1084. Solve the equations:
1) $y^{\prime}-y \cot x=\sin x$
2) $x^{2} y^{2} y^{\prime}+x y^{3}=1$
3) $y d x-(3 x+1+\ln y) d y=0$ with the condition $y\left(-\frac{1}{3}\right)=1$.
Solution: 1) Having made sure that the given equation is linear, we assume $y=u v$; then $y^{\prime}=u^{\prime} v+v^{\prime} u$ and the given equation transforms into

$$
u^{\prime} v+v^{\prime} u-u v \operatorname{ctg} x=\sin x \text { or } u^{\prime} v+u\left(v^{\prime}-v \operatorname{ctg} x\right)=\sin x
$$

Since one of the auxiliary functions $u$ or $v$ can be chosen arbitrarily, we choose as $v$ any particular integral of the equation $v^{\prime}-v \operatorname{ctg} x=0$.

Then, to find $u$, we obtain the equation $u^{\prime} v=\sin x$.

Solving the first of these equations, we find $v$; separating variables and integrating, we find its simplest, non-zero particular integral:

$$
\frac{d v}{v}=\operatorname{ctg} x d x ; \ln v=\ln \sin x ; v=\sin x .
$$

Substituting $v$ into the second equation and solving it, we find $u$ as the general integral of this equation:

$$
u^{\prime} \sin x=\sin x ; d u=d x ; u=x+C
$$

Knowing $u$ and $v$, we find the desired function $y$ :

$$
y=u v=(x+C) \sin x
$$

2) Dividing both parts of the equation by $x^{2} y^{2}$ :

$$
y^{\prime}+\frac{y}{x}=y^{-2} \cdot \frac{1}{x^{2}}
$$

we see that this is a Bernoulli equation, where $P=x^{-1}, Q=x^{-2}$.

Replacing the function $y$ by the formula $y=u v$, we have $y^{\prime}=u^{\prime} v+v^{\prime} u$,

$$
u^{\prime} v+v^{\prime} u+\frac{u v}{x}=\frac{1}{x^{2} u^{2} v^{2}}
$$

or

$$
u^{\prime} v+u\left(u^{\prime}+\frac{v}{x}\right)=\frac{1}{x^{2} u^{2} v^{2}}
$$

From here, as in the solution of the previous problem, we obtain two equations with separable variables:

$$
\text { 1) } v^{\prime}+\frac{v}{x}=0 \text { and 2) } u^{\prime} v=\frac{1}{x^{2} u^{2} v^{2}} \text {. }
$$

Solving the first equation, we find $v$ as the simplest particular integral of this equation:

$$
\frac{d v}{v}+\frac{d x}{x}=0 ; \ln v+\ln x=0 ; v x=1 ; v=\frac{1}{x}
$$

Substituting $v$ into the second equation and solving it, we find $u$ as the general integral of this equation:

$$
\frac{u^{\prime}}{x}=\frac{1}{u^{2}} ; u^{2} d u=x d x ; \frac{u^{9}}{3}=\frac{x^{2}}{2}+\frac{C}{3} ; u=\sqrt[3]{\frac{3}{2} x^{2}+C}
$$

Therefore, the desired general integral of the given equation is

$$
y=u v=\sqrt[3]{\frac{3}{2 x}+\frac{C}{x^{3}}}
$$

3) Transforming the given equation to the form

$$
\frac{d x}{d y}-\frac{3 x}{y}=\frac{1+\ln y}{y} \text { or } \frac{d x}{d y}+P(y) x=Q(y)
$$

we determine that it is a linear equation if we consider $x$ as a function of $y$.

Further, replacing the function $x$ by the formula $x=u v$, where $u$ and $v$ are functions of $y$, we have $\frac{d x}{d y}=v \frac{d u}{d y}+u \frac{d v}{d y}$ and

$$
v \frac{d u}{d y}+u \frac{d v}{d y}-\frac{3 u v}{y}=\frac{1+\ln y}{y}
$$

Or

$$
v \frac{d u}{d y}+u\left(\frac{d v}{d y}--\frac{3 v}{y}\right)=\frac{1+\ln y}{y}
$$

From here, to find $u$ and $v$, we have two equations:

$$
\text { 1) } \frac{d v}{d y}-\frac{3 v}{y}=0 \text { and 2) } v \frac{d u}{d y}=\frac{1+\ln y}{y} \text {. }
$$

From the first equation, we find $v$:

$$
\frac{d v}{v}=\frac{3 d y}{y} ; \ln v=3 \ln y ; v=y^{3}
$$

Substituting $v$ into the second equation and solving it, we find $u$:

$$
\begin{gathered}
y^{3} \frac{d u}{d y}=\frac{1+\ln y}{y} ; \quad d u=\frac{1+\ln y}{y^{4}} d y \\
u=\int y^{-4} d y+\int y^{-4} \ln y d y+C=\frac{y^{-3}}{-3}+I+C
\end{gathered}
$$

The second integral, denoted by $I$, is found separately by the formula for integration by parts. Assuming $u_{1}=\ln y, d v_{1}=y^{-4} d y$, we get $d u_{1}=\frac{d y}{y}, \quad v_{1}=\frac{y^{-3}}{-3}$ and

$$
\begin{gathered}
I=\int u_{1} d v_{1}=u_{1} v_{1}-\int v_{1} d u_{1}=-\frac{\ln y}{3 y^{3}}+\frac{1}{3} \int y^{-4} d y= \\
=-\frac{\ln y}{3 y^{3}}-\frac{1}{9 y^{3}} .
\end{gathered}
$$

Therefore, $\quad u=-\frac{1}{3 y^{3}}-\frac{\ln y}{3 y^{3}}-\frac{1}{9 y^{3}}+C$.

Multiplying $u$ by $v$, we get the general integral of the given equation:

$$
x=C y^{3}-\frac{4}{9}-\frac{1}{3} \ln y
$$

Substituting the given values of the variables $x=-\frac{\mathbf{i}}{3}$, $y=1$, we find the value of the arbitrary constant $C=\frac{1}{9}$.

Therefore, the desired particular integral will be

$$
x=\frac{y^{3}-4}{9}-\frac{1}{3} \ln y
$$

Solve the following equations: