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 November 12th, 2017, 10:59 PM #1 Newbie   Joined: Jan 2014 Posts: 18 Thanks: 0 Constant solution I want to find constant solutions of the differential equation $\frac{dy}{dx}=\frac{y^3}{x^2}-2\frac{y}{x}$. I set $y=c$, where $c$ is a constant. Then $\frac{dy}{dx}=0$ and the differential equation becomes $0=\frac{c^3}{x^2}-2\frac{c}{x}$ $\frac{c}{x}(\frac{c^2}{x}-2)=0$ $c/x=0$ or $c^2/x-2=0$ $c=0$ or $c^2=2x$ How to explain in words the equation $c^2=2x$ does not give any constant solution of the differential equation?
 November 13th, 2017, 12:06 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,720 Thanks: 598 Math Focus: Yet to find out. What do you mean you want to find a constant solution...? Are you trying to find a particular solution?
 November 13th, 2017, 06:25 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra Duplicate post.
 November 13th, 2017, 10:21 AM #4 Global Moderator   Joined: Dec 2006 Posts: 19,713 Thanks: 1806 Differentiation with respect to $x$ requires that $x$ isn't a constant, so c² = 2$x$ isn't possible. Thanks from topsquark
 November 20th, 2017, 01:16 AM #5 Senior Member   Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27 Only if $\displaystyle \exists \frac{y_1}{y_2}=C$ Example : equation $\displaystyle y'=(y-c_1 )$ has constant solution $\displaystyle y=c_1$ Last edited by idontknow; November 20th, 2017 at 01:46 AM.
 November 20th, 2017, 08:14 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,713 Thanks: 1806 What did you mean by $\displaystyle \exists \frac{y_1}{y_2} = C$?

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