Compute output given input, transfer function and initial conditions

Question

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

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

Answer 1

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.

Answer 2

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 $y'(0^-)=0$.