Example. Find the integral curves of the differential equation

$$
x d x+y d y+\frac{x d y-y d x}{x^{2}+y^{2}}=0
$$
Solution.

1. Transform the equation (3):

$$
\left(x-\frac{y}{x^{2}+y^{2}}\right) d x+\left(y+\frac{x}{x^{2}+y^{2}}\right) d y=0
$$

In this case,

$$
P(x, y)=\left(x-\frac{y}{x^{2}+y^{2}}\right), Q(x, y)=\left(y+\frac{x}{x^{2}+y^{2}}\right)
$$

These functions are continuously differentiable in the region $x^{2}+y^{2}>0$. Moreover,

$$
\frac{\partial}{\partial y}\left(x-\frac{y}{x^{2}+y^{2}}\right) \equiv \frac{\partial}{\partial x}\left(y+\frac{x}{x^{2}+y^{2}}\right)\left(x^{2}+y^{2}>0\right)
$$

Therefore, $P(x, y) d x+Q(x, y) d y$ is the differential of some function $U(x, y)$ in any simply connected region $D$ that does not contain the point $x=0$, $y=0$. Hence, equation (3) is an equation in total differentials and can be written as

$$
d U(x, y)=0 \text { for }(x, y) \in D .
$$

In this case,

$$
\frac{\partial U}{\partial x}=x-\frac{y}{x^{2}+y^{2}}, \quad \frac{\partial U}{\partial y}=y+\frac{x}{x^{2}+y^{2}}
$$

2. Integrating the first equality in (4) with respect to $x$, we get, for $y \neq 0$,

$$
U=\int\left(x-\frac{y}{x^{2}+y^{2}}\right) d x=\frac{x^{2}}{2}-\operatorname{arctg} \frac{x}{y}+\varphi(y)
$$

where $\varphi(y)$ is an unknown function that still needs to be found.

3. Differentiating $U$ with respect to $y$, we have

$$
\frac{\partial U}{\partial y}=\frac{x}{x^{2}+y^{2}}+\varphi^{\prime}(y)
$$

Using the second equality in (2), we get

$$
\frac{x}{x^{2}+y^{2}}+\varphi^{\prime}(y)=y+\frac{x}{x^{2}+y^{2}}
$$

After transformations, we have

$$
\varphi^{\prime}(y)=y
$$

4. Hence,

$$
\varphi(y)=\frac{y^{2}}{2}+C
$$

and, therefore,

$$
U(x, y)=\frac{x^{2}+y^{2}}{2}-\operatorname{arctg} \frac{x}{y}+C, \quad y>0 \text { or } y<0
$$

The integral curves in the region $y>0$ or in the region $y<0$ are determined by the equation

$$
\frac{x^{2}+y^{2}}{2}-\operatorname{arctg} \frac{x}{y}=C
$$

for all possible values of $C$.

### Conditions of the Problems. Find the integral curves of the differential equations.

1. $(x+y) d x+(x+2 y) d y=0$.
2. $\left(3 x^{2}+6 x y^{2}\right) d x+\left(6 x^{2} y+4 y^{3}\right) d y=0$.
3. $(2 x+5 y) d x+\left(5 x+3 y^{2}\right) d y=0$.
4. $\left(2 x y+3 y^{2}\right) d x+\left(x^{2}+6 x y-3 y^{2}\right) d y=0$.
5. $\left(3 y^{2}+2 x y+2 x\right) d x+\left(6 x y+x^{2}+3\right) d y=0$.
6. $\left(x^{2}+y^{2}+2 x\right) d x+2 x y d y=0$.
7. $\left(x^{3}-3 x y^{2}+2\right) d x-\left(3 x^{2} y-y^{2}\right) d y=0$.
8. $(x \operatorname{ch} y+\operatorname{sh} y) d x+(x \operatorname{ch} x+\operatorname{sh} x) d y=0$.
9. $2 x\left(1+\sqrt{x^{2}-y}\right) d x-\sqrt{x^{2}-y} d y=0$.
10. $\left(\frac{1}{x}-\frac{y^{2}}{(x-y)^{2}}\right) d x+\left(\frac{x^{2}}{(x-y)^{2}}-\frac{1}{y}\right) d y=0$.

### Answers.

1. $\frac{x^{2}}{2}+x y+y^{2}=C$.
2. $x^{3}+3 x^{2} y^{2}+y^{4}=C$.
3. $x^{2}+5 x y+y^{3}=C$.
4. $x^{2} y+3 x y^{2}-y^{3}=C$.
5. $3 x y^{2}+x^{2} y+3 y+x^{2}=C$.
6. $\frac{x^{3}}{3}+x y^{2}+x^{2}=C$.
7. $\frac{x^{4}}{4}-\frac{3}{2} x^{2} y^{2}+2 x+\frac{y^{3}}{3}=C$.
8. $y \operatorname{sh} x+x \operatorname{sh} y=C$.
9. $x^{2}+\frac{2}{3}\left(x^{2}+y^{2}\right)^{3 / 2}=C$.
10. $\frac{x y}{x-y}+\ln \frac{x}{y}=C$.

## 11.7. Equations of the Form $F\left(x, y^{(k)}, y^{(k+1)}\right)=0$

### Problem Statement. Find the general solution of the differential equation

$$
F\left(x, y^{(k)}, y^{(k+1)}\right)=0
$$

### PLAN OF SOLUTION.

1. Let $y^{(k)}=p(x)$, then we get a first-order differential equation

$$
F\left(x, p, p^{\prime}\right)=0
$$

2. Determine the type of this equation and apply the corresponding method of solution to find $p=f\left(x, C_{1}\right)$, where $C_{1}$ is an arbitrary constant.
3. Since $p=y^{(k)}(x)$, we have

$$
y^{(k)}=f\left(x, C_{1}\right)
$$

Integrating $k$ times (not forgetting the arbitrary constant at each integration), we get the answer

$$
y=\varphi\left(x, C_{1}, C_{2}, \ldots, C_{n}\right), \quad n=k+1
$$

$C_{1}, C_{2}, \ldots, C_{n}$ are arbitrary constants.