# Laplace equations w/ discont. forcing functions

1. Oct 5, 2011

### cdotter

1. The problem statement, all variables and given/known data
$$y'' + 3y' + 2y = u_2(t); y(0)=0, y'(0)=1$$

2. Relevant equations
$$L\{u_c(t)h(t-c)\}=e^{-cs}H(s)$$

3. The attempt at a solution
I've actually tried two laplace equation with discontinuous forcing function problems. I get answers somewhat close to the back of the book, but still off. Maybe I'm making a dumb algebra mistake? Could someone check my work over?

\begin{aligned} & L\{f(t)\}(s^2+3s+2)-1=\frac{e^{-2s}}{s}\\ & L\{f(t)\}=\frac{1+e^{-2s}}{(s)(s^2+3s+2)} \end{aligned}

\begin{aligned} f(t) & =L^{-1}\{\frac{1}{(s)(s^2+3s+2)}\} + L^{-1}\{\frac{e^{-2s}}{(s)(s^2+3s+2)}\} \\ f(t) & =L^{-1}\{\frac{1}{(s)(s^2+3s+2)}\} + u_2(t)h(t-2) \end{aligned}

Where $$H(s) = \{\frac{1}{(s)(s^2+3s+2)}\}$$

Partial fraction decomposition:
$$L^{-1}\{\frac{1}{(s)(s^2+3s+2)}\} = L^{-1}\{\frac{1/2}{s}\} - L^{-1}\{\frac{1}{s+1}\} + L^{-1}\{\frac{1/2}{s+2}\}$$

$$\Rightarrow f(t) = \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2} + u^2(t)h(t-2)$$

Where $$h(t)= \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}$$

My book gives the answer as:

$$f(t)=e^{-t}-e^{-2t}+u_2(t)h(t-2)$$ where

$$h(t)= \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}$$

Any ideas where I messed up? :(

Last edited: Oct 5, 2011
2. Oct 5, 2011

### cdotter

Sorry about editing the post so many times, I'm still figuring out MathJax.

3. Oct 5, 2011

### vela

Staff Emeritus
You made an algebra mistake when you moved the 1 from the LHS to the RHS. The numerator should end up as s+e-2s.

4. Oct 5, 2011

### cdotter

Thank you! I make the dumbest mistakes. :grumpy: