# Orthogonal trajectories

find the orthogonal trajectories of the following

(a) $$x^2y=c_1$$

(b) $$x^2+c_{1}y^3=1$$

for part (a) I've found $$y=\frac{1}{2}\log{|x|} + C_2$$

for part (b) if i solve this integral this should be the O.T.

$$\frac{3}{2}\int{(\frac{1}{x^2}-1)}dx= \frac{y^2}{2}$$

is this correct?

this is very unusual that nobody has posted a response to this thread.

There was about three hours between your first and second posts! Do you think we have nothing better to do? (Especially on a Saturday- I spent yesterday hiking in the Shenandoah National Park!)

Differentiating x2y= c1 with respect to x gives
2xy+ x2y'= 0: we have removed the constant c1. The point is that any member of that family of trajectories must satisfy that differential equation which is the same as y'= -y/x. We know that the slopes, m1 and m2 of two lines y= m1x+ b1 and y= m1x+ b2 must satisfy m1m2= -1 or m2= -1/m1.
Any curve orthogonal to every member of the family given must have, at each point, y'= dy/dx= -1/(-y/x)= x/y. That separates as ydy= xdx. What is the general solution to that?

For (b) $$x^2+c_{1}y^3=1$$
it's a little harder to eliminate c1 but if you rewrite this as $$x^2y^{-3}+ c_1= y^{-3}$$, differentiating with respect to x will eliminate c1.

There was about three hours between your first and second posts! Do you think we have nothing better to do? (Especially on a Saturday- I spent yesterday hiking in the Shenandoah National Park!)

I appreciate all the help that I've received on this site. Phs. forums is a great resource. But i posted my initial post yesterday and my second post today. I just thought that it was unusual for a post not to be responded to in the amount of time that my post was up. I'm glad you had a good time hiking unlike myself who has spent all of today as well as yesterday getting prepared for the upcoming week.

Oops! My bad: there were 12+ about three hours between your first and second posts! Hope my response helped.