1202. Solve the system $y^{\prime \prime} - z=0 ; z^{\prime} + 8 y=0$.
Solution. Differentiating the first equation, we determine $z'$, and substituting it into the second equation, we get $y''' + 8y = 0$. From this, we find (§ 7)

$$
y = C_1 e^{-2x} + e^x \left(C_2 \cos \sqrt{3} x + C_3 \sin \sqrt{3} x\right)
$$

Differentiating $y$ twice and substituting into the first equation, we find

$$
z = 4 C_1 e^{-2x} + 2 e^x \left[ (\sqrt{3} C_3 - C_2) \cos \sqrt{3} x - (\sqrt{3} C_2 + C_3) \sin \sqrt{3} x \right]
$$

Solve the system of equations:

$$
\text{1203. }\left\{\begin{array}{l}
\frac{d y}{d x} + 2 y + 3 z = 0, \\
\frac{d z}{d x} + y = 0.
\end{array} \text{ 1204. } \left\{\begin{array}{l}
\frac{d u}{d t} - 2 u - 4 v = \cos t \\
\frac{d v}{d t} + u + 2 v = \sin t
\end{array}\right.\right.
$$