Initial value problem, first order linear ode

  • Thread starter Jesssa
  • Start date
  • #1
51
0
Hey,

We haven't properly covered this in class yet, but I am trying to study ahead using online course notes, I manage to finish a few questions but I have gotten stuck here,

The question starts by asking for the solution to the ODE:

y' = 1 - 2xy,

When I solve this using the integrating factor technique in the final line I end up with an integral of ex^2dx which I feel isn't right, wolframalpha spits out something similar but the solution involves the complex error function which I have never studied before.

Moving past that problem I started on the rest of the question,

http://img194.imageshack.us/img194/6343/asdahi.jpg [Broken]

From reading them it looks like b implies c is that right?

If all approximations exist on |x|≤1/2 then they would converge to the solution right?

And the initial theorm in my notes states that if |x|≤h≤a then there exists a unique solution y=phi, where h = Min(a, b/M) and |f(x,y)|=|1-2xy|≤M

I'm not sure on how to show b) though I keep thinking I have found which part of the theorem proof to use but I contradict myself. Could anyone steer me in the right direction?

For d) in class we were given that the difference between any approximation and the solution is

[itex]|{{\phi }_{n}}-\phi |\,\le \frac{M}{k}\frac{{{(kh)}^{n+1}}}{(n+1)!}{{e}^{kh}}\,\,\,\,\,\,where\,\,\,\left| \frac{\partial f}{\partial y} \right|\le k[/itex]

So using this,

|f(x,y)|=|1-2xy|≤1 + 2 (1/2) 1 = 2 = M
|df(x,y)/dy|=|1-2x|≤2=k

and h = 1/2

Subbing it all into the equation i get ≤0.11

What am I doing wrong?
 
Last edited by a moderator:

Answers and Replies

Related Threads on Initial value problem, first order linear ode

  • Last Post
Replies
5
Views
5K
Replies
1
Views
1K
Replies
5
Views
1K
Replies
5
Views
4K
Replies
4
Views
8K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
640
Replies
7
Views
1K
Replies
7
Views
9K
Top