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What's the general solution of $\frac{dy}{dx} = 3y^{2/3}$ ? Im pretty sure this is a separable equation, but I'm not sure how to go forward? Just multiply by $dx$ and $\frac{1}{3y^{2/3}}$ well then I get $y^{1/3} =x+ C$, correct?

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    $\begingroup$ You are almost correct. It should be $3y^{1/3}=x+c$ $\endgroup$ – Guy Apr 1 '14 at 4:38
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    $\begingroup$ Note that the special solution $y$ identically $0$ is not picked up by the "general" solution. $\endgroup$ – André Nicolas Apr 1 '14 at 6:44
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    $\begingroup$ Also note that the solution to the initial value problem with $y(0)=0$ is non-unique, since (for any $k>0$) you can take $y(x)=0$ for $x<k$ and then $y(x)=(x-k)^3$. $\endgroup$ – Hans Lundmark Apr 1 '14 at 6:53
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On the left hand side you get , $ 3y^{\frac{1}{3}}$ and on the right hand side you get, $x+C$ $\Rightarrow 3y^{\frac{1}{3}}=x+C \Rightarrow y = (x+C)^3/27$.

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    $\begingroup$ "On the left hand side you get, $y^{\frac{1}{3}}$" Sorry but I do not, unless $y\ne0$. $\endgroup$ – Did Apr 10 '14 at 6:39
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$$\begin{align}\frac{dy}{dx}=3y^{\frac{2}{3}}\\\frac{dy}{y^{\frac{2}{3}}}=3dx\\\int\frac{dy}{y^{\frac{2}{3}}}=\int 3dx\\3y^{\frac{1}{3}}=3x+C\\y^{\frac{1}{3}}=x+C\end{align}$$

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