Compute output given input, transfer function and initial conditions
Your solution looks correct. In any case, the given solution $y(t)=5e^{-3t}-4e^{-4t}$, $t&gt;0$, must be wrong because it doesn't satisfy the initial condition on the derivative ($y'(0^+)=0$):
$$y'(t)=-15e^{-3t}+16e^{-4t},\qquad t&gt;0\tag{1}$$
So we have $y'(0^+)=1$. Your solution satisfies both initial conditions.