1067. Find the particular integral of the equation satisfying the given initial condition:
1) $y d x+\operatorname{ctg} x d y=0 ; \quad y\left(\frac{\pi}{3}\right)=-1$.
2) $s=s^{\prime} \cos ^{2} t \ln s ; s(\pi)=1$.
Solution. 1) By separating variables and integrating, we first find the general integral of the given equation:

$$
\begin{gathered}
\operatorname{tg} x d x+\frac{d y}{y}=0 ; \quad-\ln |\cos x|+\ln |y|=\ln C \\
|y|=C|\cos x| ; \quad y= \pm C \cos x=C_{1} \cos x
\end{gathered}
$$

Then, using the specified initial condition $y\left(\frac{\pi}{3}\right)=-1$, we substitute the given values of the variables $\left(x=\frac{\pi}{3}, y=-1\right)$ into the general integral and determine the corresponding value of the arbitrary constant: $-1=C_{1} \cos \frac{\pi}{3} ; C_{1}=-2$.

With this value of $C_{1}$, we obtain the desired particular integral from the general integral that satisfies the given initial condition, $y=-2 \cos x$.

2) By multiplying by $\frac{\sec ^{2} t}{s} d t$, we separate the variables $\sec ^{2} t d t=$ $=\frac{\ln s}{s} d s$ and, integrating, we find the general integral

$$
\int d(\operatorname{tg} t)=\int \ln s d \ln s+C ; \quad \operatorname{tg} t=\frac{1}{2} \ln ^{2} s+C
$$

Substituting the initial values $t=\pi, s=1$, we determine the value of $C$ :

$$
\operatorname{tg} \pi=\frac{1}{2} \ln 1+C ; \quad C=0
$$

Therefore, the desired particular integral is $\ln ^{2} s-2 \operatorname{tg} t=0$.

Solve the following differential equations (find their general integrals):