Answer to Question #265821 in Differential Equations for kamote

Question

Find the orthogonal trajectory curve of all the lines passing through the origin that is y = C

Answer 1

Let the equation be $y = Cx,$ where $C$ is an arbitrary constant.

Differentiating with respect to $x,$ we get, $y'=C.$

Eliminate $C$

$$y'=\dfrac{y}{x}$$

which is the differential equation of a family of lines. 

Replace $y'$ with $(-1/y')$

$$-\dfrac{1}{y'}=\dfrac{y}{x}$$

$$ydy=-xdx$$

Integrate

$$\int ydy=-\int xdx$$

$$\dfrac{y^2}{2}=-\dfrac{x^2}{2}+c_1$$

$$x^2+y^2=2c_1$$

By replacing $2c_1$ with $R^2$ we see that the orthogonal trajectories for the family of straight lines are concentric circles 

$$x^2+y^2=R^2$$