Orthogonal trajectories elimination of constant

Question

$$x^2+y^2-2ay=0$$

I just learnt orthogonal trajectories today. So basically we need to eliminate the constants and solve $dy/dx=f(x,y)$ right? How do I eliminate $2ay$ in this question.I tried a bunch of stuff but couldn't get it.

Answer 1

This is how to get the differential equation of the family of curves $$ x^2+y^2-2ay=0 \ \implies\ a=\frac{x^2+y^2}{2y}\\ $$ by diffrentiating the given equation we get $$ 2x +2yy' -2ay' = 0 \Rightarrow y' = \frac{x}{a-y}\\ $$ substituting the value of $a$ in the expression for $y'$ we get $$ y' = \frac{2yx}{x^2-y^2}\\ $$

Answer 2

The equation is $x^2+ y^2-2ay= 0$. Differentiate with respect to x: $2x+ 2yy'-2ay'= 0$. $2(y- a)y'= 2x$, $y'= \frac{x}{y- a}$. Now solve the original equation for a: $2ay= x^2+ y^2$ so $a= \frac{x^2+ y^2}{2y}$. Replace a in the differential equation with that: $y'= \frac{x}{y+ \frac{x^2+ y^2}{2y}}= \frac{2xy}{2y^2+ x^2+ y^2}= \frac{2xy}{x^2+ 3y^2}$.

Answer 3

This isn't an easy question for somebody like you who is beginner in these questions.

let us begin by a graphical understanding of what we are doing.

Here is the first family of circles (in red on the figure below) with equations

$$x^2+y^2-2ay=0 \ \ \ \iff \ \ \ x^2+(y-a)^2=a^2 \tag{1}$$

therefore with center in $(0,a)$ and radius $a$.

The orthogonal family, i.e., the family of curves that are orthogonal in each intersection point to the corresponding member of the other family is still a family of circles (in blue on the figure) that, moreover, are obtained from the first family by a $\pi/2$ rotation, i.e., with equations:

$$(x-b)^2+y^2=b^2 \tag{2}$$

Let us check that we can recover (2).

Let us start from (1). It is not difficult to obtain (as in the answer by Physor) the differential equation :

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

Now, the classical trick: replace in (3), $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$ giving the differential equation for the orthogonal trajectories:

$$\dfrac{dy}{dx}=\dfrac{y^2-x^2}{2xy} \ \ \ \iff \ \ \ y'=\dfrac{(\tfrac{y}{x})^2-1}{2(\tfrac{y}{x})}\tag{4}$$

which is a so-called homogeneous differential equation in $v=\tfrac{y}{x}$.

$$2(v+xv')=v-\dfrac{1}{v}\tag{5}$$

Its solution requires different steps.

In a first step, (5) is transformed into:

$$2x \dfrac{dv}{dx}=-\dfrac{1+v^2}{v}$$

then:

$$\frac{dx}{x}=-\dfrac{2vdv}{1+v^2}$$

by integration,

$$\log|x|=-\log|1+v^2|+\log |K|, \ \ (K \in \mathbb{R})$$

Therefore

$$x=\dfrac{K}{1+v^2}.$$

Remembering that $v=\dfrac{y}{x}$, we get an equation which is exactly of the desired form (2).

Remark: I just realized that 4 years ago, I had given a partly similar answer to a partly similar question here.