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Homework Statement
[tex]y'' + 3y' + 2y = u_2(t); y(0)=0, y'(0)=1[/tex]
Homework Equations
[tex]L\{u_c(t)h(t-c)\}=e^{-cs}H(s)[/tex]
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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