Example 4.6. Find a particular integral of the equation

$$
y^{\prime} \sin ^{2} x \ln y+y=0,
$$

satisfying the initial conditions $y \left(\frac{\pi}{4}\right)=1$.
Solution. Let's find the general integral of the given equation. For this, we will separate the variables:

$$
\sin ^{2} x \ln y d y+y d x=0
$$

or

$$
\frac{\ln y}{y} d y=-\frac{d x}{\sin ^{2} x} .
$$

Integrating, we get

$$
\frac{\ln ^{2} y}{2}=\operatorname{ctg} x+C
$$

This is the general integral of the given equation. Using the initial conditions \( y \left( \frac{\pi}{4} \right) = 1 \), we substitute the given values of the variables \( x = \frac{\pi}{4}, y = 1 \) into the expression for the general integral to determine the value of the arbitrary constant \( C \):

$$
\frac{\ln ^{2} 1}{2}=\operatorname{ctg} \frac{\pi}{4}+C
$$

from which \( C = -1 \). Thus, the required particular integral is

$$
\frac{\ln ^{2} y}{2}=\operatorname{ctg} x-1 \text {. }
$$

## 4.3. Homogeneous Equations

An equation of the form

$$
y^{\prime}=f \frac{y}{x}
$$

is called a homogeneous equation.

A homogeneous equation can be reduced to an equation with separable variables by the substitution \( y = u x \), where \( u \) is the new unknown function.

Differentiating the equation \( y = u x \), we get:

$$
\frac{d y}{d x}=x \frac{d u}{d x}+u
$$

Substituting the expressions for \( y \) and \( \frac{d y}{d x} \) into equation (1), we have:

$$
x \frac{d u}{d x}+u=f(u)
$$

from which

$$
\frac{d u}{f(u)-u}-\frac{d x}{x}=0
$$

This is an equation with separable variables. Finding the general solution (integral) of equation (3), we obtain the general solution (integral) of the given equation (1) by replacing \( u \) with \( \frac{y}{x} \).