Squaring a integral equation

Question

If $y:[0,\infty)\to[0,\infty)$ is a continuously differentiable function satisfying $$y(t)=y_0-\int\limits_0^t y(s)ds$$ for $t\ge0$, then

1. $y^2(t)=y_0^2+\left(\int\limits_0^t y(s)ds\right)^2-2y_0\int\limits_0^t y(s)ds$
2. $y^2(t)=y_0^2-2\int\limits_0^t y^2(s)ds$ (CSIR December 2014)

(My attempt has been converted into an answer since no one contradicted so far, see below.)

Answer 1

Trivially, 1 is true.

For 2, using Leibniz integral rule:

$\dot y(t)=-y(t)$

$\dot y^2(t)=2y(t)\dot y(t)=-2y^2(t)$

Integrating from 0 to t,

$$\int\limits_0^t d(y^2(s))=y^2(t)-y^2(0)=-2\int\limits_0^t y^2(s)ds$$

And hence 2.