# 2nd Order Inhomogeneous DE

• Apr 28th 2009, 11:02 AM
mistykz
2nd Order Inhomogeneous DE
x''-x = sin(t) x'(0)=x(0)=0
I'm supposed to solve this using both Laplace and non-Laplace methods
Using Laplace, I got that x(t) = .5cos(t)-.5sin(t)-.5e^(-t), which works with the given conditions

However, for some reason when I try to find a solution by undetermined coefficients...I can't. I got a completely different answer (I think it was something like x(t)= -.5sin(t), or something like that) which obviously doesn't work.
Solving for the homogeneous, I find that xh(t) = e^(-t) + e^t
Then I guessed that xp(t) = acos(t) + bsin(t), plugged it into the equation, but then found that a = 0 and b= -.5
Can anyone help me figure this one out? It seems silly that I can't do it...
Thanks!
• Apr 29th 2009, 01:24 AM
The Second Solution
Quote:

Originally Posted by mistykz
x''-x = sin(t) x'(0)=x(0)=0
I'm supposed to solve this using both Laplace and non-Laplace methods
Using Laplace, I got that x(t) = .5cos(t)-.5sin(t)-.5e^(-t), which works with the given conditions

However, for some reason when I try to find a solution by undetermined coefficients...I can't. I got a completely different answer (I think it was something like x(t)= -.5sin(t), or something like that) which obviously doesn't work.
Solving for the homogeneous, I find that xh(t) = e^(-t) + e^t
Then I guessed that xp(t) = acos(t) + bsin(t), plugged it into the equation, but then found that a = 0 and b= -.5
Can anyone help me figure this one out? It seems silly that I can't do it...
Thanks!

Using the 'non-Laplace method' I get $y = \frac{1}{4} e^t - \frac{1}{4} e^{-t} - \frac{1}{2} \sin t$. So you've made a mistake in using the Laplace transforms. Did you check that solution by substitution? It didn't work when I checked it.
• Apr 30th 2009, 09:17 AM
mistykz
Figured them both out, thanks! (Hi)