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Homework Help: Initial value problem, first order linear ode

  1. Apr 2, 2012 #1

    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

    and h = 1/2

    Subbing it all into the equation i get ≤0.11

    What am I doing wrong?
    Last edited by a moderator: May 5, 2017
  2. jcsd
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