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anonymous

  • 4 years ago

does the IVP dy/dt=6y^(2/3), y(1)=0 have a unique solution on an interval containing t=1?

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  1. anonymous
    • 4 years ago
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    first step, i have y=(2t+c)^3 is it right?

  2. JamesJ
    • 4 years ago
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    Yes.

  3. JamesJ
    • 4 years ago
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    and now use the initial condition to find c.

  4. anonymous
    • 4 years ago
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    that's makes it much easier, thank u

  5. anonymous
    • 4 years ago
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    is y=0 a solution to the IVP? yes isn't it, because when t=1, y=0 or is there another way to determine the solution?

  6. JamesJ
    • 4 years ago
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    Yes, the zero function is the solution of your ODE. But it's a very uninteresting solution and we don't count it as a solution for the purposes of describing uniqueness.

  7. anonymous
    • 4 years ago
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    ok thx

  8. anonymous
    • 4 years ago
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    i have another problem: show that y=tant satisfies the IVP y'=1/(1+t^2), y(0)=0. i am confused because the y'(tant) is not 1/(1+t^2). they both do equal 0 with initial condition set to 0 but what do they have to do with each other?

  9. anonymous
    • 4 years ago
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    i don't know where to go now because i'm trying to look for the limits of (t,y) to find the largest open interval but tan and arctan limits are not the same, at least for t

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