134. Solve the equation $y^{\prime \prime}-3 y^{\prime}+2 y=0$.
Solution. Substitute the function $y=e^{k x}$ and its derivatives $y^{\prime}=$ $=k e^{k x}$ and $y^{\prime \prime}=k^{2} e^{k x}$ into the given equation:

$$
k^{2} e^{k x}-3 k e^{k x}+2 e^{k x}=0
$$

Factor out $e^{k x}$: $e^{k x}\left(k^{2}-3 k+2\right)=0$. Since $e^{k x} \neq 0$, then $k^{2}-3 k+2=0$.

Solve the resulting quadratic equation for $k$:

$$
k_{1,2}=\frac{3 \pm \sqrt{9-4 \cdot 1 \cdot 2}}{2}=\frac{3 \pm 1}{2} ; k_{1}=1, k_{2}=2
$$

The particular solutions of the given equation are: $y_{1}=e^{x}, y_{2}=e^{2 x}$. The general solution of the equation is

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