# Math Help - A tricky equation of the first order

1. ## A tricky equation of the first order

My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is:

$xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$

So far I've figured that I should substitute $ln(\frac{x}{y})$ for $u(x)$, thus giving me $y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas?

2. Originally Posted by getsallad
My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is:

$xy'(ln(\frac{x}{y}) +1) = y(ln(\frac{x}{y}) - 1)$

So far I've figured that I should substitute $ln(\frac{x}{y})$ for $u(x)$, thus giving me $y = \frac{x}{e^{u(x)}}$, but from there I'm kind of lost. Any ideas?
I haven't done anything with it, but note that if you divide both sides by x you have an equation in y/x...

-Dan

3. I'm sorry, my knowledge of differential equations in itself is quite shaky, but how would that help me? If I get what you're saying, is it that I get
$y'(ln(\frac{x}{y}) + 1) = \frac{x}{y}(ln(\frac{x}{y}) - 1)$?
If that's the case, are you suggesting I could substitute u(x) for $\frac{x}{y}$ instead?

4. The substitution $u = x/y$ does render the equation separable (the original DE is homogeneous). The problem is, the resulting integral in $u$ is horrendous. I suppose you could say, at that point, that you've "reduced the DE to quadratures", but if your professor is looking for a closed-form solution, that won't do.

The substitution $u=\ln(x/y)$ is much better. You have to translate the DE over to the $u$ domain:

$u'=\dfrac{y}{x}\cdot\dfrac{y-xy'}{y^{2}}=\dfrac{y-xy'}{xy}.$

Solve this for $y',$ and plug everything into the DE. A few things should cancel, leaving you with a much nicer separable equation.

5. Very cute problem, by the way! Thanks for posting!

6. Thanks for the help. I'm trying it out right now, doing all the steps myself to see if I get it.

7. I just did the integral using Akbeet's method as well as using the homogeneous method and I'd rate them to be about the same level of difficulty.

For the "connoisseurs" out there, the homogeneous solution includes a nice little integral:
$\displaystyle \int \frac{ln(u)~du}{u}$ which I haven't seen done in a long time. (it's not particularly hard to do, just a nice little piece of work.)

-Dan

8. OK, so I've gotten somewhere. Now I'm stuck on the actual separable differential equation.

I have

$\frac{2}{u + 1} = xu'$

which leads me to

$\frac{2}{u + 1} du = x dx$

but that approach seems to give me a completely different solution from what Wolfram Alpha gives me when I feed it the equation. So I figure I must be doing something wrong, since Wolfram Alpha's solution also seems more in tune with the final solution of the problem.

9. Your first equation is correct. Your second is incorrect.

10. OK, so I've hit one last roadblock and it's mighty frustrating. In the end I get

$\frac{u^2}{2} + u = 2ln(x) + c$

and Wolfram Alpha gives the solution

$\frac{1}{2}*ln^2(\frac{x}{y}) - ln(\frac{x}{y}) = 2ln(x) + c$

Which is basically the same thing with - instead of +. Should I show you step by step what I am doing? This is driving me nuts.

11. No, WolframAlpha is not giving you that solution, it's giving you this solution. Notice that the x and y in the WolframAlpha solution are flipped from what you have. Because the left-most logarithm is squared, the change in sign goes unnoticed. That, of course, doesn't happen with the second term.

I'm saying your solution is correct.

12. Incidentally, you could, if you wanted to, use the quadratic formula on log(x/y) and then solve for y. You get a multi-valued "function" then, but it is an explicit formula for y, which is nice.

13. Ah, can't believe I missed that. Thanks a million for the help on this one! I'll be sure to include the explicit formula for y just to be on the safe side. You've been most helpful! =)

14. You're very welcome! Have a good one!