# Picard Iteration

1. Nov 19, 2012

### snesnerd

Use picard iteration 4 times to estimate the initial value problem solution:

y' = 2xy thru (0,1)

First off, when it says thru (0,1), I am assuming it means y(0) = 1. I was not sure if I was doing picard iteration right, but here it my attempt:

dy/dx = 2xy

dy/y = 2x dx

dy/1 = 2x dx - since y(0) = 1

y = x^2 + C so y = x^2 + 1. This is the first iteration

y' = x^2 + 1. So now I just solve this

........ and so on 4 times.

Or is picard iteration work like this:

dy/dx = 2xy

dy/y = 2x dx

ln|y| = x^2 + C

y = e(x^2 + C)

y = e^(x^2)e^(C)

y = Ce^(x^2)

when y(0) = 1, C = 1

So y = e^(x^2) This is the first iteration

So then I integrate and repeat this 3 more times.

I wasnt sure which one was the right way to do picard iteration or if neither way is right.

2. Nov 19, 2012

### haruspex

Having just read up on Picard Iteration (http://www.sosmath.com/diffeq/first/picard/picard.html) in order to respond:
You take as first approximation y0(x) = y(0) = 1.
Now you use y' = 2xy in the integral form to get a better approximation:
$y_1(x) = \int_{s=0}^x 2s y_0(s) ds = \int_{s=0}^x 2s (1) ds =x^2$
and to get the next one, repeat the process:
$y_2(x) = \int_{s=0}^x 2s y_1(s) ds$
etc.

3. Nov 20, 2012

### HallsofIvy

Staff Emeritus
Your first try was correct. Picard iteration gives a power series solution.
With y(0)= 1, y'= 2xy, then your first approximation is y(x)= 1 so that y'= 2x, y= x2+ 1, so $y'= 2x(x^2+ 1)= 2x^3+ 2x$ and $y(x)= (1/2)x^4+ x^2+ 1$, etc.

4. Nov 20, 2012

### snesnerd

Okay so let me see if I get this.

y' = 2xy y(0) = 1

dy/y = 2x dx

dy = 2x dx

y = x^2 + C

y = x^2 + 1 (first iteration)

integrate 2x(x^2 + 1) dx

y = (1/2)x^4 + x^2 + C

y = (1/2)x^4 + x^2 + 1 (second iteration)

integrate 2x[(1/2)x^4 + x^2 + 1] dx

y = (1/6)x^6 + (1/2)x^4 + x^2 + 1 (third iteration)

integrate 2x[(1/6)x^6 + (1/2)x^4 + x^2 + 1] dx

y = (1/24)x^8 + (1/6)x^6 + (1/2)x^4 + x^2 + 1

5. Nov 21, 2012

Looks right.