The initial value problem $y'=\sqrt {y}, \:\: y(0)=\alpha$. The initial value problem 

$$\begin{cases}y'=\sqrt {y}\\y(0)=\alpha\end{cases}$$ has

$A.$ At least two solutions if $\alpha =0$

$B.$ No solution if $\alpha &gt; 0$

$C.$ At least one solution if $\alpha &gt; 0$

$D.$ A unique solution if $\alpha = 0$

ATTEMPT

I solved the differential equation as 

$y(x) = \frac{x^2}{4} + \alpha +x\sqrt {\alpha}$

How do i choose correct options?
Answer. A true, B false, C true, D false.

If $a=0$, uniqueness is violated. 

Solution I. $\varphi(x)\equiv 0$.

Solution II. $\varphi(x)=\frac{x^2}{4}$

If $a\ne 0$, solution is unique (locally).
The general solution is
$$
y^{-1/2}dy=dx\rightarrow 2\sqrt{y}=x+c
$$
Considering initial value we have 
$$
2\sqrt{y}=x+2\sqrt{\alpha}.
$$
Hence: i) If $\alpha&gt;0$ the i.v.p has a unique solution in a neighborhood of $0$ (by existence-uniqueness theorem). ii) The i.v.p has not solution if $\alpha&lt;0$. iii) If  $\alpha=0$ then it has at least two solutions: 1) $y=0$
, 2) $\sqrt{y}=x/2$. You can now verify which of the above options is correct or not.