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$$\frac{\mathrm dw(t)}{\mathrm dt}+2w(t)=y(t)$$ $$\frac{\mathrm dy(t)}{\mathrm dt}+3y(t)=2w(t)+f(t)$$ The input to the system is $~f(t)$ and the output is $~y(t).$

The initial conditions are $~w(0)=0,\quad y(0)=1$.

Write a state space formulation of this system and solve using the

state space methods for an input $~f(t)=\delta(t)$


solving the system gives

w = -e^-t + e^t
y = -e^-t + 3e^t

This solution checks out when plugged into the first equation, but I am unsure of what to do with the second equation to check my work. Anywho, why not plug it in? I get

10e^t = f

Either I did the problem right and don't know how to check it or I did the problem wrong. In order to match the f, I need a delta function in there somewhere.

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$y(0) = -e^{0} + 3e^{0} = -1 + 3 = 2 \ne 1$. Your initial condition doesn't match. –  Tunococ Aug 24 '12 at 5:00
    
Since $10 e^t \ne f$, there must be something wrong with your solution. In this case, there should be terms in $e^{-t}$ and $e^{-4t}$, but no $e^t$. –  Robert Israel Aug 24 '12 at 5:54
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