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$$y'=1+\sqrt y\\y(0)=0 $$ Show that this IVP does not satisfy Lipschitz condition, but has a unique solution.

I have shown the first way, like this: Let $f(x,y)=1+\sqrt y $. Then $\frac {|f(x,y_1)-f(x,0)|}{|y_1-0|}=|\frac{1}{\sqrt y_1}| \rightarrow \infty $ as $y_1\rightarrow 0 $. How to show the uniqueness part?

Please help.

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Hint: You can solve this separable equation to get the general solution (in the implicit form: $x$ a function of $y$). There is only one value of the constant that makes $x \to 0$ as $y \to 0+$.

EDIT: $$\int \dfrac{dy}{1+\sqrt{y}} = 2 \sqrt{y} - 2 \ln(1+ \sqrt{y}) + C$$

EDIT: More generally, for $y' = f(y)$ where $f$ is continuous and positive for $y \ge 0$ you have the implicit solutions $x = F(y) + c$ where $F(y) = \displaystyle\int_0^y \dfrac{ds}{f(s)}$, and these solution curves never intersect.

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  • $\begingroup$ You need the fact that $1+\sqrt{y}>0$ in order to do this in the first place. It would work with $y'=\sqrt{y}$ except for the issue of division by zero. $\endgroup$ – Ian Dec 18 '14 at 17:02
  • $\begingroup$ ok I did the integration, $\int \frac{dy}{1+\sqrt y}$ which looks a big value, can you please write the full solution, or take a snapshot and paste it here? $\endgroup$ – Sankha Dec 18 '14 at 17:06

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