4.111. Find the solution to the Cauchy problem: $y^{\prime \prime}-16 y^{\prime}+20 y=0$, $y(0)=3, y^{\prime}(0)=0$.
Solution. We form the characteristic equation $r^{2}-16 r+20=0$. Its roots are complex conjugates:

$$
\begin{aligned}
r_{1,2} & =\frac{4 \pm \sqrt{(-4)^{2}-4 \cdot 1 \cdot 5}}{2 \cdot 1}=\frac{4 \pm \sqrt{-4}}{2}= \\
& =\frac{4 \pm \sqrt{4(-1)}}{2}=\frac{4 \pm 2 \cdot \sqrt{-1}}{2}=2 \pm i
\end{aligned}
$$

The general solution has the form:

$$
y(x)=\left(C_{1} \cos \beta x+C_{2} \sin \beta x\right) e^{\alpha x}=\left(C_{1} \cos x+C_{2} \sin x\right) e^{2 x}
$$

We find:

$$
\begin{aligned}
y^{\prime}(x)= & \left(-C_{1} \sin x+C_{2} \cos x\right) e^{2 x}+\left(C_{1} \cos x+C_{2} \sin x\right) e^{2 x} \cdot 2= \\
& \left.=\left(\left(2 C_{1}+C_{2}\right) \cos x+\left(-C_{1}+2 C_{2}\right) \sin x\right)\right) e^{2 x} .
\end{aligned}
$$

Using the initial conditions:

$$
\begin{gathered}
y(0)=\left(C_{1} \cos 0+C_{2} \sin 0\right) e^{2 \cdot 0}=C_{1}=3 \\
\left.y^{\prime}(0)=\left(\left(2 C_{1}+C_{2}\right) \cos 0+\left(-C_{1}+2 C_{2}\right) \sin 0\right)\right) e^{2 \cdot 0}= \\
=2 C_{1}+C_{2}=0
\end{gathered}
$$

We obtain the system of equations:

$$
\left\{\begin{array}{l}
C_{1}=3 \\
2 C_{1}+C_{2}=0
\end{array}\right.
$$

The solution to the system is: $C_{1}=3$ in the first equation, and we determine $C_{2}=-6$.

Answer. The general solution of the equation is: $y(x)=\left(C_{1} \cos x+\right.$ $\left.+C_{2} \sin x\right) e^{2 x}$. The solution to the Cauchy problem is: $y(x)=(3 \cos x-6 \sin x) e^{2 x}$.