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Show that $y^{*}(x)= \frac{x^{2}}{4}$ is the maximal solution and $y_{*}(x) = 0$ is the minimal solution when $x \geq 0$ for the IVP:

$ y^{'} = \sqrt{|y|}$ , $y(0)=0$

I know the solution is of the form $ y(x) = \frac{(x+C)^{2}}{4}$ but I am not sure how to conclude the maximal and minimal solutions.

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    Have you seen the definition of maximal and minimal solutions? – 2017-02-13
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    The only definition I have seen is $y_{*}(x) < y(x) < y^{*}(x)$ where $y(x)$ is any solution of the IVP. – 2017-02-13
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    Ok, then I think you just solved your own problem. Note that $C=0$ since $y(0)=0$. It should be clear that $y_*(x)=00$. – 2017-02-13
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    Ok, I see thanks for the comments. – 2017-02-13

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