1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/c4/3f85cf0e6820be855c6d2a21d051b71.png [Broken]
    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
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/aa/f6e0ed9c0e920a9374d9b874f69ff21.png [Broken]
    Use an appropriate substitution to solve the equation
    http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/a5/7eb1214eb5ca480f1d8188f6dd4e9a1.png [Broken]
    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?);

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 22, 2006 #2


    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


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook