First Order Differential Equations with Homogeneous Coefficients

So here is what I have done. Homogeneous has several meanings, but when I use it in this post, I will mean homogeneous functions. We only care about functions of two variables, so I will also only talk about them. A function $F(x,y)$ is called homogeneous of order $n$ if $F(tx,ty)=t^nF(x,y)$ for some $n$ . Also, if we “pull” an $x$ out, we get $F(x,y)=x^nF(1,\frac{x}{y})=x^nG(\frac{y}{x})$ , which is why the definition is often stated that the function can be written as a function of $\frac{y}{x}$ . Of course, we could do the same with $y$ and get a function of $\frac{x}{y}$ . Here is an example of a homogeneous function: $x^2+y\sqrt{x^2+y^2}\sin{\frac{x}{y}}$ . You can see that $(xt)^2+yt\sqrt{(xt)^2+(yt)^2}\sin{\frac{xt}{yt}}=t^2\left(x^2+y\sqrt{x^2+y^2}\sin{\frac{x}{y}}\right)$ .

Now, if the coefficients $P$ and $Q$ in the first order differential equation $P(x,y)dx+Q(x,y)dy=0$ are both homogeneous functions of the same order, then the substitution $x=uy$ or $y=ux$ will make the equation separable. I will show how to do it for $y=ux$ . We have $dy=xdy+udx$ by the product rule. We also have $P(x,y)=P(x,xu)$ and $Q(x,y)=Q(x,xu)$ . Because these functions are homogeneous, we can “pull out” an $x^n$ from each, giving us $x^nP(1,u)dx+x^nQ(1,u)(xdu+udx)=0$ . Because we made the substitution $u=\frac{y}{x}$ , we already have had to assume $x\neq0$ . Thus, we can divide the whole thing by $x^n$ . Doing this, and expanding the remaining terms, we get $P(1,u)dx+xQ(1,u)du+uQ(1,u)dx=0$ . Refactoring the $dx$ and $dy$ terms, we get $(P(1,u)+uQ(1,u))dx=-xQ(1,u)du$ . The equation is separable! Separating variables and integrating, we get $\int{\frac{dx}{x}}=\int{\frac{-Q(1,u)du}{P(1,u)+uQ(1,u)}}+C$ . If we had used $x=uy$ instead, we would have gotten $\int{\frac{dy}{y}}=\int{\frac{-P(u,1)du}{uP(u,1)+Q(u,1)}}+C$ . This is exactly how I was able to solve these equations in SymPy. Each homogeneous equation has two possible integrals, and often the right hand side of one equation is a much harder integral than the right hand side of the other. Therefore, I did them both, and applied a little heuristic on which one to return. It prefers expressions that can be solved for y (f(x) in the case of SymPy), expressions that have evaluated integrals, and if neither of them are solvable, the shortest one, which tends to be the simplest.

Here is an example from my text book. $2xydx+(x^2+y^2)dy=0$ . Astute readers of this blog may have noticed that this equation is exact. It is easier to solve that way, but we will try using the methods outlined above. First, we notice that the coeficients of $dx$ and $dy$ are both homogeneous of order 2, so we can make the substitution $x=uy$ or $u=yx$ and the equation would become seperable. If I were doing this by hand or for a homework assignment, I would make one of those substitutions and work it through, but we have the exact form above, so let’s use it just like SymPy would. $P(x,y)=2xy$ and $Q(x,y)=x^2+y^2$ . So we get $\int{\frac{dx}{x}}=\int{\frac{-(1+u^2)du}{2u+u(1+u^2)}}+C$ with $u=\frac{y}{x}$ and $\int{\frac{dy}{y}}=\int{\frac{-2udu}{u(2u)+(1+u^2)}}+C$ with $u=\frac{x}{y}$ (here and for the remainder of this derivation, the arbitrary constants in the two equations do not necessarily equal each other until the last step). Both integrals can be solved with a substitution of the denominator, giving, with $u$ back substituted, $\ln{x}=\frac{-\ln{(\frac{y^3}{x^3}+\frac{3y}{x})}}{3}+C$ and $\ln{y}=\frac{-\ln{(\frac{3x^3}{y^3}+1)}}{3}+C$ . If we make $C=ln(C_1)$ , and pull the constants in front of the logs on the right hand sides in as powers, we can combine everything into logarithms. After doing that, we then take the antilog of both sides and get $x=C_1{(\frac{y^3}{x^3}+\frac{3y}{x})}^{-\frac{1}{3}}$ and $y=C_1{(\frac{3x^3}{y^3}+1)}^{-\frac{1}{3}}$ . We then raise the whole equation to the $-3$ power and make $C_1^{-3}=A$ . We get
$\frac{1}{x^3}=A(\frac{y^3}{x^3}+3\frac{y}{x})$
$\frac{1}{x^3}=A\frac{y^3+3x^2y}{x^3}$
$K=y^3+3x^2y$
and
$\frac{1}{y^3}=A(\frac{3x^2}{y^2}+1)$
$\frac{1}{y}=A(3x^2y+y^2)$
$K=y^3+3x^2y$ ,
where $K=\frac{1}{A}$

Verify as an exercise that you get the same solution by the exact differential equation methods (or don’t… you know, depending on which you’d rather do).

Of course, if that was all there was to it, I would have finished in a day. There was more. First, I had to write a function that determines if an expression is homogeneous and what order if it is. That was about 150 lines of code right there plus tests (150 lines is a lot in Python).

I also wanted it to recognize that $\ln{x}-\ln{y}$ is homogeneous because $\ln{x}-\ln{y}=\ln{\frac{x}{y}}$ . SymPy was incapable of combining logarithms like this, so I had to write a logcombine function, which was another 100 lines of code plus tests. This came in handy, because I was also able to use it in the part above where I combined the logarithms of my answer to eliminate all of them.

Lastly, you may have noticed that my math looks much nicer in this post. It turns out that WordPress supports math using $latex <math>$. Unfortunately, I did not discover this until I had already uploaded images of almost all the math expressions in the post, so I had to go back and fix them.

Also, I was having a problem where the math wouldn’t render, but it turns out that this was an issue with me copying and pasting “$latex” and the expressions into the WordPress rich text editor. I changed my editor settings to plain text. I’m sorry, but web rich text editors do not work. And I’ve grown accustomed to seeing the source of stuff that I write having done everything in $\LaTeX$ for a year and having edited Wikipedia for a while.

I wish there were a free blogging site like WordPress or Blogger that used MediaWiki or $\LaTeX$ as its rendering engine. It would look nice, the formatting would be a standard thing that isn’t confusing html, and $\LaTeX$ math would come out much nicer.

This entry was posted on Sunday, May 31st, 2009 at 12:10 am and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

4 Responses to First Order Differential Equations with Homogeneous Coefficients

Meeting Ondrej in Los Alamos « GSoC 2009: SymPy ODEs says:

July 13, 2009 at 2:22 am

[…] a working Constant class, I can solve homogeneous differential equations (not to be confused with first order differential equations with homogeneous coefficients) with one case (there are several cases depending on whether the roots of the so called […]

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Testing implicit solutions to ODEs « GSoC 2009: SymPy ODEs says:

August 12, 2009 at 9:27 pm

[…] really simple ones. I ended up XFAILing most of my implicit checksol tests. I think every single homogeneous coefficients had an implicit solution, and none of them were working with […]

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2010 in review « Aaron Meurer's SymPy Blog says:

January 2, 2011 at 11:07 pm

[…] First Order Differential Equations with Homogeneous Coefficients May 20092 comments […]

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Emory says:

October 1, 2012 at 3:16 am

Wow that was odd. I just wrote an extremely long comment
but after I clicked submit my comment didn’t show up. Grrrr… well I’m not writing all that
over again. Anyway, just wanted to say excellent blog!

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