# First order ODE, orthogonal trajectories

• Panphobia
In summary, you can find the new dy/dx if you want orthogonal trajectories to the ellipse by solving the differential equation and ifentifying the orthogonal trajectories on the ellipse.

#### 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?

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?

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?

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

Panphobia said:
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.##

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

Panphobia said:
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.