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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 November 25th, 2017, 11:58 PM #1 Newbie   Joined: Apr 2014 From: Canada Posts: 4 Thanks: 0 general solution $y=0, x<0$ and $y=0, x>0$ are solutions of $y'=\frac{y^2}{x^2+xy}$. The solutions is not defined at $x=0$ because $x=0$ is impossible for the differential equation $y'=\frac{y^2}{x^2+xy}$. However, substituting $c=0$ into the general solution $y=ce^{-y/x}$, we get $y=0$ not $y=0, x<0$ and not $y=0, x>0$. How to modify the general solution $y=ce^{-y/x}$ so that the particular solution for $c=0$ gives the two solutions $y=0, x<0$ and $y=0, x>0$?
 November 26th, 2017, 04:44 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2456 Math Focus: Mainly analysis and algebra $$y=ce^{-\frac y x} \quad (x \ne 0)$$ Or, even better $$y=\begin{cases}c_1e^{-\frac y x} & (x \lt 0) \\ c_2e^{-\frac y x} & (x \gt 0) \end{cases}$$ Notice that you now need two sets of initial conditions to fix a unique solution. Another way would be to multiply the original equation by $x$ so that $y=0$ becomes a valid solution for all $x$. Thanks from eulid Last edited by v8archie; November 26th, 2017 at 04:47 AM.

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