Example 8. Find the solution of the equation

$$
x^{3} \sin y \cdot y^{\prime}=2
$$

satisfying the condition

$$
y \rightarrow \frac{\pi}{2} \quad \text { as } \quad x \rightarrow \infty
$$
Solution. By separating variables and integrating, we find the general integral of equation (6):

$$
\cos y=\frac{1}{x^{2}}+C
$$

$U_{\text {condition (7) gives }} \cos \frac{\pi}{2}=C$, i.e., $C=0$, so the particular integral will have the form $\cos y=\frac{1}{x^{2}}$. It corresponds to an infinite set of particular solutions of the form

$$
y= \pm \arccos \frac{1}{x^{2}}+2 \pi n, \quad n=0, \pm 1, \pm 2, \ldots
$$

Among these solutions, there is only one that satisfies condition (7). We find this solution by taking the limit as $x \rightarrow \infty$ in equation (8):

$$
\frac{\pi}{2}= \pm \arccos 0+2 \pi n, \quad \text { or } \quad \frac{\pi}{2}= \pm \frac{\pi}{2}+2 \pi n,
$$

from which

$$
\frac{1}{2}= \pm \frac{1}{2}+2 n
$$

It is easy to see that equation (9) has two roots: $n=0$ and $n=\frac{1}{2}$, and the root $n=\frac{1}{2}$, corresponding to the minus sign before $\arccos \left(\frac{1}{x^{2}}\right)$, is not suitable ( $n$ must be an integer or zero). Thus, the required particular solution of equation (6) will be

$$
y=\arccos \frac{1}{x^{2}}
$$

## Problems for Independent Solution

## Integrate the equations:

46. $\left(1+y^{2}\right) d x+\left(1+x^{2}\right) d y=0$.
47. $y^{\prime} \sin x-y \cos x=0,\left.y\right|_{x=\pi / 2}=1$.
48. $x \sqrt{1+y^{2}}+y y^{\prime} \sqrt{1+x^{2}}=0$.
49. $e^{-y}\left(1+y^{\prime}\right)=1$.
50. $y^{\prime}=a^{x+y} \quad(a>0, a \neq 1)$.
51. $2 x \sqrt{1-y^{2}}=y^{\prime}\left(1+x^{2}\right)$.
52. $y^{2} \sin x d x+\cos ^{2} x \ln y d y=0$.
53. $y^{\prime}=a x+b y+c \quad(a, b, c-$ const $)$.
54. $y+x y^{\prime}=a(1+x y)$,

$\left.y\right|_{x=1 / a}=-a$
47. $\left(1+y^{2}\right) d x+x y d y=0$.

49. $\left(1+y^{2}\right) d x=x d y$.
50. $x \sqrt{1-y^{2}} d x+y \sqrt{1-x^{2}}=0$,

$$
\left.y\right|_{x=0}=1
$$

53. $y \ln y d x+x d y=0,\left.\quad y\right|_{x=1}=1$.
54. $e^{y}\left(1+x^{2}\right) d y-2 x\left(1+e^{y}\right) d x=