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Bernoulli equations with trig

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of the following differential equation.
    y'+4xy=10xy2cos(x2)

    2. Relevant equations
    The usual Bernoulli equation ones:
    y'+p(x)y = g(x)ya
    u(x)=[y(x)]1-a
    u'+(1-a)pu = (1-a)g

    3. The attempt at a solution
    I got up until the general solution part.. I'll just type bits of it out because it'll take me ages (I'm a slow typer).

    So in the equation, a=2, p(x)=4x, g(x)=10xcos(x2)

    Change of variables:
    u(x)=[y(x)]1-a = y-1

    and u'+(1-a)pu = (1-a)g
    => u'-4xu = -10xcos(x2)

    Now here's where I get confused..
    General solution:
    u= e[tex]\int[/tex]p(x)dx[[tex]\int[/tex]r(x)e[tex]\int[/tex]p(x)dxdx+c
    where p=-4x, r=-10xcos(x2)

    u= e[tex]\int[/tex]-2x^2[[tex]\int[/tex](-10xcos(x2)e[tex]\int[/tex]-2x^2)dx+c

    Argh.. that's messy, I hope it makes sense.

    Anyway, I can't seem to figure out that integral... I've used parts and stuff but I don't seem to be getting anywhere.

    Any help would be appreciated.
     
    Last edited: Mar 29, 2009
  2. jcsd
  3. Mar 29, 2009 #2
    Nevermind, I figured it out.
     
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