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The problem statement is

Consider a causal LTI system whose transfer function $H(s)$ is given as $$H(s)=\frac{s+2}{(s+3)(s+4)}$$ Compute the output $y(t)$ for an input $x(t)=e^{-2t}u(t)$ when $y(0)=1$ and $y’(0)=0$.

The solution is given as $y(t)=5e^{-3t} − 4e^{-4t}$.

Here is my attempt:

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Laplace transform left side and right side and consider the initial condition enter image description here

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right side is 0 , and left side is below

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so i think answer is below but the prof's answer is not match with my answer.

enter image description here

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  • $\begingroup$ What have you tried so far $\endgroup$ – Ben Jul 1 '20 at 23:33
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Your solution looks correct. In any case, the given solution $y(t)=5e^{-3t}-4e^{-4t}$, $t>0$, must be wrong because it doesn't satisfy the initial condition on the derivative ($y'(0^+)=0$):

$$y'(t)=-15e^{-3t}+16e^{-4t},\qquad t>0\tag{1}$$

So we have $y'(0^+)=1$. Your solution satisfies both initial conditions.

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