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 September 8th, 2017, 03:20 AM #1 Member   Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0 differential equations Can someone please explain to me how to find differential equation of family of curves? Here is an example: y=Ce^((x^2+y^2)/2)
 September 8th, 2017, 04:21 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,168 Thanks: 1640 y = cg(x)$\implies$y' = cg'(x)$\implies$g(x)y' = cg(x)g'(x) = yg'(x), so the desired equation is g(x)y' - g'(x)y = 0. Sometimes, eliminating c is more difficult.
 September 8th, 2017, 04:26 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,237 Thanks: 884 Does y really occur in two places in that? A differential equation involves derivatives so we start by differentiating! $\displaystyle y(x)= Ce^{\frac{x^2+ y^2}{2}}$ so $\displaystyle y'= 2Cxe^{\frac{x^2+ y^2}{2}}+ 2Cye^{\frac{x^2+ y^2}{2}}y'$ $\displaystyle y'= 2C\left(x+ yy'\right)e^{\frac{x^2+ y^2}{2}}$ To get an equation that gives that whole family of curves we need to eliminate "C" which we can do by dividing that by the original equation: $\displaystyle \frac{y'}{y}= 2\left(x+ yy'\right)$ That answer can, of course, be rewritten in a number of different ways: perhaps as $\displaystyle y'= 2xy+ 2y^2y'$ and then $\displaystyle (1- 2y^2)y'= 2xy$ so that $\displaystyle y'= \frac{2xy}{1- 2y^2}$
September 8th, 2017, 05:32 AM   #4
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 Originally Posted by Country Boy Does y really occur in two places in that? A differential equation involves derivatives so we start by differentiating! $\displaystyle y(x)= Ce^{\frac{x^2+ y^2}{2}}$ so $\displaystyle y'= 2Cxe^{\frac{x^2+ y^2}{2}}+ 2Cye^{\frac{x^2+ y^2}{2}}y'$ $\displaystyle y'= 2C\left(x+ yy'\right)e^{\frac{x^2+ y^2}{2}}$ To get an equation that gives that whole family of curves we need to eliminate "C" which we can do by dividing that by the original equation: $\displaystyle \frac{y'}{y}= 2\left(x+ yy'\right)$ That answer can, of course, be rewritten in a number of different ways: perhaps as $\displaystyle y'= 2xy+ 2y^2y'$ and then $\displaystyle (1- 2y^2)y'= 2xy$ so that $\displaystyle y'= \frac{2xy}{1- 2y^2}$
Their answer is : -1/y' = xy/(1-y^2)

 September 8th, 2017, 08:12 AM #5 Global Moderator   Joined: Dec 2006 Posts: 19,168 Thanks: 1640 Their answer is incorrect, and the answers that Country Boy gave are also incorrect. When replying, please don't quote the whole previous post unnecessarily.

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