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Math Help - Separable Differential Equation

  1. #1
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    Separable Differential Equation

    I am having trouble with this equation:

     dy/dx = y + 4/y


    I multiplied by y to get a common denominator, but then got stucked there. Can anyone help me?
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  2. #2
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    Quote Originally Posted by littlesohi View Post
    I am having trouble with this equation:

     dy/dx = y + 4/y


    I multiplied by y to get a common denominator, but then got stucked there. Can anyone help me?
     \frac{dy}{dx} = y + \frac{4}{y}

     \frac{dy} {y + \frac{4}{y}} = 1~dx

     \frac{dy} { \frac{y^2+4}{y}} = 1~dx

     \frac{y}{y^2+4} ~dy = 1~dx

     \int \frac{y}{y^2+4} ~dy = \int 1~dx

    Use u = y^2+4 on the LHS
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  3. #3
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    Okay I did that and got:

     ln y^2 + 4 = t + c

    took the power of e to cancel ln :

    y^2 + 4 = e ^{t+c}


    since this is an initial-value problem y(0) = 3 .. you solved for C and got that C is approx. 2.56 .. but my answer is not quite similar to the one of the back of the book which says

     y=\sqrt{13e^{2t}-4}
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  4. #4
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     \int \frac{y}{y^2+4} ~dy = \int 1~dx

     \frac{1}{2}\int \frac{2y}{y^2+4} ~dy = \int 1~dx

     \frac{1}{2}\int \frac{du}{u} = x+C

     \frac{1}{2}\ln u = x+C

     \frac{1}{2}\ln (y^2+4) = x+C

     \ln (y^2+4) = 2x+C

     y^2+4 = e^{2x+C}

     y^2+4 = e^{2x}e^{C}

     y^2+4 = Ae^{2x}

     y^2 = Ae^{2x}-4

     y = \sqrt{Ae^{2x}-4}

    Now use y(0) = 3 to solve for A

     3 = \sqrt{Ae^{2\times 0 }-4} = \dots
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  5. #5
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    Using alternative (and better) notation:

    \frac{dy}{dx} = y + \frac{4}{y}

    \frac{dy}{dx} = \frac{y^2 + 4}{y}

    \left(\frac{y}{y^2 + 4}\right)\frac{dy}{dx} = 1

    \int{\left(\frac{y}{y^2 + 4}\right)\frac{dy}{dx}\,dx} = \int{1\,dx}

    \int{\frac{y}{y^2 + 4}\,dy} = x + C_1

    \frac{1}{2}\int{\left(\frac{1}{y^2 + 4}\right)2y\,dy} = x + C_1

    Now let u = y^2 + 4 so that \frac{du}{dy} = 2y


    \frac{1}{2}\int{\frac{1}{u}\,\frac{du}{dy}\,dy} = x + C_1

    \frac{1}{2}\int{\frac{1}{u}\,du} = x + C_1

    \frac{1}{2}\ln{|u|} + C_2 = x + C_1

    \frac{1}{2}\ln{|y^2 + 4|} + C_2 = x + C_1

    \frac{1}{2}\ln{(y^2 + 4)} = x + C_1 - C_2 (notice we can get rid of the modulus because y^2 + 4 is always positive

    \ln{(y^2 + 4)} = 2x + C, where C = 2(C_1 - C_2)

    y^2 + 4 = e^{2x + C}

    y^2 + 4 = e^{2x}e^{C}

    y^2 + 4 = Ae^{2x} where A = e^{C}

    y^2 = Ae^{2x} - 4

    y = \pm \sqrt{Ae^{2x} - 4}.


    And now using your initial condition that y(0) = 3

    3 = \pm \sqrt{Ae^{2(0)} - 4}

    3 = \pm \sqrt{A - 4}

    9= A - 4

    A = 13.


    So finally, your solution is y = \pm \sqrt{13e^{2x} - 4}.
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