4.109. Find the solution to the Cauchy problem: $y^{\prime \prime}-3 y^{\prime}+2 y=0$, $y(0)=1, y^{\prime}(0)=0$.
Solution. We form the characteristic equation $r^{2}-3 r+2=0$. Its roots are real and distinct: $r_{1}=1, r_{2}=2$. The general solution is:

$$
y(x)=C_{1} e^{\eta x}+C_{2} e^{r_{2} x}=C_{1} e^{x}+C_{2} e^{2 x}
$$

We find $y^{\prime}(x)=C_{1} e^{x}+2 C_{2} e^{2 x}$. Using the initial conditions:

$$
\begin{gathered}
y(0)=C_{1} e^{0}+C_{2} e^{2 \cdot 0}=C_{1}+C_{2}=1 \\
y^{\prime}(0)=C_{1} e^{0}+2 C_{2} e^{2 \cdot 0} \Rightarrow C_{1}+2 C_{2}=0
\end{gathered}
$$

We obtain the system of equations $\left\{\begin{array}{l}C_{1}+C_{2}=1 ; \\ C_{1}+2 C_{2}=0 .\end{array}\right.$ Subtract the second equation from the first and write the obtained equation instead of the second. Keep the first equation: $\left\{\begin{array}{l}C_{1}+C_{2}=1 ; \\ C_{2}=-1 .\end{array}\right.$ Substitute the found $C_{2}=-1$ into the first equation and determine: $C_{1}=2$.

Answer. The general solution of the equation: $y(x)=C_{1} e^{x}-C_{2} e^{2 x}$. The solution to the Cauchy problem: $y(x)=2 e^{x}+e^{2 x}$.