Use the Picard approximation to find a solution for $y'=2t(1+y),y(0)=0$.
$\varphi_{0}=0$
$\varphi_{1}=0+\int_{0}^{t}f(s,\varphi_{0})ds=t^{2}$
$\varphi_{2}=0+\int_{0}^{t}f(s,\varphi_{1})ds=t^{2} +\frac{1}{2}t^{4}$
$\varphi_{3}=0+\int_{0}^{t}f(s,\varphi_{2})ds=t^{2} +\frac{1}{2}t^{4}+\frac{1}{6}t^{6}$
My claim is that $\varphi_{m}(t)=\sum_{k=1}^{m}\frac{t^{2k}}{k!}$. I am not sure if this is a right claim though. I want to use induction to prove this claim, but I am not sure how to calculate $\varphi_{m+1}(t)$, the inductive step.
Also, I cannot find the limit of this claim to find the solution. Do I use a convergence test and find radius of convergence?
The claim
$$\varphi_{m}(t)=\sum_{k=1}^{m}\frac{t^{2k}}{k!}$$
is correct. The exact solution is given by the limit
$$\varphi(t)=\lim_{m\to\infty}\varphi_{m}(t)=\sum_{k=1}^{\infty}\frac{(t^{2})^k}{k!}=\sum_{k=0}^{\infty}\frac{(t^{2})^k}{k!}-1=\text{exp}(t^2)-1$$
and the ratio test can be used to show that the series converges (and to find the radius of convergence). For the inductive step assume
$$\varphi_{m}(t)=\sum_{k=1}^{m}\frac{t^{2k}}{k!}$$
holds for all $m\in\mathbb N$. Then, you need to show that
$$\varphi_{m+1}(t)=\sum_{k=1}^{m+1}\frac{t^{2k}}{k!}$$
if I am not mistaken this follows directly from the definition.