Results 1 to 2 of 2

Thread: Bernoulli's equation

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Bernoulli's equation

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

    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:
    $\displaystyle (3\sin{y} - 5x)dx +2x^2\frac{\cos{y}}{\sin{y}}dy=0$

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

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

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

    $\displaystyle 2x^2dw - 5xwdx = -3w^2dx$

    $\displaystyle \frac{dw}{dx} - \frac{5w}{2x} = \frac{-3w^2}{2x^2}$
    ....
    now im stuck
    Last edited by ^_^Engineer_Adam^_^; Jan 9th 2008 at 04:56 AM.
    Follow Math Help Forum on Facebook and Google+

  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 $\displaystyle u=\frac{5}{3w}$.
    Then $\displaystyle w=\frac{5}{3u}$, $\displaystyle dw=-\frac{5}{3u^2}du$, and thus our equation becomes
    $\displaystyle -\frac{5}{3u^2}\frac{du}{dx} = \frac{25}{6ux} - \frac{25}{6u^2x^2}$

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

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

    --Kevin C.
    Last edited by TwistedOne151; Jan 9th 2008 at 03:56 PM. Reason: typo in math
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Bernoulli's Equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Sep 6th 2011, 04:44 AM
  2. bernoulli's equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Jul 25th 2011, 06:00 PM
  3. Bernoulli equation particular
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 26th 2008, 09:51 AM
  4. bernoulli equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 10th 2008, 09:52 AM
  5. Bernoulli's Equation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 23rd 2007, 06:54 AM

Search Tags


/mathhelpforum @mathhelpforum