WW2 - Problem 1

[Diane Sicsic]

Hey! I'm having difficulties for the Problem 1 of WW2.
The answer I found is Phi = yx + x + ln[(xy)^-2], but it seems like it's not the correct one.
Anyone knows where I got wrong/how I can find the answer?
Thankss

[Jasonisgood]

How did you arrive at this answer?

I got phi = xy + x + y - 5ln(y) - 5ln(x)

I got this by integrating 1 - 5/x + y with respect to x and then differentiating this result with respect to y and setting it equal to 1 - 5/y + x.

[Victor Ivrii]

Diane is correct and Jason is almost correct (wrong coefficient). Using MathJax http://www.math.toronto.edu/courses/mat244h1/20131/about_mathjax.html one can write an answer much nicer
\begin{equation*}
xy+x+y-2\ln (xy)=C.
\end{equation*}
Input is

Code: [Select]

\begin{equation*}
xy+x+y-2\ln (xy)=C.
\end{equation*}

[Arash Jalili]

Wouldn't the correct answer be yx + x +y -4ln|xy|?

[Victor Ivrii]

Wouldn't the correct answer be yx + x +y -4ln|xy|?

Nope. I already confirmed that Diane is correct