??Why don't you just integrate?
This does not look right to me: if you take differential of the right hand side it won't match the original. You seem to integrate over dx assuming y to be constant, and over dy assuming x to be constant. But x and y are interdependent.I would approach it naively and see where I get:
[imath] \int (y \log x)dx + \int(x - \log y) dy =x y + y (x \log(x) - x) + y - y \log(y) [/imath] according to WA. Assuming [imath]xy=yx[/imath] would further simplify it so that we could divide by [imath] y. [/imath]
A bit of a physicist's approach and more context would be helpful.
Hello! I was unable to solve. But the worksheet I downloaded says the solution to the D.E.This does not look right to me: if you take differential of the right hand side it won't match the original. You seem to integrate over dx assuming y to be constant, and over dy assuming x to be constant. But x and y are interdependent.
If you have a complete and verified solution you can post it tomorrow after 1 week since the original post.
Could you post pictures or screenshots of both the problem statement and the answer?Hello! I was unable to solve. But the worksheet I downloaded says the solution to the D.E.
(y log x) dx + (x - log y) dy = 0 is
(x log x) - 1/2(log y)^2 = C