So the differential equation is given like this: $x'=t+x^2, x(0)=0$. I know the formula to find the next result for the equation:
$$x(t)=x(t_0)+\int^t_{t_0}f(s,x(s))ds,$$ where $x(t)$, but I don't understand what $s$ notifies in this case. I know that the next solution is $\frac{t^2}{2}$ and I assume it is by integrating the $x(t)$ by $t$, only then, where does the $x^2t$ goes? And do we simply rewrite the equation as $x'=s+x^2$?
For Picard's approximation, the recurrence formula is
$$x_{n+1}(t)=x_0+\int_{t_0}^t f(s,x(s))\,ds$$
where $x_0=x(0)=0$ and $t_0=0$. Therefore the associated integral equation is
$$x(t)=\int_0^t \Big(s+x(s)^2\Big)\,ds$$
so for any $n\ge 1$, we find the recurrence formula
$$x_{n+1}(t)=x_0+\int_0^t \Big(s+x_n(s)^2\Big)\,ds=\int_0^t \Big(s+x_n(s)^2\Big)\,ds$$
hence $$x_0(t)=0$$ $$x_1(t)=\int_0^t \Big(s+0\Big)\,ds=\frac{t^2}{2}$$ $$x_2(t)=\int_0^t \left(s+\frac{s^4}{4}\right)\,ds=\frac{t^2}{2}+\frac{t^5}{20}$$ $$x_3(t)=\int_0^t \left(s+\frac{s^4}{4}+\frac{s^7}{20}+\frac{s^{10}}{400}\right)\,ds=\frac{t^2}{2}+\frac{t^5}{20}+\frac{t^{8}}{160}+\frac{t^{11}}{4400}$$ $$\vdots$$