Ordinary differential equation/ successive approximation

Question

My assignment is to solve this initial value problem by successive approximation:

$y'=2xy$

$y(0)=1$

I have calculated the first three iterations:

$y_0=1$

1. $y_1=1+x^2$
2. $y_2=1+x^2+x^{2x^2}$
3. $y_3=1+x^2+x^{2x^2}+x^{2x^{2x^2}}$

Well, the pattern is recognizable already. The only problem is:

How do I construct an infinite series with this?

I think I would need to find an explicit formula for the sequence first (see my other question regarding this).

But what if there isn't any explicit formula? Are there any other methods to turn the sum into an infinite series?

Answer 1

First of all, you can solve the equation explicitly by separation of variables, the solution is $e^{x^2}$.

Second, you should not have those tetrations in your approximations. The initial approximation is a polynomial, and the integral of a polynomial is a polynomial, so your approximations should all be polynomials. For instance

$$y_0 = 1 \\ y_1 = 1+\int_0^x 2z(1) dz= 1+x^2 \\ y_2 = 1+\int_0^x 2z(1+z^2) dz = 1+x^2+\frac{x^4}{2}$$

etc. (Here $z$ is a dummy integration variable.) What you get should be easy to recognize, especially given the explicit solution.

Answer 2

You should check your calculations for $y_2$ and $y_3$: $$\begin{align} y_2(x)&=1+\int_0^x2\,t\,(1+t^2)\,dt=1+x^2+\frac{x^4}{2},\\ y_3(x)&=1+\int_0^x2\,t\,(1+t^2+\frac{t^4}{2})\,dt=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}. \end{align}$$ Can you see the pattern?