140. Find the particular solution of the equations $y^{\prime \prime}-2 y^{\prime}-3 y=0$, if $y=8$ and $y^{\prime}=0$ when $x=0$.
Solution. We have $k^{2}-2 k-3=0, D>0, k_{1}=-1, k_{2}=3$; hence, $y_{1}=e^{-x}, y_{2}=e^{3 x}$ are particular linearly independent solutions of the given equation. Its general solution is of the form

$$
y=C_{1} y_{1}+C_{2} y_{2}=C_{1} e^{-x}+C_{2} e^{3 x}
$$

Next, we find

$$
y^{\prime}=-C_{1} e^{-x}+3 C_{2} e^{3 x}
$$

Substituting the initial conditions:

$$
8=C_{1} e^{-1 \cdot 0}+C_{2} e^{3 \cdot 0}=C_{1}+C_{2} ; 0=-C_{1} e^{-1 \cdot 0}+3 C_{2} e^{3 \cdot 0}=-C_{1}+3 C_{2}
$$

i.e., $8=C_{1}+C_{2} ; 0=-C_{1}+3 C_{2}$, from which $C_{1}=6$ and $C_{2}=2$.

Thus, the particular solution satisfying the given initial conditions is $y=6 e^{-x}+2 e^{3 x}$.