Solve differential equation $y(t)=1-\int_{0}^t s\cdot y^2(s)~ds$

Question

I found an exercise from an exam on differential equations. The task was to find a differentiable function $y:[0, \infty) \rightarrow \mathbb{R}$, such that

$$y(t)=1-\int_{0}^t sy^2(s)ds \quad t \geqslant 0$$

I found one solution with $y(t)=\frac{2}{t^2+2}$. Now I have to decide wether any continuous solution of the equation above is also differentiable. However, I have no idea how to even approach that question.

Answer 1

Note this equation implies initial condition: $$y(0)=1$$

Next, we use Fundamental Theorem of Calculus, take derivative on both sides,

$$\frac{dy}{dt}=-t\cdot y^2$$ This differential equation is separable, so we get $$\int-\frac{1}{y^2}~dy=\int t ~dt\Rightarrow \frac{1}y=\frac12t^2+C$$ Plug in the initial condition, we get $C=1$, therefore

$$y=\frac{2}{2+t^2}$$