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Bernoulli differential equation, i got the form down, where did i f it up? :-x

  1. Jan 22, 2006 #1
    Hello everyone! ITS ME! i'm having a good time with some Bernoulli differential equations, and yet it didn't work. Here is the directions:
    A Bernoulli differential equation is one of the form:
    [​IMG]
    Observe that, if n=0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u=y^{1-n} transforms the Bernoulli equation into the linear equation
    [​IMG]
    Use an appropriate substitution to solve the equation
    [​IMG]
    and find the solution that satisfies y(1)=3.

    y(x) =

    Okay here is my work:
    P(x) = 1/x;
    Q(x) = 9
    n = 2

    u' + (-1)(1/x)*u = (-1)*9
    so integrating facotr would be:
    I(t) = e^(-1/x)
    I = -x;

    -x*u' + u = 9x;
    integration gives:
    u = (-9/2)*x -c/x

    applying intial condition: y(1) = 3;
    3 = -9/2 - C/1
    C = -15/2;

    Plugging in constant of integration and solving for u gives me:
    u = (-9/2)*x +(15/2);
    which is wrong of course! any idea why? Thank you! (this is the first time doing a problem of this form so im' not sure if i'm suppose to plug somthing back in or what?);

    THANKS
     
  2. jcsd
  3. Jan 22, 2006 #2

    TD

    User Avatar
    Homework Helper

    From a first look, your Q(x) is wrong. As you see, Q may be a function of x and you just took 9, but the coefficient of y² was 9x.

    By using the Bernouilli substitution, i.e. z = 1/y, your DE becomes

    [tex]xy' + y = 9xy^2 \mathop \to \limits_{z' = - y'/y^2 }^{z = y^{ - 1} } - xz' + z = 9x[/tex]
     
  4. Jan 23, 2006 #3
    Thanks TD, i'm kinda confusd on your notation of
    z = y^-1
    ->
    z' = -y'/y^2

    what do u mean? Also I fallowed an example program my professor did and he just used the number, but i agree i think it should be 9x
     
  5. Jan 23, 2006 #4

    TD

    User Avatar
    Homework Helper

    Well my z is just your u, for the Bernouilli equation of degree n, the substitution is - as you said - u = y^(1-n). With n being 2 here, and your u being my z, that gives us z = y^(-1), no? Then z' is just dz/dx.
     
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