Example 3. Form a differential equation for the family of lines that are at a distance of one unit from the origin.

 

The text provided is a task to form a differential equation for a family of lines that are equidistant from the origin, with the distance being one unit.
Solution. We will start from the normal equation of a straight line

$$
x \cos \alpha + y \sin \alpha - 1 = 0
$$

where $\alpha$ is a parameter.

Differentiating (6) with respect to $x$, we get $\cos \alpha + y' \sin \alpha = 0$, from which $y' = -\operatorname{ctg} \alpha$, hence,

$$
\sin \alpha = \frac{1}{\sqrt{1 + (y')^2}}, \quad \cos \alpha = -\frac{y'}{\sqrt{1 + (y')^2}}
$$

Substituting $\sin \alpha$ and $\cos \alpha$ into (6), we obtain

$$
\frac{-x y'}{\sqrt{1 + (y')^2}} + \frac{y}{\sqrt{1 + (y')^2}} - 1 = 0, \quad \text{or} \quad y = x y' + \sqrt{1 + (y')^2}
$$

## Problems for Independent Solution

Form the differential equations of the following families of curves:
238. $y = \frac{a}{x}$.
239. $x^2 - y^2 = a x$.
240. $y = a e^{z / a}$.

241. $y = C x - C - C^2$.
242. $y = e^x (a x + b)$.
243. $y^2 = 2 C x + C^2$.
244. $y = a x^2 + b x + c$.
245. $y = C_1 x + \frac{C_2}{x} + C_3$. 246. $(x - a)^2 + (y - b)^2 = 1$.
246. $y = C_1 e^x + C_2 e^{-x}$.
247. $y = a \sin (x + \alpha)$.

$2^{\circ}$. Problems on Trajectories. Let a family of plane curves be given by

$$
\Phi(x, y, a) = 0
$$

depending on one parameter $a$.

A curve that forms a constant angle $\alpha$ with the curve of the family passing through each of its points is called an isogonal trajectory of this family; if, in particular, $\alpha = \frac{\pi}{2}$, it is called an orthogonal trajectory.

Assuming the family (7) is given, we will seek its isogonal trajectories.

A. Orthogonal Trajectories. We form the differential equation of the given family of curves (see section $1^{\circ}$). Suppose it has the form

$$
F(x, y, y') = 0
$$

The differential equation of the orthogonal trajectories has the form

$$
F\left(x, y, -\frac{1}{y'}\right) = 0
$$

The general integral of this equation

$$
\Phi_1(x, y, C) = 0
$$

gives the family of orthogonal trajectories.

Let a family of plane curves be given by the equation in polar coordinates

$$
\Phi(\rho, \varphi, a) = 0
$$

where $a$ is a parameter. Eliminating the parameter $a$ from (8) and $\frac{\partial \Phi}{\partial \varphi} = 0$, we obtain the differential equation of the family (8): $F(\rho, \varphi, \rho') = 0$. Replacing $\rho'$ with $-\frac{\rho^2}{\rho'}$ in it, we obtain the differential equation of the family of orthogonal trajectories

$$
F\left(\rho, \varphi, -\frac{\rho^2}{\rho'}\right) = 0
$$

B. Isogonal Trajectories. Suppose the trajectories intersect the curves of the given family at an angle $\alpha$, where $\operatorname{tg} \alpha = k$. It can be shown that the differential equation of the isogonal trajectories has the form

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