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 Differential Equations Ordinary and Partial Differential Equations Math Forum

January 19th, 2015, 02:25 AM   #11
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Quote:
 Originally Posted by jiasyuen $\displaystyle (2x-y)dx+(x+y)dy=0$ How to continue?
Go back to the beginning and divide by $y^2$ properly. That means every term.

January 19th, 2015, 10:59 AM   #12
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Quote:
 Originally Posted by jiasyuen $\displaystyle (2x-y)dx+(x+y)dy=0$ How to continue?
My previous reply took this to be a new question. If you still wish to solve the original equation, you should apply v8archie's advice so that your division by y² is correct.

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