Special case of nonlinear first order ordinary differential equation.

  1. Hi there,

    I've having problems solving a particular nonlinear ODE. Any help/suggestions will be highly appreciated.

    The nonlinear ODE is:

    v[t]*v'[t] + (4*v[t])/(t^2 - 1) = t/(t^2 - 1)

    Thank you.
     
  2. jcsd
  3. If:

    [tex]z=\int f(x)dx[/tex]

    then:

    [tex]\frac{dz}{dx}=f(x)[/tex]

    or:

    [tex]\frac{dx}{dz}=\frac{1}{f}[/tex]

    now just substitute back into the original DE
     
  4. Thanks for your reply, I understand your point but if you look at the example in the link I posted, you will see that the substitution is not as straightforward as you put it.

    Seems to me that there's some sort of parametrization to be done.
     
  5. . . . what did I get myself into. But that's ok, need to be willing to try things in math to succeed and not let the risk of failure stop you but I'd rather not make two mistakes in one day. So I think it's this below but I'd have to work on it more with real problems to verify it ok.


    Suppose you have:

    [tex]vv'=f(t)v+g(t)[/tex]

    and you let:

    [tex]z=\int f dt[/tex]

    to obtain:

    [tex]v\frac{dv}{dz}=v+\Phi(t)[/tex]

    where:

    [tex]\Phi(t)=\frac{g(t)}{f(t)}[/tex]

    and suppose we're able to solve:

    [tex]v\frac{dv}{dz}=v+\Phi(z)=v+\frac{g(z)}{f(z)}[/tex]

    say for example the solution is:

    [tex]v(z)=z^3-3\sin(z)[/tex]

    but z is paramaterized by t so that we can write the solution for the original problem as:

    [tex]v(t)=z^3-3\sin(z);\quad z=\int f(t)[/tex]

    I'm not sure at this point but it's what I'm going with for now until I can verify that or someone can correct me.

    Also, I should mention this holds for the simple case:

    [tex]vv'=tv+t[/tex]

    which can be solved exactly but that's still no guarantee my explanation above is correct for the general case. Never worked on this type of problem before
     
    Last edited: Sep 12, 2011
  6. Ok, after reviewing I don't think that's correct. If we have:

    [tex]y\frac{dy}{dz}=y+\Phi(t)=y+\frac{g(t)}{f(t)}[/tex]

    and:

    [tex]z=\int f(t)=h(t)[/tex]

    then I believe we'd have to compute the inverse:

    [tex]t=h^{-1}(z)[/tex]

    then solve:

    [tex]y\frac{dy}{dz}=y+\Phi\big(h^{-1}(z)\big)[/tex]

    I'm probably making a mess out of this but I'd at least like to demonstrate how sometimes real math is not so neat and pretty and quick and often fought with many wrong turns and maybe I'm still going wrong.
     
  7. Pheeww! Mathematics indeed could be rattling, been on the problem for sometime now.
    The problem I foresee with your last review is computing:

    [tex]t=h^{-1}(z)[/tex]

    Can't see how that will be done.
     
  8. Well I probably don't have it exactly right yet. However if this was my problem, I would go to the library and find one of those Russian texts cited in the Eqnworld reference and get some simple test cases with the answers and study them and maybe the references would explain how the parameterization is used. I'll look at it more later.
     
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