I'm asked to find the first few terms of the solution to a differential equation by using Picard iteration, then "guessing" the general form of the solution.
The differential equation is $\frac{dx}{dt} = -2x$ with initial condition $x(0) = 3$.
Using Picard iteration I got the first three terms: $$ x_{1}(t) = -6t+3\\ x_{2}(t) = 6t^2-6t+3\\ x_{3}(3) = -4t^3+6t^2-6t+3 $$
I need help finding the general form of $x_{k}(t)$. I know there's going to be $(-t)^k$ of some form but I can't figure how to include the coefficients.
I have two more exercises like this one, and I'm unable to guess the general form. I know it's not really just guessing, but it's finding a pattern. However, I just can't see it. So I was wondering if there are some steps to take or specific popular formulas that I should keep in my mind?
Edit: I know now that it has to be some form of $\displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!} $. For one of the other exercises, the third iteration is $x_{3}(t) = \frac{1}{2}t^3+\frac{3}{2}t^2+3t+1$ . I found that $x_{k}(t) = 3e^{t}$ works for all terms except the 1 at the end and I'm not sure how else to make it work?
Notice that \begin{align}3-6t+6t^2-4t^3&=3 \left( 1 -2t+2t^2-\frac{4}{3}t^3\right)\\&=3\left( 1 +(-2t) + \frac{(-2t)^2}{2!}+\frac{(-2t)^3}{3!}\right) \end{align} I would guess the general form is: $$x_k(t) = 3\sum_{j=0}^k \frac{(-2t)^j}{j!}$$ So it seems that $x_k(t) \to 3e^{-2t}$. Can you verify this?