# Solving a differential equation using Picard's iteration method

1. Feb 27, 2012

### stripes

1. The problem statement, all variables and given/known data

Use Picard's iteration method to solve the initial value problem y' = t + y, y(0) = 0. Determine $\phi_{n}(t)$ for an arbitrary value of n, and take the limit as n goes to infinity. Of course, you know the series for e^t, so you will recognize the limit function $\phi_{n}(t)$. Also evaluate $\phi_{3}(t)$ at t=0, 0.1, and 0.2.

2. Relevant equations

none

3. The attempt at a solution

I set up the problem by letting $\phi_{n+1}(t) = \int^{0}_{t}f(s, \phi_{n}(s))ds$

What I get is $\phi_{0}(t) = 0$, $\phi_{1}(t) = \frac{t^{2}}{2}$, $\phi_{2}(t) = \frac{t^{2}}{2} + \frac{t^{3}}{6}$, $\phi_{3}(t) = \frac{t^{2}}{2} + \frac{t^{3}}{6} + \frac{t^{4}}{24}$. So clearly,

$\phi_{n}(t) = \frac{t^{2}}{2!} + \frac{t^{3}}{3!} + \frac{t^{4}}{4!} + ... + \frac{t^{n}}{n!} = e^{t} - e -1$.

But I know that the differential equation y' = t + y has the unique solution $y = \frac{-te^{-t} - e^{-t} +1}{e^{-t}}$:

Rewrite the de as follows:
$y' + (-1)y = t$
$\mu(t) = e^{\int(-1)dt} = e^{-t}$
$e^{-t}y' + e^{-t}(-1)y = e^{-t}t$
$\frac{d}{dt}[ye^{-t}] = e^{-t}t$
$ye^{-t} = -te^{-t} + \int e^{-t} = -te^{-t} - e^{-t}$
$y =\frac{-te^{-t} - e^{-t} + C}{e^{-t}}$

And we can solve for C easily...but i'll save you that and tell you C = 1 (I think)

These two are not equal...

I'm really confused...as for the evaluation part, I can do that once I've solved the DE.

Last edited: Feb 27, 2012
2. Feb 27, 2012

### kai_sikorski

Series for et is
$$e^t =1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}+...$$
$e^t - 1 - t$

Thats the same thing as your second solution.

3. Feb 27, 2012

### stripes

So $y = \frac{-te^{-t} - e^{-t} +1}{e^{-t}} = e^{t} - e -1$?

4. Feb 27, 2012

### kai_sikorski

No, where do you keep getting this -e from?

$y = \frac{-te^{-t} - e^{-t} +1}{e^{-t}} = e^{t} - t -1$

5. Feb 27, 2012

### stripes

Right. I kept thinking e^1. But it's t^1. Thank you.