4.110. Find the solution to the Cauchy problem: $y^{\prime \prime}-4 y^{\prime}+4 y=0$, $y(0)=2, y^{\prime}(0)=1$.
Solution. We form the characteristic equation: $r^{2}-4 r+4=0$. Its roots are real and equal to each other: $r_{1}=r_{2}=r=2$. The general solution has the form:

$$
\begin{aligned}
& y(x)\left(C_{1}+x C_{2}\right) e^{r_{x}}=\left(C_{1}+x C_{2}\right) e^{2 x} \\
& y(x)=C_{1} e^{\eta x}+C_{2} e^{r_{2} x}=C_{1} e^{x}+C_{2} e^{2 x}
\end{aligned}
$$

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

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

We obtain the system of equations:

$$
\left\{\begin{array}{l}
C_{1}=2 \\
2 C_{1}+C_{2}=1
\end{array}\right.
$$

Substitute $C_{1}=2$ into the second equation and determine $C_{2}=-3$.

Answer. The general solution of the equation: $y(x)=\left(C_{1}+C_{2} x\right) e^{2 x}$. The solution to the Cauchy problem: $y(x)=(1-3 x) e^{2 x}$.