Example 5. Solve the equation $y^{\prime \prime}+4 y=-8 \sin 2 x+32 \cos 2 x+$ $+4 e^{2 x}$.
Solution. $k^{2}+4=0, k_{1,2}= \pm 2 i$. Next (see Theorem 2) we find two particular solutions (solve as if two equations, which is equivalent to one equation with two independent right-hand sides).

a) We find the particular solution corresponding to the trigonometric part, keeping in mind that the numbers $\gamma \pm i \delta= \pm 2 i-$ are roots of the characteristic equation.

4: $y_{4}=(A \cos 2 x+B \sin 2 x) x$

0: $y_{4}^{\prime}=(-2 A \sin 2 x+2 B \cos 2 x) x+A \cos 2 x+B \sin 2 x$;

1: $y_{4}^{\prime \prime}=(-4 A \cos 2 x-4 B \sin 2 x) x+2(-2 A \sin 2 x+2 B \cos 2 x)$;

$x \cos 2 x: 4 A-4 A=0 ; \quad \cos 2 x: 4 B=32, \quad b=8$;

$x \sin 2 x: \quad 4 B-4 B=0 ; \quad \sin 2 x:-4 A=-8, \quad A=2$.

b) We find the particular solution corresponding to the algebraic part: $y_{4}=C e^{2 x}, \quad y_{4}^{\prime}=2 C e^{2 x}, \quad y_{4}^{\prime \prime}=4 C e^{2 x}, \quad 8 C=4, \quad C=\frac{1}{2}$.

Answer. $y_{\text {gen }}=C_{1} \cos 2 x+C_{2} \sin 2 x+(2 \cos 2 x+8 \sin 2 x) x+\frac{1}{2} e^{2 x}$.