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Math Help - how to expres in terms of y

  1. #1
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    how to expres in terms of y

    dy/dt = t/2y+1. Find the general solution.

    I get that y^2 +y = 1/2t^2 + C. How do I solve for Y?
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  2. #2
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    Quote Originally Posted by lord12 View Post
    dy/dt = t/2y+1. Find the general solution.

    I get that y^2 +y = 1/2t^2 + C. How do I solve for Y?
    Complete the square.

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

    \left(y + \frac{1}{2}\right)^2 - \frac{1}{4} = \frac{1}{2}t^2 + C

    \left(y + \frac{1}{2}\right)^2 = \frac{1}{2}t^2 + \frac{1}{4} + C

    y + \frac{1}{2} = \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C}

    y= \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C} - \frac{1}{2}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    [snip]

    y + \frac{1}{2} = {\color{red}\pm} \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C}

    y= {\color{red}\pm} \sqrt{\frac{1}{2}t^2 + \frac{1}{4} +C} - \frac{1}{2}.
    ..
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    ..
    Yes - I was hoping that would be implied...
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  5. #5
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    how can i tell if solution is unique

    dy/dt = t/2y+1


    y = sqrt(1/2t^2 + 1/4 +k)

    y(1) = 3/2
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  6. #6
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    Quote Originally Posted by lord12 View Post
    dy/dt = t/2y+1


    y = sqrt(1/2t^2 + 1/4 +k)

    y(1) = 3/2
    First, your equation for y is incorrect.

    y = \pm\sqrt{\frac{1}{2}t^2 + \frac{1}{4} + C} - \frac{1}{2}.

    If y(1) = \frac{3}{2}, then


    \frac{3}{2} = \pm \sqrt{\frac{1}{2}(1)^2 + \frac{1}{4} +C} - \frac{1}{2}

    2 = \pm \sqrt{\frac{1}{2} + \frac{1}{4} + C}

    4 = \frac{3}{4} + C

    \frac{13}{4} = C.


    I think you'll find the solution is NOT unique...

    y = \pm \sqrt{\frac{1}{2}t^2+ \frac{1}{4} + \frac{13}{4}} - \frac{1}{2}

     = \pm \sqrt{\frac{1}{2}t^2 + \frac{7}{2}} - \frac{1}{2}.
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