Ordinary Differential Equations by Tenenbaum & Pollard

ODE by Tenenbaum & Pollard, Exercise 2, Problem 12

Leave a comment

Problem

The following is a standard type of exercise in the calculus.

If x^3 y^3 - 3xy = 0, then 3x^2 + 3y^2 \frac{dy}{dx} - 3x \frac{dy}{dx} - 3y = 0. Therefore

\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}, y^2 \neq x

Explain by the use of Fig. 2.77 what this means geometrically.

Solution

As discussed in problem 11, the Folium of Descartes is not a function but an equation. We can graphically section off parts of the plot to transform the equation into a function.

The manipulation above is simply the derivative so geometrically \frac{y - x^2}{y^2 - x} is the slope of the equation.

Consider the case where x = 0.4.

There are three possible values of y for which the Folium relationship holds true.

  • y = -1.1212
  • y = 0.05346
  • y = 1.06774

These are the three functions.

Function 3
For the interval 0 < x < 2^(2/3), we choose the bottom path.

Here (x, y) = (0.4, -1.1212) and the slope is \frac{dy}{dx} = -1.49483.

 

Function 2
For the interval 0 < x < 2^(2/3), we choose the middle path.

Here (x, y) = (0.4, 0.05346) and the slope is \frac{dy}{dx} = 0.268265.

 

Function 1
For the interval 0 < x < 2^(2/3), we choose the top path.

Here (x, y) = (0.4, 1.06774) and the slope is \frac{dy}{dx} = 1.22656.

 

Leave a comment