Differential Equations with Discontinuous Forcing Functions

1. Nov 24, 2014

_N3WTON_

1. The problem statement, all variables and given/known data
Solve the given initial value problem:
$y'' + y = u(t-\pi) - u(t-2 \pi)$
$y(0) = 0$
$y'(0) = 1$

2. Relevant equations

3. The attempt at a solution
First I took the Laplace transform of both sides:
$\mathcal{L}[y'' + y ] = \mathcal{L}(u(t-\pi)) - \mathcal{L}(u(t-2 \pi))$
$(s^{2}Y(s) - sy(0) - y'(0)) + Y(s) = \frac{e^{-\pi s}}{s} - \frac{e^{-2 \pi s}}{s}$
$s^{2}Y(s) - 1 + Y(s) = \frac{e^{-\pi s}-e^{-2 \pi s}}{s} + 1$
$Y(s)(s^{2}+1) = \frac{e^{-\pi s}-e^{-2 \pi s}+s}{s}$
$Y(s) = \frac{e^{-\pi s}-e^{-2 \pi s} + s}{s(s^{2}+1)}$
At this point I get stuck, I tried doing a partial fraction decomp, but that didn't seem to get me anywhere closer to the solution..

2. Nov 24, 2014

LCKurtz

I haven't checked your algebra, but assuming it is correct you need to use the formula$$\mathcal Lf(t-a)u(t-a) = e^{-as}\mathcal L(f(t)) = e^{-as}F(s)$$So if you have a transform $F(s)$ that is multiplied by $e^{-as}$ all you have to do is invert the $F(s)$ to $f(t)$ then truncate and translate it go get $f(t-a)u(t-a)$. So just break up your problem into three fractions and work them separately.

3. Nov 24, 2014

_N3WTON_

ah, ok.. So I'll have something like:
$Y(s) = \frac{e^{-\pi s}}{s(s^{2}+1)} - \frac{e^{-2\pi s}}{s(s^{2}+1)} + \frac{1}{s^{2}+1}$
Then I would do a decomposition on this last fraction and then apply the inverse Laplace?

4. Nov 24, 2014

LCKurtz

Yes. Use the formula on each of the first two and do the third one by itself. Then add the three answers.

5. Nov 24, 2014

_N3WTON_

ok...I'm getting a lot of terms so I wanted to make sure I'm doing this correctly. For my decomposition I found:
$\frac{1}{s(s^{2}+1)} = \frac{1}{s} - \frac{s}{s^{2}+1}$
Therefore, I found the following:
$Y(s) = \frac{e^{-\pi s}}{s} - \frac{se^{-\pi s}}{s^{2}+1} - \frac{e^{-2\pi s}}{s} + \frac{se^{-2\pi s}}{s^{2}+1} - \frac{1}{s^{2}+1}$
So after applying the inverse Laplace I get:
$y(t) = u(t - \pi)-\cos (t-\pi)u(t-\pi) - u(t -2\pi)+\cos (t-2\pi)u(t-2\pi)+\sin t$
The above answer can probably be simplified, but I'll worry about that later :)

Last edited: Nov 24, 2014
6. Nov 24, 2014

LCKurtz

That's the idea. Hopefully it's starting to look more like your given solution.

7. Nov 24, 2014

_N3WTON_

It definitely looks close, the given solution is:
$\sin t + u(t-\pi)(1-\cos (t-\pi)) - u(t-2\pi)(1-\cos (t -2\pi))$
I need to check and make sure my signs are okay, but I think I've got the hang of it, thanks for the help!