Answer to Question #152208 in Differential Equations for Wingelyn Dangarang

Question

Find the orthogonal trajectories of the given family of curves y= Ce^-mx; with m held fixed

Answer 1

$y = Ce^{-mx}$

First, this family of curves satisfies the differential equation

$\begin{aligned}y = Ce^{-mx} \quad \implies \quad y' = -mCe^{-mx} \end{aligned}$

Since $C = ye^{mx}$ we then have

$y' = -my$

Letting $f(x,y) = -my$, we know the orthogonal trajectories are the curves which satisfy a differential equation

$y' = \dfrac{-1}{f(x,y)} \implies y' = \dfrac{1}{my}$

Hence, the orthogonal trajectories that satisfy the differential equation are the curves

$\begin{aligned} y' = \frac{1}{my} \qquad \implies \quad & myy' = 1\\ \implies \quad & m\int{ydy} = dx\\ \implies \quad & \dfrac{my^2}{2} = x + C\\ \implies \quad & my^2 - 2x = 2C \end{aligned}$

where C is an arbitrary constant.

The orthogonal trajectories for the family of straight lines are parabolas (y² = ax + c)