# Compute output given input, transfer function and initial conditions

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:

Laplace transform left side and right side and consider the initial condition

right side is 0 , and left side is below

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.