Let $y:[0,\infty)\to[0,\infty)$ be a continuously differentiable function satisfying $y(t)=y(0)+ \int_0^t y(s)ds$ for $t\ge0$ Let $y:[0,\infty)\to[0,\infty)$ be a continuously differentiable function satisfying $y(t)=y(0)+ \int_0^t y(s)ds$ for $t\ge0$. Then


A.  $y^2(t)=y^2(0)+\int_0^ty^2(s)ds.$
B. $y^2(t)=y^2(0)+2\int_0^ty^2(s)ds.$
C.  $y^2(t)=y^2(0)+\int_0^ty(s)ds.$
D.$y^2(t)=y^2(0)+(\int_0^ty(s)ds)^2+2y(0)\int_0^ty(s)ds.$


It is very much clear that option D is correct. But I am not sure about whether any other option is correct or not. Please help. 
 A: You can just take the derivative of the equation to obtain

$$y'(t)=y(t)$$

and thus

$$(y(t)^2)'=2y(t)y'(t)=2y(t)^2$$

Applying the fundamental theorem gives version B.

In general, every initial value problem $y'=f(t,y)$, $y(t_0)=y_0$ is equivalent to the Picard integral equation

$$
y(t)=y_0+\int_{t_0}^t f(s,y(s))\,ds
$$

which shows how to schematically identify  the IVP from your given equation.
A: All four options are possible. 
However, A,C,D are only possible with $y(t)=0$. At any rate, $A,C,D$ fail for $y(t)=ce^t$ with $c\ne 0$.
A: This is probably not the intended way of doing it, but your equation actually determines $y$ uniquely: you can stick the equation back into itself to find
$$ y(t) = y(0)+\int_0^t \left( y(0)+\int_0^s y(s') \, ds' \right) ds \\
= y(0)+y(0)t+ \int_0^t\int_0^s y(s) \, ds' ds. $$
Repeating this procedure gives
$$ y(t) = y(0) \sum_{r=0}^n \frac{t^r}{r!} + \int_0^t \dotsi \int_0^{s_n} y(s_n) \, ds_n \dots ds_0. $$
Since $y$ is obviously continuous, the trivial bound of (volume of integration set) $\times$ (maximum absolute value of function) shows that the error term in this is smaller than a multiple of $t^{n+1}$, and hence this series converges to $y(0)e^t$.
We don't need the whole series here, though: look at the first few terms. We have
$$ y^2(t) = y^2(0)(1+t+t^2/2 +o(t^2))^2 = y^2(0)(1+2t+2t^2+o(t^2)). $$
If $y(0)=0$, these are all obviously satisfied.
But if $y(0)\neq 0$, this immediately allows you to eliminate A and C, since the right-hand side of each of these produces the wrong $t$ term. D is obviously true, and that leaves B, which you can check works by inserting $y(t)=y(0)e^t$.

On the other hand, the better way is probably to differentiate: the right-hand sides of all five equations have derivatives by the fundamental theorem of calculus. D is obviously true, so we only need to examine the first four. The defining equation gives $y=y'$. Then we have
$$ \begin{align}
2yy' = y^2 \tag{A'} \\
2yy' = 2y^2 \tag{B'} \\
2yy' = y \tag{C'}.
\end{align} $$
If $y \equiv 0$, these are all true as expected. If not, cancelling a $y$ gives a contradiction with the defining equation in the case of A' and C'. We can get B' by multiplying the defining equation by $2y$, and then integrating gives B, so B is also true in general.
