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for chad dv/dp=-v/p
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dv/v=-dp/p integrating both sides ln|v|=-ln|p|+C
ln|v|+ln|p|=C
Uzma
yes?
is that all the full solution
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Uzma wer are you
http://s783.photobucket.com/albums/yy112/Chad_25_2009/?action=view¤t=DE.jpg
That could be enough or you can simplify it even more. \(\ln |v|=-\ln |p|+c \implies v=ke^{-p}\), where \(k\) is another constant \(k=e^c\).
I am a little busy now, I will help you when I get time.
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