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

[MarcusG]

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.

 

[blamocur]

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

 

[fresh_42]

Why don't you just integrate?

 

[blamocur]

Why don't you just integrate?

??

 

[fresh_42]

I would approach it naively and see where I get:
$\int (y \log x)dx + \int(x - \log y) dy =x y + y (x \log(x) - x) + y - y \log(y)$ according to WA. Assuming $xy=yx$ would further simplify it so that we could divide by $y.$

A bit of a physicist's approach and more context would be helpful.

 

[blamocur]

I would approach it naively and see where I get:
$\int (y \log x)dx + \int(x - \log y) dy =x y + y (x \log(x) - x) + y - y \log(y)$ according to WA. Assuming $xy=yx$ would further simplify it so that we could divide by $y.$

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.

 

[MarcusG]

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

 

[fresh_42]

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 $0=(y \log x) dx + xdy - (\log y) dy$

 

[blamocur]

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.