Answer
See below
Work Step by Step
Since $y'+ay=f(t)\\
y(0)=y_0$
We have $L[y'+ay]=(s+a)L[y]-y_0\\
\rightarrow (s+a)L[y]= L[f]+y_0\\
\rightarrow L(y)= \frac{L[f]}{s+a}+\frac{f_0}{s+a}$
Using the inverse transformation:
$y(t)=y_0L^{-1}[\frac{1}{s+a}]+L^{-1}[\frac{L(f)}{s+a}]=y_0e^{-at}+L^{-1}[L(f).\frac{1}{s+a}]\\
=y_0e^{-at}+\int^t_0 e^{-a(t-\omega)}f(\omega)d\omega$
