# Help solve (y log x)dx + (x - log y) dy = 0: I have used integrating factor log(y/x)

#### MarcusG

##### New member
Having difficult time with non-exact type DE. I have used integrating factor: log(y/x). The problem is: (y log x)dx + (x - log y) dy = 0. Re-arranged to
dx/dy + x(1/y log x) =y log(y/x). Multiply by log(y/x) and keep getting incorrect solution.

SymPy and I both failed to solve this ODE. But replacing '-' with '*' (i.e. [imath]y \log x dx + x \log y dy = 0[/imath]), makes an easily solvable equation. Are you sure your post does not have a typo?

Why don't you just integrate?

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.

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.
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.

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.
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

You have xdy in your differential equation. If I differentiate your solution, I get ydx.
$\begin{gathered} C=(x \log x) - \dfrac{1}{2}(\log y)^2 \\ -2C+2x\log x=(\log y)^2\\ 2(\log y)\dfrac{y'}{y}=2\log(x)+\dfrac{2x}{x}\\ (\log y)dy=(y+y\log x)dx\\ 0=(y\log x)dx + ydx - (\log y )dy \end{gathered}$whereas you asked about [imath] 0=(y \log x) dx + xdy - (\log y) dy [/imath]

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
Could you post pictures or screenshots of both the problem statement and the answer?
Thank you.