My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is:
So far I've figured that I should substitute for , thus giving me , but from there I'm kind of lost. Any ideas?
My teacher gave me a "bonus" assignment the other day, and I've been racking my brain trying to solve it. Here it is:
So far I've figured that I should substitute for , thus giving me , but from there I'm kind of lost. Any ideas?
The substitution does render the equation separable (the original DE is homogeneous). The problem is, the resulting integral in 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 is much better. You have to translate the DE over to the domain:
Solve this for and plug everything into the DE. A few things should cancel, leaving you with a much nicer separable equation.
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:
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
OK, so I've gotten somewhere. Now I'm stuck on the actual separable differential equation.
I have
which leads me to
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.
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.