First order ODE, orthogonal trajectories

[Panphobia]

1. The problem statement, all variables and given/known da
$\frac{x^{2}}{k^{2}} + \frac{y^{2}}{\frac{k^{4}}{4}} = 1$ with k != 0
this can be simplified to
$x^{2} + 4y^{2} = k^{2}$
Find dy/dx implicitly, then find the new dy/dx if you want orthogonal trajectories to the ellipse. Lastly solve the differential equation and ifentify the orthogonal trajectories on the ellipse.

The Attempt at a Solution

$2x + 8y\frac{dy}{dx} = 0$

now I know that for it to be orthogonal you need -dx/dy, so rearranging for dy/dx would give

$\frac{dy}{dx} = \frac{-x}{4y}$

and -dx/dy is

$\frac{-dx}{dy} = \frac{4y}{x}$

but my question is wouldn't you be losing solutions for y=0 and x=0? How would you account for that in the solution?

 

[RUber]

You are not necessarily losing those solutions. They are still there. For example, if $-\frac{dx}{dy} = \frac{4y}{x}$, then what does that mean when x = 0? Normally an ellipse (centered at the origin) might have a vertical tangent line as it crosses the x axis, how is that slope represented? What is perpendicular to that?

 

[Panphobia]

Yea I get what you mean, I solved the differential equation and got $y = Cx^{4}$ and one of the questions is to draw four members of each family of curves, so it should be pretty simple with the ellipse, but then with the $y = Cx^{4}$, is it still possible for me to draw the solution through x=0?

 

[HallsofIvy]

An obvious point is that taking C= 0 gives y= 0 as a solution- a horizontal line through (0, 0).

 

[Ray Vickson]

1. The problem statement, all variables and given/known da
$\frac{x^{2}}{k^{2}} + \frac{y^{2}}{\frac{k^{4}}{4}} = 1$ with k != 0
this can be simplified to
$x^{2} + 4y^{2} = k^{2}$
Find dy/dx implicitly, then find the new dy/dx if you want orthogonal trajectories to the ellipse. Lastly solve the differential equation and ifentify the orthogonal trajectories on the ellipse.

The Attempt at a Solution

$2x + 8y\frac{dy}{dx} = 0$

now I know that for it to be orthogonal you need -dx/dy, so rearranging for dy/dx would give

$\frac{dy}{dx} = \frac{-x}{4y}$

and -dx/dy is

$\frac{-dx}{dy} = \frac{4y}{x}$

but my question is wouldn't you be losing solutions for y=0 and x=0? How would you account for that in the solution?

You have the wrong differential equation: the solutions of $2x + 8y\frac{dy}{dx} = 0$ just give you the ellipse itself, because when $2x dx + 8y dy = 0$ that means that the points $(x,y)$ and $(x + dx, y + dy)$ lie in the ellipse; in other words, $(dx,dy)$ lies in the tangent line to the ellipse.

You want your $(dx,dy)$ to be perpendicular to the tangent, not parallel to it. Thus, you want $(dx,dy)$ to point along the gradient of the function $f = x^2 + 4 y^2$. Thus, you need
$dx = 2x \, dt, \;\; dy = 8y \, dt$, so ##dy/dx = 4y/x.##

 

[Panphobia]

I did solve the differential equation for $dy/dx = 4y/x$ it is $y = Cx^{4}$

 

[Ray Vickson]

I did solve the differential equation for $dy/dx = 4y/x$ it is $y = Cx^{4}$

Nowhere in your writeup in post #1 did you write that DE and that solution. I looked and looked, but could not find it.