Results 1 to 2 of 2

Math Help - Solutions to ODE y'(t) = t^2 +y(t)^2

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    11

    Solutions to ODE y'(t) = t^2 +y(t)^2

    Hi all,

    The problem assigned to me was to show that the equation y'(t)==t^2+y(t)^2 has a solution for initial conditions y(0) = 0 over the interval 0<= t <= min(a, b/(a^2+b^2)), where we are considering a rectangle R, 0<= t <= a, -b<=y<=b.

    What puzzles me is that the uniqueness and existence theorem says that if the DE is of the form y'(t) = f(t,y), where f and df/dy (partial derivative) are continuous over some rectangle containing the initial conditions (t0,y0) then the solution exists and is unique in that rectangle. In this case, f(t,y) is simply t^2 + y^2, and df/dy = 2 y. Both are clearly continuous for all values of t and y, so what is up with this question?

    Also, trying to solve both numerically and analytically in Mathematica state that y(0)=0 is a funky point (the error statest that singularity or stiff system suspected).

    Thanks in advance,
    Julian
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by aznmaven View Post
    Hi all,

    The problem assigned to me was to show that the equation y'(t)==t^2+y(t)^2 has a solution for initial conditions y(0) = 0 over the interval 0<= t <= min(a, b/(a^2+b^2)), where we are considering a rectangle R, 0<= t <= a, -b<=y<=b.

    What puzzles me is that the uniqueness and existence theorem says that if the DE is of the form y'(t) = f(t,y), where f and df/dy (partial derivative) are continuous over some rectangle containing the initial conditions (t0,y0) then the solution exists and is unique in that rectangle. In this case, f(t,y) is simply t^2 + y^2, and df/dy = 2 y. Both are clearly continuous for all values of t and y, so what is up with this question?
    I think that the clause in blue is the key to this. The question is asking for a solution y(t) that is defined in the interval 0 ≤ t ≤ a and satisfies the condition |y(t)| ≤ b in that interval. In order to prevent the solution from getting too large, it may be necessary to restrict the interval so that instead of going up to t=a it has to stop short at t = b/(a^2+b^2).

    There is a very clear statement of the existence and uniqueness theorem here.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 4th 2011, 08:21 PM
  2. solutions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 11th 2010, 05:07 AM
  3. Replies: 6
    Last Post: July 26th 2010, 11:45 AM
  4. help with ODE. *i got no solutions
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 10th 2009, 09:12 AM
  5. No. of solutions..
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 19th 2008, 09:32 AM

Search Tags


/mathhelpforum @mathhelpforum