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This comes from a real analysis class, and I currently cannot assume any knowledge about integration.

I want to find all solutions to the initial value problem $y' = y^{\frac{1}{2}}$, $y(0) = 0$.

I know that $f(x) = 0$ and $f(x) = \frac{x^2}{4}$ both satisfy the conditions, as well as any function of the form $f(x) = 0$ if $x < c$, $f(x) = \frac{(x-c)^2}{4}$ if $x \geq c$, for $c > 0$.

However, now I want to prove that these are the only solutions to the initial-value problem, and am unsure how to go about doing so. Any pointers in the right direction would be greatly appreciated!

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You have to be a bit careful with $f(x)=\frac{x^2}{4}$ because $f'(x) = \frac{x}{2}$ but $\sqrt{f(x)} = \frac{|x|}{2}$. –  SomeEE Feb 28 at 5:27

1 Answer 1

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Hints: Suppose $y$ is not identically zero.

Let $c = \sup\{x| y(x) = 0\}$ then

1) $c$ exists (is not $\infty$).

2) $\{x | y(x) = 0\} = [-\infty, c]$.

3) $\sqrt{y}$ is differentiable on $(c, \infty)$, hence so is $y'$ and we have

$$y'' = \frac{1}{2}\frac{y'}{\sqrt{y}} = \frac{1}{2}\frac{\sqrt{y}}{\sqrt{y}} = \frac{1}{2}.$$

4) $y'(c) = 0$.

Now you can uniquely classify $y'$ on $(c, \infty)$.

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