(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]y'' + 3y' + 2y = u_2(t); y(0)=0, y'(0)=1[/tex]

2. Relevant equations

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

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?

[tex]

\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}

[/tex]

[tex]

\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}

[/tex]

Where [tex]H(s) = \{\frac{1}{(s)(s^2+3s+2)}\}[/tex]

Partial fraction decomposition:

[tex]

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}\}[/tex]

[tex]

\Rightarrow f(t) = \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2} + u^2(t)h(t-2)[/tex]

Where [tex]h(t)= \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}[/tex]

My book gives the answer as:

[tex]f(t)=e^{-t}-e^{-2t}+u_2(t)h(t-2)[/tex] where

[tex]h(t)= \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}[/tex]

Any ideas where I messed up? :(

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# Homework Help: Laplace equations w/ discont. forcing functions

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