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Math Help - Differential Equations-Separation of variables

  1. #1
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    Differential Equations-Separation of variables

    I need some help with this problem please:

    solve by separation of variables.

    dydx=y(1-y^2)

    1/(y(1-y^2)dy=dx

    How do I get it into y= form after taking the integral?
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  2. #2
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    Quote Originally Posted by yellowrose View Post
    I need some help with this problem please:

    solve by separation of variables.

    dydx=y(1-y^2)

    1/(y(1-y^2)dy=dx

    How do I get it into y= form after taking the integral?
    y(1 - y^2) = y(1 - y)(1 + y), so you have a case for partial fraction decomposition:

    \frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y} ........
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    y(1 - y^2) = y(1 - y)(1 + y), so you have a case for partial fraction decomposition:

    \frac{1}{y(1 - y)(1 + y)} = \frac{A}{y} + \frac{B}{1 - y} + \frac{C}{1 + y} ........
    Could you tell me how I solve for A, B, and C? Thanks!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yellowrose View Post
    Could you tell me how I solve for A, B, and C? Thanks!
    multiply through by y(1 - y)(1 + y) to get 1 = A(1 - y)(1 + y) + By(1 + y) + Cy(1 - y) .......(1)

    then, plug in y = 0 into (1) and solve for A

    then, plug in y = 1 into (1) and solve for B

    then plug in y = -1 into (1) and solve for C
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  5. #5
    Math Engineering Student
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    \frac{1}<br />
{{y\left( {1 - y^2 } \right)}} = \frac{{\left( {1 - y^2 } \right) + y^2 }}<br />
{{y\left( {1 - y^2 } \right)}} = \frac{1}<br />
{y} + \frac{y}<br />
{{1 - y^2 }}.

    Easily integrable.
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