Applying conditions

Question

I have this $\displaystyle 2(y')^2=2\cdot\sqrt y+c$ equation and $y(0)=1$, $y'(0)=1$. How do you tell from the information given, that $c$ here is equal to $0$? I just can not figure it out, thanks for any help.

Answer 1

$$2(y'(x))^2=2\sqrt {y(x)}+c$$

Put x=0:

$$2(y'(0))^2=2\sqrt{ y(0)}+c$$

$$2*1^2=2*\sqrt1+c\implies c=0$$