Example 7. $\rho^{2}=a \cos 2 \varphi$. Find the orthogonal trajectories of the family of lemniscates
Solution. We have

$\rho^{2}=a \cos 2 \varphi, \quad \rho \rho^{\prime}=-a \sin 2 \varphi$.

By eliminating the parameter $a$, we obtain the differential equation of this family of curves

$$
\rho^{\prime}=-\rho \operatorname{tg} 2 \varphi
$$

Replacing $\rho^{\prime}$ with $-\frac{\rho^{2}}{\rho^{\prime}}$, we find the differential equation of the family of orthogonal trajectories

$$
-\frac{\rho^{2}}{\rho^{\prime}}=-\rho \operatorname{tg} 2 \varphi
$$

from which $\frac{d \rho}{\rho} \operatorname{ctg} 2 \varphi d \varphi$. Integrating, we find the equation of the orthogonal trajectories

$$
\rho^{2}=C \sin 2 \varphi
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-058.jpg?height=560&width=505&top_left_y=379&top_left_x=749)

Fig. 18

The orthogonal trajectories of the family of lemniscates are lemniscates whose axis of symmetry forms an angle of $\pm 45^{\circ}$ with the polar axis (Fig. 18).

## Problems for Independent Solution

Find the orthogonal trajectories for the given families of curves:
249. $y^{2}+2 a x=0$,
250. $y=a x^{n}$,
251. $y=a e^{\sigma x}$, $a>0$.
$a$ - parameter.

$$
\sigma=\text { const }
$$

252. $\cos y=a e^{-x}$.
253. $x^{2}+\frac{1}{2} y^{2}=a^{2}$.
254. $x^{2}-y^{2}=a^{2}$.
255. $x^{k}+y^{k}=a^{k}$.
256. $x^{2}+y^{2}=2 a y$.
257. $x^{2}-\frac{1}{3} y^{2}=a^{2}$.
258. $\rho=a(1+\cos \varphi)$.
259. $y^{2}=4(x-a)$.

## § 11. Singular Solutions of Differential Equations

A solution $y=\varphi(x)$ of the differential equation

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

is called singular if at each of its points the property of uniqueness is violated, i.e., if through each of its points ( $x_{0}, y_{0}$ ) there passes another solution, which has the same tangent at the point ( $x_{0}, y_{0}$ ) as the solution $y=\varphi(x)$, but does not coincide with it in any arbitrarily small neighborhood of ( $x_{0}, y_{0}$ ). The graph of a singular solution will be called a singular integral curve of the equation (1). If the function $F\left(x, y, y^{\prime}\right)$ and its partial derivatives $\frac{\partial F}{\partial y}$ and $\frac{\partial F}{\partial y^{\prime}}$ are continuous in all arguments $x, y, y^{\prime}$, then any singular solution of equation (1) also satisfies the equation

$$
\frac{\partial F\left(x, y, y^{\prime}\right)}{\partial y^{\prime}}=0
$$

Thus, to find the singular solutions (1), one needs to eliminate $y^{\prime}$ from equations (1) and (2).

The equation obtained after eliminating $y^{\prime}$ from (1) and (2)

$$
\psi(x, y)=0
$$

is called the $p$-discriminant of the equation (1), and the curve defined by the equation (3) is called the $p$-discriminant curve (briefly PDC).

Often, the PDC splits into several branches. Then, it is necessary to determine whether each branch is a solution of equation (1), and if it is, whether it is a singular solution, i.e., whether the uniqueness is violated at each of its points.