136. Solve the equation $y^{\prime \prime}+2 y^{\prime}-8 y=0$.
Solution. Let's form the characteristic equation $k^{2}+2 k$ $-8=0$. Here $D=p^{2}-4 q=2^{2}-4 \cdot(-8)=4+32=36>0$. Therefore, the characteristic equation has two distinct real roots. Let's determine them: $k_{1}=-4, k_{2}=2$.

We find the particular solutions of the given differential equation: $y_{1}=e^{-4 x}, y_{2}=e^{2 x}$.

The general solution of the equation is

$$
y=C_{1} e^{-4 x}+C_{2} e^{2 x}
$$