Example 6. Find the general solution of the equation $y^{\prime \prime}+4 y=\sin 2 x$.
Solution. Consider the equation $z^{\prime \prime}+4 z=e^{2 i x}$.

We have

$$
\sin 2 x=\operatorname{Im} e^{2 i x}
$$

therefore

$$
y_{4, \mu}=\operatorname{Im} z_{4, H} .
$$

The characteristic equation $\lambda^{2}+4=0$ has simple roots $\lambda_{1,2}= \pm 2 i$. Therefore, we seek a particular solution in the form (see Table 1, case III (2)):

$$
z_{Y, H}=A x e^{2 i x}
$$

then

$$
z_{y, 11}^{\prime \prime}=-4 A x e^{2 i x}+4 A i e^{2 i x}
$$

Substituting the expressions for $z_{\text {у.н }}$ and $z_{\text {ч, , }}^{\prime \prime}$ into the equation and canceling by $e^{2 i x}$, we get $4 A i=1$, hence $A=-\frac{i}{4}$, and thus

$$
z_{1, \text { H }}=-\frac{1}{4} i x e^{2 i x}=\frac{1}{4} x \sin 2 x-i \frac{1}{4} x \cos 2 x
$$

The particular solution of the given non-homogeneous equation is

$$
y_{4,11}=\operatorname{Im} z_{4,14}=-\frac{1}{4} x \cos 2 x
$$