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: 18,721 Thanks: 1536 
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: 2,976 Thanks: 807 
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  
Member Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0  Quote:
 
September 8th, 2017, 08:12 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,721 Thanks: 1536 
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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