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Math Help - ODE-Lipschitz&Picard–Lindelöf theorem

  1. #1
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    ODE-Lipschitz&Picard–Lindelöf theorem

    Given This ODE:

    y' = (y-2) (x^2+y)^5
    y(0)=5

    A. Show that this problem has one solution that is defined in an open segment that contains 0.

    B. Let y(x) be a solution for this problem. Prove that y(x)>2 for every x in I and conclude that y'(x)>0 in I.
    Hint: You can use the solution of the problem: y'=(y-2)(x^2+y)^5 , y(x0)=2


    Help is needed !

    TNX!
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  2. #2
    Super Member Rebesques's Avatar
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    For part a, as you mentioned,
    we can use the...

    Picard–Lindelöf theorem

    ...which says that, as f(x,y)=(y-2)(x^2+y)^5 is continuous in x and Lipschitz continuous in y, there is a unique solution to the initial value problem, which is defined in a neighbourhood J of 0.

    Now for part b, note that y_0=2 is a solution of the equation. If y was to meet y_0 somewhere over J, the uniqueness of the initial value problem \{z'=f(x,z), \ z(0)=2, x\in \mathbb{R}\} would be violated. Since y(0)>2, we must have y>2 throughout J. Now the ode gives us a constant sign for y' over J.
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    Thanks a lot
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