Given the following initial value problem $x'(t) = x(t) + e^t$ with $x(0) = 0$. I've calculated its picard's iteration, but having trouble guessing the pattern: $$\begin{cases} x_1 = e^t - 1, \\[3pt] x_2 = 2e^t - t - 2, \\[3pt] x_3 = 3e^t - 2t - 3 - \frac{t^2}{2}, \\[3pt] x_4 = 4e^t - 3t - 4 - t^2 - \frac{t^3}{6}, \\[3pt] x_5 = 5e^t - 4t - 5 - \frac{3t^2}{4} - \frac{t^3}{3} - \frac{t^4}{24}, \\[3pt] x_6 = 6e^t - 5t - 6 - 2t^2 - \frac{t^3}{2} - \frac{t^4}{12} - \frac{t^5}{120},... \end{cases} $$
So maybe we have something like $$x_n(t) = n(e^t - 1) - (n-1)t - \sum_{k = 2}^{n-1} \frac{t^k}{??}$$ Since the solution is actually $x(t) = t e^t$, I also don't see how the $n$ in the first two elements should disappear as $n \to \infty$, because otherwise, it diverges.
Any hint/help would be appreciated. Thanks.
If you expand with Maclaurin $$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}+O\left(t^6\right)$$ and plug this in the, let's say, $6$th step, you get $$6 \left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+\frac{t^5}{120}\right)-\frac{t^5}{120}-\frac{t^4}{12}-\frac{t^3}{2}-2 t^2-5 t-6+O\left(t^6\right)=\\=t+t^2+\frac{t^3}{2}+\frac{t^4}{6}+\frac{t^5}{24}+O\left(t^6\right)$$ the last one is the expansion of $te^t$