1201. Find the particular solution of the system of differential equations

$$
\frac{d x}{d t}+2 x+y=\sin t ; \quad \frac{d y}{d t}-4 x-2 y=\cos t
$$

satisfying the initial conditions: $x(\pi)=1, y(\pi)=2$.
Solution. First, we find the general solution of the given system. Differentiating the first equation with respect to $t$: $x^{\prime \prime}+2 x^{\prime}+y^{\prime}=\cos t$ and, substituting in the result the derivative $y^{\prime}$ with its expression through $t$ and $x^{\prime}$, determined from the given system, we obtain a second-order equation with one unknown function $x$:

$$
x^{\prime \prime}+2 \sin t=0
$$

Solving it as a simple higher-order equation by integrating both parts twice (§ 6) or as a linear inhomogeneous equation with constant coefficients (§ 8), we find

$$
x=C_{1} t+C_{2}+2 \sin t
$$

Substituting the found function $x$ and its derivative $x^{\prime}=C_{1}+2 \cos t$ into the first equation of the given system, we find

$$
y=-2 C_{2}-C_{1}(2 t+1)-2 \cos t-3 \sin t
$$

The set of functions $x$ and $y$ is the general solution of the given system.

Next, based on the given initial conditions, we determine the values of the constants $C_{1}$ and $C_{2}$. From the first condition: $x=1$ when $t=\pi$ and equality (a), we have the equation

$$
1=C_{1} \pi+C_{2}
$$

and from the condition: $y=2$ when $t=\pi$ and equality (b), we have the second equation with unknowns $C_{1}$ and $C_{2}$

$$
2=-2 C_{2}-C_{1}(1+2 \pi)+2 .
$$

Solving these equations as a system, we find: $C_{1}=-2, C_{2}=1+2 \pi$.

Finally, substituting these values of $C_{1}$ and $C_{2}$ into the general solution, we obtain the desired particular solution of the given system, satisfying the given initial conditions:

$$
x=1-2(t-\pi)+2 \sin t ; \quad y=4(t-\pi)-2 \cos t-3 \sin t .
$$