Answer to Question #285292 in Differential Equations for Jericho

Question

orthogonal trajectories of y' = - (x/y)

Answer 1

$y' = -\dfrac{x}{y} \\ \text{To get the orthogonal trajectory}\\ \text{We replace} \, -\dfrac{x}{y} \, \text{by its negative reciprocal}\\ \text{Hence, we have}\\ y' = \dfrac{y}{x} \\ \text{We solve this D.E using separation of variables}\\ \text{Hence, we have}\\ \dfrac{dy}{dx} = \dfrac{y}{x} \\ \text{Integrating both sides, we have}\\ \int \dfrac{dy}{y} = \int \dfrac{dx}{x} \\ \text{Integrating, we obtain}\\ y = cx \\ \text{which is a family of the orthogonal trajectory}\\ \text{where C is an arbitrary constant}$