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Math Help - Bernoulli's equation

  1. #1
    Senior Member
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    Bernoulli's equation

    solve the equation:
    (3\sin{y} - 5x)dx +2x^2\cot{y}dy=0

    this is a tough problem for me that i cant start...
    i can't find any separable nor any linearity in this solution

    ill try to solve
    solution:
    (3\sin{y} - 5x)dx +2x^2\frac{\cos{y}}{\sin{y}}dy=0

    substitute let w = sin(y)
    dw = cos(y)dy so substitute

    (3w - 5x)dx + 2x^2\frac{dw}{w} = 0

    \frac{3w - 5x}{2x^2}dx = \frac{dw}{w}

    2x^2dw - 5xwdx = -3w^2dx

    \frac{dw}{dx} - \frac{5w}{2x} = \frac{-3w^2}{2x^2}
    ....
    now im stuck
    Last edited by ^_^Engineer_Adam^_^; January 9th 2008 at 04:56 AM.
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  2. #2
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276

    Further transform

    Let us define another change of variable for our dependent variable: choose u=\frac{5}{3w}.
    Then w=\frac{5}{3u}, dw=-\frac{5}{3u^2}du, and thus our equation becomes
    -\frac{5}{3u^2}\frac{du}{dx} = \frac{25}{6ux} - \frac{25}{6u^2x^2}

    \frac{du}{dx} = -\frac{5}{2x}u + \frac{5}{2x^2}

    \frac{du}{dx} + \frac{5}{2x}u = \frac{5}{2x^2}
    This is a nonhomogenous first order linear differential equation. We can see by substitution that u=\frac{5}{3x} is a solution. Thus, we find and add to that the general solution for the corresponding homogenous equation
    \frac{du}{dx} + \frac{5}{2x}u = 0, which is separable:
    \frac{du}{u} = \frac{-5}{2x}
    \int\frac{du}{u} = \int\frac{-5}{2x}
    \ln{u} = \frac{-5}{2}\ln{x}+c
    u = \frac{C}{x^{\frac{5}{2}}}
    Thus the solution to \frac{du}{dx} + \frac{5}{2x}u = \frac{5}{2x^2} is u=\frac{5}{3x}+\frac{C}{x^{\frac{5}{2}}}, and from there you can convert back into w, and then into y=\arcsin{w}.

    --Kevin C.
    Last edited by TwistedOne151; January 9th 2008 at 03:56 PM. Reason: typo in math
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