Solving a differential equation using Picard's iteration method

[stripes]

Homework Statement

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.

Homework Equations

none

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.

 

[kai_sikorski]

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

Thats the same thing as your second solution.

 

[stripes]

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

 

[kai_sikorski]

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

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

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

 

[stripes]

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