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It is asked to analyze the existence and uniqueness of the solutions of

$$y'=x\sqrt[5]{y}$$

The answer given is that there exists one unique solution of this equation in $\mathbf R^2-\{(x,0):x \in \mathbf R\} $ and two solutions in every point $(x_0,0)$. I couldn't find those two solutions (for the second case, I mean. I understand the uniqueness in the first case). I thought that $y=0$ was unique in this case. What am I doing wrong?

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Think first of $x_0=0$. Then $y=0$ is clearly a solution. But so is $$ y=\Bigl(\frac{2\,x^2}{5}\Bigr)^{5/4}, $$ obtained by separation of variables. In fact there are are infinite solutions. Given $a\ge0$, the function $$ y_a(x)=\begin{cases} 0 & 0\le x\le a\\ \Bigl(\dfrac{2\,(x-a)^2}{5}\Bigr)^{5/4} & x>a \end{cases} $$ is also a solution. Non uniqueness occurs because $y^{1/5}$ is not Lipschitz at $y=0$.

You can argue similarly for any $x_0>0$.

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  • $\begingroup$ I cant understand why for $0 \leq x \leq a$ the y must be zero. $\endgroup$ – Giiovanna Nov 5 '14 at 18:51
  • $\begingroup$ It must not, but then it would not be a solution of the differential equation. All I'm saying is that the function $y_a$ is a solution, and you have a different solution for each $a$. You only have to check. $\endgroup$ – Julián Aguirre Nov 5 '14 at 21:29
  • $\begingroup$ ok,thanks for your help! $\endgroup$ – Giiovanna Nov 5 '14 at 21:42

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