1364. Integrate the following equations and find the particular solutions that satisfy the given initial conditions:
1) $(x+2) d y+(y-1) d x=0, y(1)=2$
2) $y^{\prime}=(2 y-1) \operatorname{ctg} x, y\left(\frac{\pi}{6}\right)=5$.
Solution. 1) First, we find the general integral:

$$
\begin{aligned}
& \frac{d x}{x+2}+\frac{d y}{y-1}=0, \int \frac{d x}{x+2}+\int \frac{d y}{y-1}=\ln |C|(C \neq 0) \\
& \ln |x+2|+\ln |y-1|=\ln |C|, \ln |(x+2)(y-1)|=\ln |C|
\end{aligned}
$$

from which

$$
|(x+2)(y-1)|=|C| \text { and }(x+2)(y-1)= \pm C
$$

or

$$
(x+2)(y-1)=C_{1}\left(C_{1}= \pm C\right)
$$

Substituting $x=1, y=2$ into equation (a), we get $3(2-1) = C_{1}$, from which $C_{1}=3$.

Substituting the found value $C_{1}=3$ into the general integral (a), we find the particular integral $(x+2)(y-1)=3$, satisfying the initial conditions.

There are also two particular solutions $x=-2$ and $y=1$, which do not arise from the general solution. The integral curves $x=-2$ and $y=1$ will be asymptotes of the family of integral curves (plot this family and their asymptotes).

2) We find the general solution:

$$
\begin{gathered}
\frac{d y}{d x}=(2 y-1) \operatorname{ctg} x, \frac{d y}{2 y-1}=\operatorname{ctg} x d x \\
\int \frac{d y}{2 y-1}=\int \operatorname{ctg} x d x, \frac{1}{2} \ln |2 y-1|=\ln |\sin x|+\ln |C|(C \neq 0) \\
\ln |2 y-1|=2 \ln |C \sin x|=\ln C^{2} \sin ^{2} x
\end{gathered}
$$

from which

$$
2 y-1= \pm C^{2} \sin ^{2} x, 2 y-1=C_{1} \sin ^{2} x\left(C_{1}= \pm C^{2}\right)
$$

and

$$
y=\frac{1+C_{1} \sin ^{2} x}{2}
$$

Substituting $x=\frac{\pi}{6}, y=5$ into equation (b), we get

$$
5=\frac{1+C_{1} \cdot \frac{1}{4}}{2}
$$

from which $C_{1}=36$. Substituting this value of $C_{1}$ into the general solution (b), we obtain the particular solution $y=18 \sin ^{2} x+\frac{1}{2}$, satisfying the initial conditions.

Verify that $y=\frac{1}{2}$ is a particular solution that does not arise from the general solution.

Integrate the equations: