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Help with nonlinear ordinary differential equation

  1. Apr 22, 2008 #1
    Any help with solving this first-order nonlinear ODE would be greatly appreciated! I do believe that an explicit solution exists.

    1. The problem statement, all variables and given/known data

    dy/dt = 1/(4t^2) + 1/2 + 1/2*y/t - 1/(2t)*((1+4ty)^(1/2))

    I was led to believe that it could be solved by turning it into a linear differential equation. Perhaps by an appropriate substitution, although I have not had any luck with finding something that works here.
     
  2. jcsd
  3. Apr 22, 2008 #2
    I can separate the rewrite the equation to look like:

    dy/dt = 1/2 - 1/2*y/t + [1/(4t^2) + y/t] - [1/(4t^2) + y/t]^1/2

    so if I substitute v = [1/(4t^2) + y/t] I get

    dy/dt = 1/2 - 1/2*y/t + v - v^1/2

    after more re-arranging i then get

    dy/dt = 1/2 + 1/(8t^2) + v/2 - v^(1/2)

    and dv/dt = -1/(2t^3) - y/t^2 + 1/t*dy/dt

    which can be rewritten as dv/dt = -1/(4t^3) - v/t + 1/t*dy/dt

    but substituting in for dy/dt gives me big mess with v's and t's but nothing I know what to do with

    Please help!!!
     
    Last edited: Apr 22, 2008
  4. Apr 22, 2008 #3
    [tex]
    \frac{dy}{dt} = \frac{1}{4t^2} + \frac{1}{2} + \frac{y}{2t} - \frac{\sqrt{1+4ty}}{2t}
    [/tex]

    It looks better this way...


    Using [tex] v = \frac{1}{4t^2} + \frac{y}{t} [/tex]

    I can reduce this equation to the following:

    [tex]
    \frac{dv}{dt} = \frac{1}{2t} - \frac{1}{8t^3} - \frac{v}{2t} - \frac{\sqrt{v}}{t}
    [/tex]

    but from there I'm stuck...

    Any ideas?? Maybe I didn't pick the right expression for v...
     
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