# Isolating Constants in Orthogonal Trajectories

The original function is $y^2 = kx^3$, and I'm being asked to find the orthogonal trajectory.

$$2y\frac{dy}{dx} = 3kx^2$$ $$\frac{dy}{dx} = \frac{3kx^2}{2y}$$ $$\text{New }\frac{dy}{dx} = \frac{2y}{3kx^2}$$ $$\frac{1}{2y}\,dy = \frac{1}{3kx^2}\,dx$$ $$\frac{\ln|y|}{2} = \frac{-1}{3kx} + c$$ $$\ln|y| = \frac{-2}{3kx} + c$$ $$y = e^{\frac{-2}{3kx} + c}$$ $$y = e^ce^\frac{-2}{3kx}$$

I'm told that the final solution is $2x^2+3y^2 = d$, where d is a constant. However, I can't find a way to get there from where I currently am. I suppose you'd have to combine c and k in some way, but I don't see it. Any hints?

-

Note that

$\dfrac{dy}{dx} = \dfrac{3kx^2}{2y} = \dfrac{3kx^3}{2xy} = \dfrac{3y^2}{2xy} = \dfrac{3y}{2x}$

so the orthogonal curves satisfy

$\dfrac{dy}{dx} = -\dfrac{2x}{3y}.$

Rewrite this equation as

$2x\ dx + 3y\ dy = 0$

and integrate both sides.

-