When I see your ODE this is what I try
so
so
so
Problem 2.10.6 Test the following DE for exactness. If it is not exact, try to find an integrating factor. After the equation is made exact, solve by looking for integrable combinations. If you cannot find any, use the standard exact method.
Answer.
We test for exactness. Is
No. We try to find an integrating factor.
1. Based on the symmetry of the problem, I'm going to guess a function . Then we have
The exactness condition implies
I'm not sure this is going anywhere, because we'd have to write all the non- stuff as a function of , which I don't think is going to happen.
2. Let's try again, this time with a function . The DE is
The exactness condition yields
Again, I'm not seeing how the non- stuff can be written as a function of .
3. Let's go for a simpler one: . Then we have
The exactness condition implies
Again, no prospect of writing it the way I want. By symmetry, I don't think a function of the type will work.
4. Let's try the type:
Exactness implies
Still not seeing how this could work.
Does anyone see a mistake in my computations anywhere? See any new ideas?
Thanks!
In my opinion - Yes! If your ODE is
and is not already in exact form them you seek an integrating factor such that
or
.
It is natural to assume certain form of this PDE
etc
but it really depends on what the PDE (1) looks like. Really to solve (1) is to solve the ODE itself.
Incidentally, I just found a way to encapsulate all five of Tenenbaum and Pollard's five guesses for integrating factors. They say that you should guess and You can unify all this in one place if you just guess and you get more possibilities besides.
What a time-saver! I don't think this method will work for the DE in this thread, but at the very least, I won't have to trudge through guessing all five of those options above.
It just occurred to me that there are possibilities that my little time-saver leaves out: for example, would be a possible integrating factor you might find if you tried but you couldn't find that with Still, for DE's where it looks like powers of the variables will work, this could still save some time.