# Orthogonal Trajectory

1. Oct 14, 2006

### piano.lisa

I have to find the orthogonal trajectory to the family of circles tangents to the x-axis.

I eventually found that the derivative of the orthogonal trajectory would be:
$$\frac{dy}{dx}= \frac{-x^2 + y^2}{2xy}$$

But how do i find the equation for "y" explicitly?

(ps. This is not an assignment problem. I'm simply solving problems to get ready for my test)
ps#2: I recently discovered that this may be a solvable differential equation. Perhaps Bernoulli?

2. Oct 14, 2006

$$\frac{dy}{dx} + \frac{x^{2}+y^{2}}{2xy} = 0$$

So $$\frac{dy}{dx} + \frac{x^{2}}{2xy} + \frac{y^{2}}{2xy} = 0$$

$$\frac{dy}{dx} + \frac{x}{2y} = -\frac{y}{2x}$$ which is the form: $$\frac{dy}{dx} + P(x)y = Q(x)y$$ (i.e. the Bernouli equation is linear)

Last edited: Oct 14, 2006
3. Oct 15, 2006

### HallsofIvy

Staff Emeritus
Do you mean with center on the y-axis also? You need another condition to have a one parameter family.

Yes, that is correct for the orthogonal trajectories to the family of circles with center on the y-axis tangent to the x-axis.

This is a "homogeneous" equation (note that "homogeneous" for first order equations is not the same as "homogeneous" for linear equations!): If you divide both numerator and denominator by x2 you get
$$\frac{dy}{dx}= \frac{\frac{y^2}{x^2}- 1}{2\frac{y}{x}}[/itex] Now let v= y/x so that y= xv and y'= xv'+ v. Your equation becomes [tex]x\frac{dv}{dx}+ v= \frac{v^2-1}{v^2}= 1- \frac{1}{v^2}$$ That's a separable equation.

(Courtigrad's suggestion that this is linear is completely wrong: 1/y is not linear.)

4. Oct 15, 2006

was I correct on what I did before that then? It is Bernoulli right? And its courtrigrad, not courtigrad.

Last edited: Oct 15, 2006
5. Oct 15, 2006

$$\frac{dy}{dx} - \frac{x^{2}-y^{2}}{2xy} = 0$$