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Thread: Separable Differential Equation

  1. #1
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    Separable Differential Equation

    I am having trouble with this equation:

    $\displaystyle 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:

    $\displaystyle dy/dx = y + 4/y $


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

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

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

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

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

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

    $\displaystyle ln y^2 + 4 = t + c $

    took the power of e to cancel ln :

    $\displaystyle 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

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

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

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

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

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

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

    $\displaystyle y^2+4 = e^{2x+C}$

    $\displaystyle y^2+4 = e^{2x}e^{C}$

    $\displaystyle y^2+4 = Ae^{2x}$

    $\displaystyle y^2 = Ae^{2x}-4$

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

    $\displaystyle y^2 + 4 = e^{2x + C}$

    $\displaystyle y^2 + 4 = e^{2x}e^{C}$

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

    $\displaystyle y^2 = Ae^{2x} - 4$

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


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

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

    $\displaystyle 3 = \pm \sqrt{A - 4}$

    $\displaystyle 9= A - 4$

    $\displaystyle A = 13$.


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