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Math Help - Find y'' given x + xy + y =2

  1. #1
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    Find y'' given x + xy + y =2

    I'm learning Calculus on my own this summer and I've encountered a simple problem which i can't seem to get the right answer to. Here is the problem and my solution:

    Find y'' given x + xy + y =2
    first derivative
    1 + y + x(y') + y' = 0 took derivative of each term
    1 + y + y'(x+1) = 0 simplified a bit
    (-1-y)/(x+1) = y' simple algebra.
    second derivative
    [-y'(x+1) - 1(-1-y)]/[(x+1)^2] quotient rule
    [-y'(x+1)]/[(x+1)^2] - (-1-y)/[(x+1)^2] = 0 restated in a more useful manner.
    [-y'(x+1)]/[(x+1)^2] = (-1-y)/[(x+1)^2] simple algebra
    -y'(x+1) = -1-y simple algebra
    y'(x+1) = 1+y multiplied both sides by -1
    y' = (1+y) / (x+1)

    The answer the books gives is: [2(1+y)]/[(1=x)^2]

    Thanks in advance for help.
    Last edited by mr fantastic; August 8th 2011 at 10:11 PM. Reason: Re-titled.
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  2. #2
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    Re: Basic calculus problem.

    Quote Originally Posted by RussianScience View Post

    Find y'' given x + xy + y =2
    Have a go this way,

    x + xy + y =2

     xy + y =2-x

     y(x + 1) =2-x

     y =\frac{2-x}{x+1}

    Now find y' and y''
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  3. #3
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    Re: Basic calculus problem.

    Quote Originally Posted by RussianScience View Post
    [-y'(x+1)]/[(x+1)^2] - (-1-y)/[(x+1)^2] = 0 restated in a more useful manner.
    Very bad. You have just decreed that the second derivative is everywhere zero (0).

    Try = y" rather than = 0.
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  4. #4
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    Re: Basic calculus problem.

    Quote Originally Posted by TKHunny View Post
    Very bad. You have just decreed that the second derivative is everywhere zero (0).

    Try = y" rather than = 0.
    That clears so much up, thanks for finding my stupid mistake!
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Basic calculus problem.

    An alternative (implicit differentiation):

    x+xy+y=2\Rightarrow 1+y+xy'+y'=0\Rightarrow y'+y'+xy''+y''=0\Rightarrow

    y''=\dfrac{-2y'}{x+1}

    Of course, in this case we can substitute y'=\ldots and y=\ldots .
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