Example 4.8. Consider the following example:

$$
\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}
$$
Solution. By making the substitution $y=x u$, we reduce the equation to the form:

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

From this,

$$
x \frac{d u}{d x}=\frac{1+u^{2}}{2 u}-u=\frac{1-u^{2}}{2 u}
$$

and therefore

$$
\begin{gathered}
\frac{d x}{x}=\frac{2 u d u}{1-u^{2}} ; \ln |x|=-\ln \left|1-u^{2}\right|+\ln |C| \\
x=\frac{C}{1-u^{2}} ; 1=\frac{C x}{x^{2}-y^{2}} ; x^{2}-y^{2}=C x \\
y^{2}=x^{2}-C x ; \quad y= \pm \sqrt{x^{2}-C x}
\end{gathered}
$$

Problem.

In the theory of cutting, the following problem arises: find a curve such that the tangent at each point forms a constant angle $\alpha$ with the radius vector of that point (Fig. 52). According to the problem, we have $\phi=\alpha+\psi$ and thus

$$
\operatorname{tg} \phi=\operatorname{tg}(\alpha+\psi)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \psi}{1-\operatorname{tg} \alpha \operatorname{tg} \psi}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-189.jpg?height=458&width=551&top_left_y=1332&top_left_x=509)

Fig. 52

But from Fig. 52, it is clear that $\operatorname{tg} \psi=\frac{y}{x}$, and as is known, $\operatorname{tg} \phi=y^{\prime}$. Therefore, equation (1) can be written as:

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

where for brevity, $k=\operatorname{tg} \alpha$. This is a homogeneous equation. Making the substitution $y=x u$, after simple transformations, we get

$$
x \frac{d u}{d x}=\frac{k\left(1+u^{2}\right)}{1-k u}
$$

from which we find

$$
\frac{1}{k} \operatorname{arctg} u-\frac{1}{2} \ln \left(1+u^{2}\right)+\ln C=\ln x
$$

Substituting $u$ with $\frac{y}{x}$, we get the equation:

$$
\frac{1}{k} \operatorname{arctg} \frac{y}{x}=\ln \frac{\sqrt{x^{2}+y^{2}}}{C}
$$

This equation is easier to write in polar coordinates, setting $x^{2}+y^{2}=r^{2}, \frac{y}{x}=\operatorname{tg} \phi$. We get

$$
r=C e^{\frac{\phi}{k}}
$$

This curve is called a logarithmic spiral.

## 4.4. Linear Equations

An equation of the form

$$
\frac{d y}{d x}+P(x) \cdot y=Q(x)
$$

is called a linear equation.

A linear equation can be reduced to two equations with separable variables by substituting the unknown function $y$ with the product of two auxiliary functions $u$ and $U$, i.e., $y=u U$.

Then

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

and equation (1) takes the form

$$
v \frac{d u}{d x}+u \frac{d v}{d x}+P(x) v=Q(x)
$$

Using the fact that one of the auxiliary variables, for example $v$, is chosen arbitrarily, we select it so that the expression in square brackets becomes zero, i.e., we take $U$ as one of the particular solutions $v=U(x)$ of the separable equation

$$
\frac{d v}{d x}+P(x) v=0
$$

Substituting the expression $v=U(x)$ into equation (2), we get an equation for the function $u$:

$$
u \frac{d u}{d x}=Q(x)
$$

This is also a separable equation. Finding the general solution of this equation $u=u(x, C)$, we get the general solution of equation (1):

$$
y=u(x, C) \cup(x).
$$