m |
|||
| Line 1: | Line 1: | ||
[[MA366|Back to the MA366 course wiki]] | [[MA366|Back to the MA366 course wiki]] | ||
| − | == Homogeneous Equations == | + | == Homogeneous (first order) Equations == |
| − | ( | + | Given a differential equation of the form |
| + | |||
| + | <math>\frac{dy}{dx}=f(x,y)</math> | ||
| + | |||
| + | If ''f'' can be written ''solely'' as a function of the ratio y/x, then your equation is said to be '''homogeneous'''. Take, for example, the differential equation: | ||
| + | |||
| + | <math>\frac{dy}{dx}=\frac{y-4x}{x-y}\qquad \qquad (*)</math> | ||
| + | |||
| + | To change ''f'' into a function of ''just'' y/x, we do a little trick: use the substitution | ||
| + | <math>v=\frac{y}{x}</math> | ||
| + | |||
| + | Ideally, then, we want ''f'' (the right side of the equation) to be a function of '''v'''. If the equation is homogeneous, then we can accomplish this by noting that | ||
| + | |||
| + | <math>y=vx</math> | ||
| + | |||
| + | and substituting '''vx''' for '''y''' in (*). We obtain: | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | \frac{dy}{dx}&=\frac{vx-4x}{x-vx}\\ | ||
| + | &=\frac{x(v-4)}{x(1-v)}\\ | ||
| + | \frac{dy}{dx}&=\frac{v-4}{1-v}\qquad \qquad (**) | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | This would be a nice equation if it weren't for the dy/dx on the left side of the equation. In order to do any kind of integration, we need to change that dy/dx into a dv/dx. How do we do this? Well, we take the relation we used a moment ago, that y=vx, and differentiate with respect to x. Here's what happens: | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | y&=vx\\ | ||
| + | \frac{d}{dx}(y)&=\frac{d}{dx}(vx)\\ | ||
| + | \frac{dy}{dx}&=\frac{dv}{dx}x+v | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Notice here we used the multiplication rule for derivatives. We now have a new identity for dy/dx. To use it, we just go back to (**) and replace dy/dx with this new expression. | ||
| + | |||
| + | <math> | ||
| + | \frac{dv}{dx}x+v=\frac{v-4}{1-v} | ||
| + | </math> | ||
[[MA366|Back to the MA366 course wiki]] | [[MA366|Back to the MA366 course wiki]] | ||
Revision as of 09:00, 6 October 2009
Homogeneous (first order) Equations
Given a differential equation of the form
$ \frac{dy}{dx}=f(x,y) $
If f can be written solely as a function of the ratio y/x, then your equation is said to be homogeneous. Take, for example, the differential equation:
$ \frac{dy}{dx}=\frac{y-4x}{x-y}\qquad \qquad (*) $
To change f into a function of just y/x, we do a little trick: use the substitution $ v=\frac{y}{x} $
Ideally, then, we want f (the right side of the equation) to be a function of v. If the equation is homogeneous, then we can accomplish this by noting that
$ y=vx $
and substituting vx for y in (*). We obtain:
$ \begin{align} \frac{dy}{dx}&=\frac{vx-4x}{x-vx}\\ &=\frac{x(v-4)}{x(1-v)}\\ \frac{dy}{dx}&=\frac{v-4}{1-v}\qquad \qquad (**) \end{align} $
This would be a nice equation if it weren't for the dy/dx on the left side of the equation. In order to do any kind of integration, we need to change that dy/dx into a dv/dx. How do we do this? Well, we take the relation we used a moment ago, that y=vx, and differentiate with respect to x. Here's what happens:
$ \begin{align} y&=vx\\ \frac{d}{dx}(y)&=\frac{d}{dx}(vx)\\ \frac{dy}{dx}&=\frac{dv}{dx}x+v \end{align} $
Notice here we used the multiplication rule for derivatives. We now have a new identity for dy/dx. To use it, we just go back to (**) and replace dy/dx with this new expression.
$ \frac{dv}{dx}x+v=\frac{v-4}{1-v} $
