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: Solve bernoulli differential equation with extra constant

  1. Nov 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Find a solution for:
    [tex]u'(t)=c*u(t)^2-c*(a+b)*u(t)+c*a*b[/tex]

    3. The attempt at a solution
    I've found the solution for the homogeneous equation:
    [tex]u_0(t)=(\frac{1}{a+b}+d*e^{c(a+b)t)})^{-1}[/tex]
    Where c is a random constant.

    Now I've tried the solution [tex]u(t)=x(t)*u_0(t)[/tex], when I fill this in it gets a mess and i can't figure out what x(t) should be.
    Can someone help me?
     
  2. jcsd
  3. Nov 16, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This isn't a linear equation, so I'm not sure what you consider the "homogeneous" equation. If you write it as$$
    u'+ c(a+b)u =cu^2+cab$$at least the left side of the equation is linear. That means if you can find a solution ##u_1## of ##u'+ c(a+b)u =cu^2## and a solution ##u_2## of ##u'+ c(a+b)u =cab## then ##u_1+u_2## should be a solution of your original DE.
     
  4. Nov 16, 2012 #3
    With homogeneous I just ment the differential equation without the constant part.

    But your approach won't work because [tex]u'(t)=u_1'(t)+u_2'(t)=-c(a+b)u_1(t)+cu_1^2(t)-c(a+b)u_2(t)+cab=-c(a+b)u(t)+cu_1^2(t)+cab[/tex]

    Or am I seeing it wrong?
     
  5. Nov 16, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I would forget Bernoulli's equation and just try separation of variables. The right side factors. Just use partial fractions to integrate it.
     
  6. Nov 16, 2012 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Right. You can't combine them that way. Just separate.
     
  7. Nov 16, 2012 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are seeing it wrong. If you let$$
    L(u) = u' + c(a+b)u$$be the left side linear part then ##L## is a linear operator, which means ##L(u_1+u_2) = L(u_1)+L(u_2)##. So if ##L(u_1) = cu_1^2## and ##L(u_2)=abc##, what do you get for ##L(u_1+u_2)## ?
     
  8. Nov 16, 2012 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    But you want to get ##L(u_1+u_2) = c(u_1+u_2)^2+abc##. You don't get that.
     
  9. Nov 16, 2012 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No wonder I was feeling a little uncertainty about my argument. :blushing: And it seemed so neat at the time...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook