### Author Topic: WW2 - Problem 1  (Read 914 times)

#### Diane Sicsic

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##### WW2 - Problem 1
« on: September 17, 2014, 07:50:59 PM »
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

#### Jason Lequyer

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##### Re: WW2 - Problem 1
« Reply #1 on: September 17, 2014, 09:07:33 PM »
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.
« Last Edit: September 17, 2014, 09:09:38 PM by Jason Lequyer »

#### Victor Ivrii

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##### Re: WW2 - Problem 1
« Reply #2 on: September 18, 2014, 05:55:28 AM »
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

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##### Re: WW2 - Problem 1
« Reply #3 on: September 23, 2014, 11:33:14 AM »
Wouldn't the correct answer be yx + x +y -4ln|xy|?