Solution
= Solution
Taking Laplace transforms and using $y(0)=0$ gives
$$
(\alpha s+1)Y(s)=F(s).
$$
Hence the transfer <function> is
$$
\boxed{G(s)=\frac{Y(s)}{F(s)}=\frac1{\alpha s+1}}.
$$
Its pole is at $-1/\alpha$, and its causal impulse response is
$$
g(t)=\frac1\alpha e^{-t/\alpha}H(t).
$$
It is absolutely integrable exactly when
$$
\boxed{\alpha>0}.
$$