
Orthogonal trajectories
Show that the orthogonal trajectories for the family of curves $\displaystyle y = ae^{x}$ (a constant) are a family of parabolas.
Attempt:
$\displaystyle y = ae^{x}$
$\displaystyle \frac{dy}{dx} = ae^{x}$
$\displaystyle \frac{dy}{dx} = \frac{1}{ae^{x}}$
$\displaystyle \int dy = \int \frac{1}{ae^{x}}dx$
$\displaystyle y = \frac{1}{a}\int e^{x}dx$
$\displaystyle y = \frac{1}{a}e^{x} + C$
but the correct answer was:
$\displaystyle \frac{1}{2}y^2 = x +C$
I cant seem to see where I went wrong though.

Welll, maybe you could just change this part:
$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{ae^{x}}$
$\displaystyle \dfrac{dy}{dx} = \dfrac{1}{y}$
$\displaystyle \int y \ dy = \int \ dx$
$\displaystyle \frac12 y^2 = x + C$
From what you did though.. maybe you can do this:
$\displaystyle y = \frac{1}{a}e^{x} + C$
$\displaystyle y = \frac{1}{ae^{x}} + C$
$\displaystyle y = \frac{1}{y} + C$
$\displaystyle y^2 = 1 + yC$
$\displaystyle y^2 = 1 + ae^{x}C$
But then, it gets complicated for me... the constant becomes another constant... (Thinking)