| Line 5: | Line 5: | ||
Given a differential equation of the form | Given a differential equation of the form | ||
| + | <center> | ||
<math>\frac{dy}{dx}=f(x,y)</math> | <math>\frac{dy}{dx}=f(x,y)</math> | ||
| + | </center> | ||
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: | 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: | ||
| + | <center> | ||
<math>\frac{dy}{dx}=\frac{y-4x}{x-y}\qquad \qquad (*)</math> | <math>\frac{dy}{dx}=\frac{y-4x}{x-y}\qquad \qquad (*)</math> | ||
| + | </center> | ||
To change ''f'' into a function of ''just'' y/x, we do a little trick: use the substitution | To change ''f'' into a function of ''just'' y/x, we do a little trick: use the substitution | ||
| + | |||
| + | <center> | ||
<math>v=\frac{y}{x}</math> | <math>v=\frac{y}{x}</math> | ||
| + | </center> | ||
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 | 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 | ||
| + | <center> | ||
<math>y=vx</math> | <math>y=vx</math> | ||
| + | </center> | ||
and substituting '''vx''' for '''y''' in (*). We obtain: | and substituting '''vx''' for '''y''' in (*). We obtain: | ||
| + | <center> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| Line 27: | Line 37: | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
| + | </center> | ||
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: | 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: | ||
| + | <center> | ||
<math> | <math> | ||
\begin{align} | \begin{align} | ||
| Line 37: | Line 49: | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
| + | </center> | ||
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. | 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. | ||
| + | <center> | ||
<math> | <math> | ||
\frac{dv}{dx}x+v=\frac{v-4}{1-v} | \frac{dv}{dx}x+v=\frac{v-4}{1-v} | ||
</math> | </math> | ||
| + | </center> | ||
| + | |||
| + | Until now, you've probably been wondering what the point of all this substitution and rearrangement is. Here is the payoff: this new differential equation involving v and x is ''separable'', and we know how to solve separable equations. Get all of the v's and dv's on one side of the equation, and all the x's and dx's on the other. | ||
| + | |||
| + | <center> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | \frac{dv}{dx}x+v&=\frac{v-4}{1-v}\\ | ||
| + | \frac{dv}{dx}x&=\frac{v-4}{1-v}-v\\ | ||
| + | \frac{dv}{dx}x&=\frac{v-4}{1-v}-\frac{v(1-v)}{1-v}\\ | ||
| + | \frac{dv}{dx}x&=\frac{v-4}{1-v}-\frac{v-v^2}{1-v}\\ | ||
| + | \frac{dv}{dx}x&=\frac{v^2-4}{1-v}\\ | ||
| + | \frac{1-v}{v^2-4}dv&=\frac{1}{x}dx | ||
| + | \end{align} | ||
| + | </math> | ||
| + | </center> | ||
| + | |||
| + | So after some algebra, we have this nice equation ready for integration. You know what to do: integrate the left side with respect to v and the right side with respect to x. | ||
| + | |||
| + | <center> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | \int \frac{1-v}{v^2-4}dv&=\int \frac{1}{x}dx | ||
| + | \end{align} | ||
| + | </math> | ||
| + | </center> | ||
| + | |||
| + | Now, it turns out that we get: | ||
| + | |||
| + | <center> | ||
| + | <math> | ||
| + | \frac{-1}{4}ln(v-2)-\frac{3}{4}ln(v+2)=lnx+C | ||
| + | </math> | ||
| + | </center> | ||
| + | |||
| + | I used Mathematica for the integration; if you recall Calculus II well enough to do it on your own, then more power to you (I think you'd use partial fractions). But if you're like the rest of us, then you should ask your professor whether she expects you to be able to perform such integration analytically--some will be indifferent, and will allow you to use software or an integral table, but others might be less merciful. In any case, this is not a class on techniques of integration, so I am just going to do some hand waving here and skip over the issue. | ||
| + | |||
| + | The last step in solving this equation would be to change all of the v's back into y/x. Doing this gives us: | ||
| + | |||
| + | <center> | ||
| + | <math> | ||
| + | \frac{-1}{4}ln(\frac{y}{x}-2)-\frac{3}{4}ln(\frac{y}{x}+2)=lnx+C | ||
| + | </math> | ||
| + | </center> | ||
| + | |||
| + | And this is the solution. | ||
| + | |||
| + | == How do I know whether an equation is vulnerable to this approach? == | ||
| + | |||
| + | There is an easy way to determine whether this approach will work or not. If your equation is of the form | ||
| + | |||
| + | <center> | ||
| + | <math> | ||
| + | \frac{dy}{dx}=f(x,y) | ||
| + | </math> | ||
| + | </center> | ||
| + | |||
| + | and f(kx,ky)=f(x,y), then this approach will work. In other words, replace x with kx and y with ky in your differential equation. If the k's "cancel out", then your equation is homogeneous. In the example we did a moment ago, this would give us: | ||
| + | |||
| + | <center> | ||
| + | <math> | ||
| + | \begin{align} | ||
| + | f(kx,ky)&=\frac{ky-4kx}{kx-ky}\\ | ||
| + | &=\frac{k(y-4x)}{k(x-y)}\\ | ||
| + | &=\frac{y-4x}{x-y}\\ | ||
| + | &=f(x,y) | ||
| + | \end{align} | ||
| + | </math> | ||
| + | </center> | ||
| + | |||
| + | Since we have seen that f(kx,ky)=f(x,y), then we know the approach works. | ||
[[MA366|Back to the MA366 course wiki]] | [[MA366|Back to the MA366 course wiki]] | ||
Revision as of 09:46, 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} $
Until now, you've probably been wondering what the point of all this substitution and rearrangement is. Here is the payoff: this new differential equation involving v and x is separable, and we know how to solve separable equations. Get all of the v's and dv's on one side of the equation, and all the x's and dx's on the other.
$ \begin{align} \frac{dv}{dx}x+v&=\frac{v-4}{1-v}\\ \frac{dv}{dx}x&=\frac{v-4}{1-v}-v\\ \frac{dv}{dx}x&=\frac{v-4}{1-v}-\frac{v(1-v)}{1-v}\\ \frac{dv}{dx}x&=\frac{v-4}{1-v}-\frac{v-v^2}{1-v}\\ \frac{dv}{dx}x&=\frac{v^2-4}{1-v}\\ \frac{1-v}{v^2-4}dv&=\frac{1}{x}dx \end{align} $
So after some algebra, we have this nice equation ready for integration. You know what to do: integrate the left side with respect to v and the right side with respect to x.
$ \begin{align} \int \frac{1-v}{v^2-4}dv&=\int \frac{1}{x}dx \end{align} $
Now, it turns out that we get:
$ \frac{-1}{4}ln(v-2)-\frac{3}{4}ln(v+2)=lnx+C $
I used Mathematica for the integration; if you recall Calculus II well enough to do it on your own, then more power to you (I think you'd use partial fractions). But if you're like the rest of us, then you should ask your professor whether she expects you to be able to perform such integration analytically--some will be indifferent, and will allow you to use software or an integral table, but others might be less merciful. In any case, this is not a class on techniques of integration, so I am just going to do some hand waving here and skip over the issue.
The last step in solving this equation would be to change all of the v's back into y/x. Doing this gives us:
$ \frac{-1}{4}ln(\frac{y}{x}-2)-\frac{3}{4}ln(\frac{y}{x}+2)=lnx+C $
And this is the solution.
How do I know whether an equation is vulnerable to this approach?
There is an easy way to determine whether this approach will work or not. If your equation is of the form
$ \frac{dy}{dx}=f(x,y) $
and f(kx,ky)=f(x,y), then this approach will work. In other words, replace x with kx and y with ky in your differential equation. If the k's "cancel out", then your equation is homogeneous. In the example we did a moment ago, this would give us:
$ \begin{align} f(kx,ky)&=\frac{ky-4kx}{kx-ky}\\ &=\frac{k(y-4x)}{k(x-y)}\\ &=\frac{y-4x}{x-y}\\ &=f(x,y) \end{align} $
Since we have seen that f(kx,ky)=f(x,y), then we know the approach works.
