I am trying solve the differential equation $y''+3y'+2y=u(t-1)-u(t-2), y(0)=y'(0)=0$, by calculating the convolution of $f(t)=1$ and $g(t)=e^{-t}-e^{-2t}$. The problem is that I get two different answers for $(f \ast g)$ and $(g \ast f)$. Does anyone see if the mistake is in my reasoning or in my calculations?
for $1<t<2$:
$\int_0^t f(t-\tau)g(\tau) = \int_1^t [e^{-\tau}-e^{-2\tau}]d\tau = \frac{1}{2}e^{-2t}-e^{-t}-\frac{1}{2}e^{-2}+e^{-1}$,
$\int_0^t f(\tau)g(t-\tau) = \int_1^t [e^{-(t-\tau)}-e^{-2(t-\tau)}]d\tau = \frac{1}{2}e^{-2(t-1)}-e^{-(t-1)}+\frac{1}{2}$