3.44. Solve the differential equation $y^{\prime \prime}-2 y^{\prime}-$ $-3 y=e^{3 t}$, if $y(0)=0, y^{\prime}(0)=0$.
Solution. We proceed to the images:

$$
p^{2} \bar{y}(p)-p y(0)-y^{\prime}(0)-2(p \bar{y}(p)-y(0))-3 \bar{y}(p)=\frac{1}{p-3}
$$

or

$$
\bar{y}(p)=\frac{1}{(p-3)\left(p^{2}-2 p-3\right)}=\frac{1}{(p+1)(p-3)^{2}} .
$$

We decompose this rational fraction into partial fractions:

$$
\frac{1}{(p+1)(p-3)^{2}}=\frac{A}{(p-3)^{2}}+\frac{B}{p-3}+\frac{C}{p+1}
$$

Acting as above, we get that

Therefore,

$$
A=\frac{1}{4}, \quad B=-\frac{1}{16}, C=\frac{1}{16}
$$

$$
\vec{y}(p)=\frac{1}{4} \frac{1}{(p-3)^{2}}-\frac{1}{16} \frac{1}{p-3}+\frac{1}{16} \frac{1}{p+1}
$$

from which, by transitioning to the originals, we obtain:

$$
y(t)=\frac{1}{4} t e^{3 t}-\frac{1}{16} e^{3 t}+\frac{1}{16} e^{-t}
$$