Step Function Laplace w/ piecewise

[Destroxia]

Homework Statement

$y'' + 2y' + 2y = h(t); y(0)=0, y'(0)=1, h(t) = \begin{cases} t &, \pi \leq t < 2\pi \\ 0 &, 3\pi \leq t < \infty \\ \end{cases}$



2. Homework Equations

The Attempt at a Solution

Take the Laplace of both sides:[/B]

$\mathcal{L}(y'' + 2y' + 2y) = \mathcal{L}(h(t))$

$s^2 F(s) - sf(0) - f'(0) + 2sF(s) - 2f(0) + 2F(s) = \mathcal{L}(h(t))$

$(s^2 + 2s +2)F(s) - 1 = \mathcal{L}(h(t)) = \int_{\pi}^{2\pi} te^{-st}$

$(s^2 + 2s +2)F(s) - 1 =\frac {\pi e^{-\pi s}} {s} + \frac {e^{-\pi s}} {s^2} - \frac {2\pi e^{-2\pi s}} {s} - \frac {e^{2\pi s}} {s^2}$

Solve for F(s):

This is where I always screw up, If I even dare to add the 1 to the other side and divide by $(s^2 + 2s + 2)$ I get a ridiculously long expansion, that I don't even think is plausible for a practice problem, or any other problem.

I have no idea where to continue once I get to this point.

 

[Ray Vickson]

Homework Statement

$y'' + 2y' + 2y = h(t); y(0)=0, y'(0)=1, h(t) = \begin{cases} t &, \pi \leq t < 2\pi \\ 0 &, 3\pi \leq t < \infty \\ \end{cases}$
2. Homework Equations

The Attempt at a Solution

Take the Laplace of both sides:[/B]

$\mathcal{L}(y'' + 2y' + 2y) = \mathcal{L}(h(t))$

$s^2 F(s) - sf(0) - f'(0) + 2sF(s) - 2f(0) + 2F(s) = \mathcal{L}(h(t))$

$(s^2 + 2s +2)F(s) - 1 = \mathcal{L}(h(t)) = \int_{\pi}^{2\pi} te^{-st}$

$(s^2 + 2s +2)F(s) - 1 =\frac {\pi e^{-\pi s}} {s} + \frac {e^{-\pi s}} {s^2} - \frac {2\pi e^{-2\pi s}} {s} - \frac {e^{2\pi s}} {s^2}$

Solve for F(s):

This is where I always screw up, If I even dare to add the 1 to the other side and divide by $(s^2 + 2s + 2)$ I get a ridiculously long expansion, that I don't even think is plausible for a practice problem, or any other problem.

I have no idea where to continue once I get to this point.

It is actually not so bad, once you simplify things a bit. Start with
F(s) = \frac{e^{-\pi s}}{D_2(s)} + \frac{\pi e^{-\pi s}}{D_1(s)} - \frac{e^{-2 \pi s}}{D_2(s)} - \frac{2 \pi e^{-2 \pi s}}{D_1(s)} + \frac{1}{D_0(s)},
where $D_2(s) = s^2 (s^2 + 2 s + 2), \: D_1(s) = s (s^2 + 2 s + 2), \; D_0(s) = s^2 + 2s + 2$.

It is convenient to first take the inverse Laplace transforms of the $1/D_k(s)$ functions:
\frac{1}{D_0(s)} \leftrightarrow e^{-t} \sin(t) \equiv e_0(t) \\<br /> \frac{1}{D_1(s)} = \frac{1}{s} \frac{1}{D_0(s)} \leftrightarrow \int_0^t e_0(\tau) \, d \tau = \frac{1}{2} - \frac{1}{2} e^{-t} (\sin(t) + \cos(t)) \equiv e_1(t) \\<br /> \frac{1}{D_2(s)} = \frac{1}{s} \frac{1}{D_1(s)} \leftrightarrow \int_0^t e_1 (\tau) \, d \tau = \frac{1}{2}(e^{-t} \cos(t) + t-1) \equiv e_2(t)

Now we have
F(t) = [e_2(t-\pi)+\pi e_1(t-\pi)]u(t-\pi) -[e_2(t-2 \pi) + 2 \pi e_1(t - 2 \pi)]u(t - 2 \pi) + e_0(t)

These are obtained using the following two standard properties, wherein below we have $f(t) \leftrightarrow g(s)$:
\begin{array}{rl}1. &amp; \int_0^t f(\tau) d \tau \leftrightarrow \frac{1}{s} g(s) \\<br /> 2. &amp; f(t-a)u(t-a) \leftrightarrow e^{-as} g(s) \end{array}