Example 3. Integrate the differential equation $y^{\prime}=\frac{y}{2 y \ln y+y-x}$.
Solution. We swap the roles of the function $y$ and the argument $x$:

$$
x^{\prime}=\frac{2 y \ln y+y-x}{y}, \quad x^{\prime}+\frac{x}{y}=2 \ln y+1
$$

We obtain a linear equation with respect to the unknown function $x=x(y)$. Assuming $x=u \cdot v, x^{\prime}=u^{\prime} v+u v^{\prime}$, we get the system

$$
\left\{\begin{array}{l}
v^{\prime}+\frac{v}{y}=0 \\
u^{\prime} v=2 \ln y+1
\end{array} \Leftrightarrow \left\{\begin{array}{l}
v=\frac{1}{y} \\
u=y^{2} \ln y+C
\end{array}\right.\right.
$$

Answer. $x=y \ln y+\frac{C}{y}$.

## Exercises

Solve the differential equations.

1. $y^{\prime}+y \cos x=\sin 2 x$. 2. $y^{\prime}-y \operatorname{tg} x=\operatorname{ctg} x$. 3. $y^{\prime}=\frac{4}{x} y+x \sqrt{y}$.
2. $y^{\prime}+\frac{y}{x}+x y^{2}=0$. 5. $x \cdot y^{\prime}+y=y^{2} \ln x$.
3. $x \ln x \cdot y^{\prime}=y+x^{3}(3 \ln x-1) .7 .3 x y^{\prime}-2 y=\frac{x^{3}}{y^{2}} \cdot 8 \cdot y^{\prime}=\frac{3 x^{2}}{x^{3}+y+1}$.
4. $y^{\prime}=\frac{y}{2 y \ln y+y-x}$. 10. $y^{\prime}=-\frac{y}{x+1}-\frac{1}{2}(x+1)^{3} \cdot y^{3}$.

## Answers

1. $y=C e^{-\sin x}+2(\sin x-1)$. 2. $y=1+\frac{\ln C \operatorname{tg} \frac{x}{2}}{\cos x}$.
2. $y=x^{4}\left(\ln x^{2}+C\right)^{2}$. 4. $y\left(x^{2}+C x\right)=1$. 5. $y=\frac{x}{1+C x+\ln x}$.
3. $y=C \ln x+x^{3}$. 7. $y^{3}=C x^{2}+x^{3}$. 8. $x^{3}=C e^{y}-y-2$.
4. $x=y \ln y+\frac{C}{y}$. 10. $\frac{1}{y^{2}}=\frac{1}{2}(x+1)^{4}+C(x+1)^{2}$.

## § 5. Equations in Total Differentials

$1^{\circ}$. A differential equation of the form

$$
P(x, y) d x+Q(x, y) d y=0
$$

is called an equation in total differentials if the functions $P(x, y)$ and $Q(x, y)$ satisfy the condition $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$.

In this case, the left-hand side of the equation represents the total differential of some function of two variables $u(x, y)$, and if this function is found, the general solution of the equation is $u(x, y)=C$.

$2^{\circ}$. Equations for which the condition $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$ is not satisfied (they are not equations in total differentials) are reduced to equations in total differentials by term-by-term multiplication by an appropriate factor $\mu$, called the integrating factor:

1) $\ln \mu=\ln \mu(x)=\int \frac{1}{Q}\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\right) d x$, if $\frac{1}{Q}\left(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x}\right)$ depends only on $x$ 2) $\ln \mu=\ln \mu(y)=\int \frac{1}{P}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d y$, if $\frac{1}{P}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)$ depends
only on $y$.

## Examples with Solutions