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Math Help - Diff Eq - Existence/uniqueness theorem

  1. #1
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    Diff Eq - Existence/uniqueness theorem

    So the question is:
    Using the existence and uniqueness theorem, show that the differential equation dy/dt = sqrt(y-1), with initial condition that y(0)=2, has a unique solution. Furthermore, solve it.

    I verified that it both existed and was unique, and I also got a solution but I'm not quite sure if it's right...

    I separated it into 1/sqrt(y-1) * dy = 1 dt
    integrated: 2 sqrt(y-1) = t + C
    divide by 2: sqrt(y-1) = t/2 + C

    now this is where I'm not sure if I messed up or not. Silly, since it's simple algebra, but when i remove the square root from the y-1, would the right side turn into (t^2)/4 + C, or (t^2)/4 + Ct + C? I chose the former and got a final answer of y(t) = (t^2)/4 + 2
    I'm totally braindead right now, haha, but should I have done it the second way?
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  2. #2
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    Quote Originally Posted by mistykz View Post
    So the question is:
    Using the existence and uniqueness theorem, show that the differential equation dy/dt = sqrt(y-1), with initial condition that y(0)=2, has a unique solution. Furthermore, solve it.

    I verified that it both existed and was unique, and I also got a solution but I'm not quite sure if it's right...

    I separated it into 1/sqrt(y-1) * dy = 1 dt
    integrated: 2 sqrt(y-1) = t + C (**)
    divide by 2: sqrt(y-1) = t/2 + C

    now this is where I'm not sure if I messed up or not. Silly, since it's simple algebra, but when i remove the square root from the y-1, would the right side turn into (t^2)/4 + C, or (t^2)/4 + Ct + C? I chose the former and got a final answer of y(t) = (t^2)/4 + 2
    I'm totally braindead right now, haha, but should I have done it the second way?
    If

    \sqrt{y-1} = \frac{t}{2} + c then y - 1 = \left( \frac{t}{2} + c\right)^2

    but I would personally bring in the IC at the step (**)
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