# 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  • What have you tried so far
– Ben
Jul 1, 2020 at 23:33

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.
You probably should not assume $$x(0)=1$$ for the initial condition, but rather use that $$x(0^-)=0$$ due to the step function (See Initial Conditions, Generalized Functions, and the Laplace Transform).
And instead use $$\displaystyle \mathcal{L}\left\{\frac{\text{d}}{\text{d}t}x(t)\right\}=sX(s)-x(0^-)$$. Then your right-hand side will not be zero. Maybe the question would have been better formulated if it would have stated the initial conditions as $$y(0^-)=1$$ and $$y'(0^-)=0$$.