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anonymous

  • 4 years ago

does anyone know why y'=2x, y(0)=0, y(1)=100 has no solution?

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  1. JamesJ
    • 4 years ago
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    If dy/dx = 2x, what is the general solution of that equation? It's y(x) = x^2 + C. Now is there ANY choice of the number C such that y(0) = 0 and y(1) = 100?

  2. anonymous
    • 4 years ago
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    Ya general solution y=x^2 + c1 no problem. but the book says "Show that the problem has no solution. Is this an Initial value problem?" I don't how this statement can be correct when c can take on any value since it is arbitrary.

  3. myininaya
    • 4 years ago
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    right you have a contradiction so there is no y with those conditions that satisfy the differential equation

  4. JamesJ
    • 4 years ago
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    Given that y(x) = x^2 + c, you now what to find ONE value of c such that it both true that y(0) = 0 AND y(1) = 100 Does there exist one value of c such that both of those conditions are satisfied?

  5. anonymous
    • 4 years ago
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    nope

  6. JamesJ
    • 4 years ago
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    Therefore the initial value problem y' = 2x, y(0) = 0, y(1) = 100 has no solution.

  7. anonymous
    • 4 years ago
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    why does it have to be so opaque, but yet so simple.

  8. JamesJ
    • 4 years ago
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    it's not opaque now. That's why we do exercises! :-)

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