Apply Laplace Transform to each term:
L{y'(t)}+3L{y(t)}=L{e⁻ᵗ}, so sY(s)-y(0)+3Y(s)=1/(s+1).
Y(s)(s+3)=1+1/(s+1) because y(0)=1 as the initial condition.
Y(s)=(s+2)/((s+1)(s+3)).
(s+2)/((s+1)(s+3))≡A/(s+1)+B/(s+3), so s+2=As+3A+Bs+B.
Therefore, A+B=1, 3A+B=2; 3A+B-(A+B)=2-1=1, 2A=1, A=½=B.
Y(s)=½(1/(s+1))+½(1/(s+3)).
Now we apply inverse Laplace:
y(t)=½e⁻ᵗ+½e⁻³ᵗ.