Example 2. Check that the given functions are solutions to the given differential equations, and form the general solutions of these equations:

1) $y_{1}=3 \sin 5 x, \quad y_{2}=\frac{1}{5} \cos 5 x ; \quad y^{\prime \prime}+25 y=0 ;$
2) $y_{\mathrm{I}}=e^{3 x}, \quad y_{2}=x e^{3 x} ; \quad y^{\prime \prime}-6 y^{\prime}+9 y=0$
3) $y_{1}=e^{2 x}, \quad y_{2}=e^{-3 x} ; \quad y^{\prime \prime}+y^{\prime}-6 y=0$
Solution. 1) We find the first and second derivatives of the first function:

$$
y_{1}^{\prime}=15 \cos 5 x, \quad y_{1}^{\prime \prime}=-75 \sin 5 x
$$

Substituting these values into the given equation, we obtain the identity:

$$
y_{\mathrm{I}}^{\prime \prime}+25 y_{1}=-75 \sin 5 x+75 \sin 5 x \equiv 0, \quad x \in \mathbb{R}
$$

Therefore, $y_{\mathrm{i}}=3 \sin 5 x$ is a particular solution of the equation $y^{\prime \prime}+25 y=0$. Similarly, it can be verified that the function $y_{2}=\frac{1}{5} \cos 5 x$ also satisfies this equation (we suggest you do this on your own). These functions are linearly independent, since the ratio $\frac{y_{1}}{y_{2}}=15 \operatorname{tg} 5 x \neq$ const.

According to Theorem 4, the general solution of the given equation is $y_{00}=C_{1} \cdot 3 \sin 5 x+C_{2} \cdot \frac{1}{5} \cos 5 x$, or $y_{00}=\widetilde{C}_{1} \sin 5 x+\widetilde{C}_{2} \cos 5 x$.

2) Similarly, this part of the example is solved. We have (we will limit ourselves to solving $y_{2}$ of the given equation):

$$
\begin{aligned}
& y_{2}=x e^{3 x}, y_{2}^{\prime}=e^{3 x}+3 x e^{3 x}, y_{2}^{\prime \prime}=3 e^{3 x}+3 e^{3 x}+9 x e^{3 x} \\
& y_{2}^{\prime \prime}-6 y_{2}^{\prime}+9 y=6 e^{3 x}+9 x e^{3 x}-6\left(e^{3 x}+3 x e^{3 x}\right)+9 x e^{3 x} \equiv 0, x \in \mathbb{R} \\
& y_{00}=C_{1} e^{3 x}+C_{2} \cdot x e^{3 x}
\end{aligned}
$$

3) We will limit ourselves to checking the function $y_{2}=e^{-3 x}$ (complete this on your own):

$$
\begin{aligned}
& y_{2}^{\prime}=-3 e^{-3 x}, y_{2}^{\prime \prime}=9 e^{-3 x}, y_{2}^{\prime \prime}+y_{2}^{\prime}-6 \\
& y_{2}=9 e^{-3 x}-3 e^{-3 x}-6 e^{-3 x} \equiv 0, x \in \mathbb{R} \\
& y_{00}=C_{1} e^{2 x}+C_{2} e^{-3 x}
\end{aligned}
$$