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  • 1 Post By v8archie
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November 26th, 2017, 12:58 AM   #1
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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 $y=0, x>0$?
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November 26th, 2017, 05:44 AM   #2
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$$y=ce^{-\frac y x} \quad (x \ne 0)$$
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 $x$.
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Last edited by v8archie; November 26th, 2017 at 05:47 AM.
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