but i imagine if you try some things you will figure it out. and i didnt check to see if its correct so i hope it is lol

This post was edited by khemist on Aug 27 2013 09:57pm