Laplace equations w/ discont. forcing functions

[cdotter]

Homework Statement

y'' + 3y' + 2y = u_2(t); y(0)=0, y'(0)=1

Homework Equations

L\{u_c(t)h(t-c)\}=e^{-cs}H(s)

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?

<br /> \begin{aligned}<br /> &amp; L\{f(t)\}(s^2+3s+2)-1=\frac{e^{-2s}}{s}\\<br /> &amp; L\{f(t)\}=\frac{1+e^{-2s}}{(s)(s^2+3s+2)}<br /> \end{aligned}<br />

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

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

Partial fraction decomposition:
<br /> 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}\}

<br /> \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? :(

 

[cdotter]

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

 

[vela]

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.

 

[cdotter]

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.

Thank you! I make the dumbest mistakes.