3.42. Solve the differential equation (with initial condition):

\[
\left\{\begin{array}{l}
y^{\prime}(t)-2 y(t)=e^{t} \\
y(0)=0 \\
t>0
\end{array}\right.
\]
Solution. We proceed to the images:

$$
p \bar{y}(p)-2 \bar{y}(p)=\frac{1}{p-1} \text { or } \bar{y}(p)=\frac{1}{(p-1)(p-2)}
$$

We decompose this rational fraction into partial fractions:

$$
\frac{1}{(p-1)(p-2)}=\frac{A}{p-1}+\frac{B}{p-2} \Rightarrow 1 \equiv A(p-2)+B(p-1)
$$

Setting $p=1$, we get $A=-1$; for $p=2$ we have $B=1$. Therefore,

$$
\bar{y}(p)=\frac{1}{p-2}-\frac{1}{p-1}
$$

from which:

$$
y(t)=e^{2 t}-e^{t}
$$