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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?

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up vote 3 down vote accepted

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.

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