Picard's method solution vs exact solution

[phosgene]

Homework Statement

I have to use Picard's method to find an approximate solution to

\frac{dy}{dx}=2xy

and then compare it to the exact solution, given that y(0)=2

Homework Equations

Picard's method:

y_{n}=y_{0} + ∫^{x}_{x_{0}} f(\phi,y_{n-1}(\phi))d\phi

The Attempt at a Solution

So I go over Picard's method about 3 times and get the result:

2(1+x^2 + \frac{x^4}{2} + \frac{x^6}{6})

which looks like it miiiight be going towards:

2(\sum ^{\infty}_{n=0} \frac{x^{2n}}{n!}) = 2e^{x^2}I then solve the DE exactly by noting that the DE is separable, so I separate and integrate to obtain:

y= e^{x^2}+C

Using my initial condition, I find that C=1. So, my question is, have I done this correctly? I have a feeling that I'm supposed to get the same answer for both.

 

[Dick]

Homework Statement

I have to use Picard's method to find an approximate solution to

\frac{dy}{dx}=2xy

and then compare it to the exact solution, given that y(0)=2

Homework Equations

Picard's method:

y_{n}=y_{0} + ∫^{x}_{x_{0}} f(\phi,y_{n-1}(\phi))d\phi

The Attempt at a Solution

So I go over Picard's method about 3 times and get the result:

2(1+x^2 + \frac{x^4}{2} + \frac{x^6}{6})

which looks like it miiiight be going towards:

2(\sum ^{\infty}_{n=0} \frac{x^{2n}}{n!}) = 2e^{x^2}


I then solve the DE exactly by noting that the DE is separable, so I separate and integrate to obtain:

y= e^{x^2}+C

Using my initial condition, I find that C=1. So, my question is, have I done this correctly? I have a feeling that I'm supposed to get the same answer for both.

e^(x^2)+1 is NOT a solution to the ODE. Try it. You messed up where the constant C belongs.

 

[phosgene]

Argh! Silly mistake. Corrected it and now they are equal, thanks :)