EXAMPLE. Find the integral curves of the differential equation

$$
6 x d x-6 y d y=2 x^{2} y d y-3 x y^{2} d x
$$
Solution.

1. Rewrite the original equation as

$$
3 x\left(2+y^{2}\right) d x=2 y\left(x^{2}+3\right) d y
$$

Since $x^{2}+3>0$ and $2+y^{2}>0$, we separate the variables, i.e., represent equation (2) in the form

$$
\frac{3 x}{x^{2}+3} d x=\frac{2 y}{2+y^{2}} d y
$$

2. Compute the integrals in the equation

$$
\int \frac{3 x}{x^{2}+3} d x=\int \frac{2 y}{2+y^{2}} d y
$$

We have

$$
\int \frac{3 x}{x^{2}+3} d x=\frac{3}{2} \ln \left(x^{2}+3\right)+C_{1}
$$

$$
\int \frac{2 y}{2+y^{2}} d y=\ln \left(2+y^{2}\right)+C_{2}
$$

Therefore,

$$
\frac{3}{2} \ln \left(x^{2}+3\right)=\ln \left(2+y^{2}\right)+C_{0}
$$

where $C_{0}=C_{2}-C_{1}$.

3. Simplifying this equality, we get

$$
\frac{\left(x^{2}+3\right)^{3}}{\left(2+y^{2}\right)}=C
$$

Answer. The integral curves are defined by the equation

$$
\frac{\left(x^{2}+3\right)^{3}}{\left(2+y^{2}\right)}=C
$$

for all possible values of $C$.

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

1. $y d x+\left(1+x^{2}\right) d y=0$.
2. $y^{\prime}=\operatorname{tg} x \operatorname{tg} y$.
3. $y^{\prime} y=-2 x \sec y$.
4. $y^{\prime}+\sin (x+y)=\sin (x-y)$.
5. $y\left(1+x^{2}\right) y^{\prime}+x\left(1+y^{2}\right)=0$.
6. $e^{x} d x-\left(1+e^{y}\right) y d y=0$.
7. $y^{\prime}=2 e^{x} \cos x$.
8. $y^{\prime}=y \ln y$.
9. $y^{\prime}=\frac{2 x}{1+x^{2}}$.
10. $y^{\prime}=\frac{1}{\sqrt{x^{2}-1}}$.

Answers.

1. $\ln |y|+\operatorname{arctg} x=C$.
2. $\sin y \cos x=C$.
3. $x^{2}+y \sin y+\cos y=C$.
4. $\ln |\operatorname{tg}(y / 2)|+2 \sin x=C$.
5. $\left(x^{2}+1\right)\left(1+y^{2}\right)=C$.
6. $2 \ln \left(1+e^{x}\right)-y^{2}=C$.
7. $y=e^{x}(\cos x+\sin x)+C$.
8. $\ln |\ln y|-x=C$.
9. $y=\ln \left(1+x^{2}\right)+C$.
10. $y=\ln \left|x+\sqrt{x^{2}-1}\right|+C$.

## 11.3. Homogeneous Equations

Problem Statement. Find the integral curves of a homogeneous differential equation of the first order, i.e., a differential equation of the form

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

where $P(x, y)$ and $Q(x, y)$ are homogeneous functions of the same order, i.e., $P(t x, t y)=t^{n} P(x, y)$ and $Q(t x, t y)=t^{n} Q(x, y)$.

## PLAN OF SOLUTION.

1. Transform equation (1) to the form

$$
y^{\prime}=f\left(\frac{y}{x}\right)
$$

2. Make the substitution $y(x)=x z(x)$, where $z(x)$ is the new unknown function. Then $y^{\prime}=z+x z^{\prime}$ and equation ($1^{\prime}$) is transformed to the form

$$
x \frac{d z}{d x}=f(z)-z
$$

i.e., to an equation with separable variables.

Note that the substitution $y(x)=x z(x)$ can be made directly in equation (1), without transforming it to the form ($1^{\prime}$).

3. Separate the variables in the domain where $f(z)-z \neq 0$:

$$
\frac{d z}{f(z)-z}=\frac{d x}{x}
$$

4. Integrate the resulting equation with separated variables and substitute $z(x)=y(x) / x$. Write the answer.

Remarks.

1. If $z=z_{0}$ is a root of the equation $f(z)-z=0$, then $y=z_{0} x$ is also a solution of equation (1).
2. The integral curves of a homogeneous equation can also be sought in polar coordinates.