The answer is $-6x^2y^2 + x^4 + y^4 = \beta$, but I get $-3x^2y^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta$ and I can't seem to find my mistake. My work:

Differentiaing the given equation, we get $$3x^2y + x^3\frac{dy}{dx} - y^3 - 3y^2x\frac{dy}{dx} = 0. $$
So $$3x^2y - y^3 = (3y^2x -x^3)\frac{dy}{dx}, $$
and hence $$ \frac{dy}{dx} = \frac{3x^2y - y^3}{3y^2x - x^3}. $$
Now, the slope of the other family of curves will be the negative inverse, so for the orthogonal trajectory, we have
$$\frac{dy}{dx} = -\frac{3y^2x-x^3}{3x^2y-y^3}, $$
which we can write as $$-(3y^2x-x^3)dx = (3x^2y-y^3)dy. $$
Integration of the last differential equation yields $$-\frac{3}{2}y^2x^2 + \frac{x^4}{4} = \frac{3}{2}x^2y^2 - \frac{y^4}{4} + \beta, $$
$\beta$ being an arbitrary constant. This simplifies to $$-3y^2x^2 + \frac{x^4}{4} + \frac{y^4}{4} = \beta. $$ which is not correct.

Where am I going wrong?

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    $\begingroup$ You mishandled the equation $$-(3y^2x-x^3)dx=(3x^2y-y^3)dy.$$ You can't just integrate the two sides w.r.t. different variables (one side w.r.t. $x$, the other w.r.t $y$). To see what went wrong differentiate your answer implicitly. You will not get back that same equation. $\endgroup$ – Jyrki Lahtonen Oct 21 '13 at 10:36
  • $\begingroup$ Thanks, @JyrkiLahtonen, I figured. I now see that. I'm not too sure how to tackle this then. $\endgroup$ – David Oct 21 '13 at 10:38
  • $\begingroup$ David, I guess your integration is not correct. Try back-tracking. I mean differentiate the relation you've obtained and see if you get the differential equation for the orthogonal trajectory back. $\endgroup$ – Saaqib Mahmood Oct 21 '13 at 11:28
  • $\begingroup$ @SaaqibMahmuud, it seems I am indeed incorrect. I understand the solution when going through the complex plane, however, I am having trouble integrating $-(3y^2x-x^3)dx=(3x^2y-y^3)dy$ for this particular approach. $\endgroup$ – David Oct 21 '13 at 11:29

Since this is tagged complex analysis,

let $z = x+iy$ and let $f(x+iy) = x^3y - xy^3$. Expressed in terms of $z$, this is $f(z) = \frac 1 {16i}\left((z+\bar{z})^3(z - \bar z) + (z + \bar z)(z - \bar z)^3\right) = \frac 1{8i}(z^4 - \bar z^4) = \frac 1 4 Im(z^4) = Im(z^4/4)$.

So the curves $x^3y-xy^3 = A$ are the pullback of the horizontal lines by the holomorphic function $z \mapsto z^4/4$. Since holomorphic functions preserve angles (except when the derivative vanish, here it is at $z=0$), the curves that are always orthogonal to those, are the pullback of the vertical lines by $z \mapsto z^4/4$, i.e. the curves $Re(z^4/4) = B$.

And finally, $Re(z^4/4) = (x^4-6x^2y^2+y^4)/4$

At $z=0$, $f$ multiplies angles by $4$ and is $4$-to $1$ around $z=0$, so the two curves $Re(f(z)) = 0$ and $Im(f(z)) = 0$ both have a star-like singular point (an $8$-branched star), and the angle between the two stars is $\pi/8$

  • $\begingroup$ How did you come across $\frac{1}{16i}$? $\endgroup$ – David Oct 21 '13 at 10:47
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    $\begingroup$ I just replaced $x$ with $(z+\bar z)/2$ and $y$ with $(z- \bar z)/(2i)$ everywhere, then factored $16i$ out. $\endgroup$ – mercio Oct 21 '13 at 10:48
  • $\begingroup$ I notice that $z^4 - z^{*4} = 8ix^3y - 8ixy^3$, hence we cancel out the $8i$, but how did you get from there to $\frac{1}{4}Im(z^4)$? $\endgroup$ – David Oct 21 '13 at 11:03
  • $\begingroup$ $(z^4 - (\bar z)^4)/2i = (z^4 - \overline{z^4})/2i = Im(z^4)$ $\endgroup$ – mercio Oct 21 '13 at 11:11
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    $\begingroup$ if you meant $Re(f(z))$ and $Im(f(z))$, then yes. $\endgroup$ – mercio Oct 21 '13 at 11:16

I t seems that we have $$y'=\frac{3xy^2-x^3}{y^3-3x^2y}$$ instead, which is an homogenous OE. We have $$\frac{du}{\frac{3u^2-1}{u^3-3u^2}-u}=\frac{dx}{x}~\equiv~\frac{(-u^3+3u^2)du}{u^4+3u^2-3u^2+1}=\frac{dx}{x},~~~~~~u=y/x,x\neq 0$$ in which the right ODE is a separable one.

  • $\begingroup$ That is what I got, but my next step is incorrect. I'm not sure how to proceed from there. $\endgroup$ – David Oct 21 '13 at 11:10
  • $\begingroup$ Nice work, @Babak! +1 $\endgroup$ – Namaste Oct 21 '13 at 14:18

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