The number of differentiable functions y: (-i | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = 2 \sqrt{y}$ with the initial condition $y(0) = 0$. Case $I$: The trivial solution $y(x) = 0$ for

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infinite

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(D) Given the differential equation $\frac{dy}{dx} = 2 \sqrt{y}$ with the initial condition $y(0) = 0$. Case $I$: The trivial solution $y(x) = 0$ for all $x \in \mathbb{R}$ satisfies the equation and the initial condition. Case $II$: For any constant $a \geq 0$,we can define a family of functions: $y(x) = \begin{cases} 0 & x Let us check the differentiability at $x = a$: Left-hand derivative: $\lim_{h 	o 0^-} \frac{y(a+h) - y(a)}{h} = \lim_{h 	o 0^-} \frac{0 - 0}{h} = 0$. Right-hand derivative: $\lim_{h 	o 0^+} \frac{y(a+h) - y(a)}{h} = \lim_{h 	o 0^+} \frac{(a+h-a)^2 - 0}{h} = \lim_{h 	o 0^+} \frac{h^2}{h} = 0$. Since the left-hand derivative equals the right-hand derivative,the function is differentiable at $x = a$. For $x > a$,$y'(x) = 2(x-a) = 2\sqrt{(x-a)^2} = 2\sqrt{y(x)}$. Since $a$ can be any non-negative real number,there are infinitely many such functions.

The number of differentiable functions $y: (-\infty, \infty) \rightarrow [0, \infty)$ satisfying $y^{\prime} = 2 \sqrt{y}$ and $y(0) = 0$ is

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