Jump to content


  • Posts

  • Joined

  • Last visited

Posts posted by allen_83

  1. 1. Since the equation is non-linear, what makes you think that the solution is expressible as the sum of "the general homogeneous solution" plus an "inhomogeneous solution" ? When the differential operator is linear, any two solutions of the inhomogeneous equation differ by a solution of the homogeneous equation. Not so for non-linear equations.

    alright, wanted to give "the tree"'s idea a try.

    got no clue. am stuck.

    this is a type : [math] x'' = f(t,x,x') [/math] with both x and t missing if I were to think of it as a non-linear DE.

    [math] x(t) = - \frac {t^2}{2} - at + c [/math]

    how ?

  2. Sorry, brain fart.


    None of the techniques discussed are going to work, because this ODE is non-linear.


    [math] \dfrac {d^2 \ x}{dt^2} - e^x = -1[/math]


    Given the OP's statement that a solution in terms of a general solution to the homogeneous equation plus a particular solution to the inhomogeneous equation is expected (which applies to a linear ODE), I suspect that the equation written is not the equation of the real problem.


    Should it be [math] \dfrac {d^2 \ x}{dt^2} = e^t -1[/math] ?


    If so, then integrate twice.

    That's right. This ODE is non-linear.

    The problem is as it is.

    [math] x''(t) = e^{ x(t)} - 1 for x'(0)= 0 and x(0) = 1 [/math]


    [math]x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0 [/math]

    solvin for D1 and D2 :

    D1 = 1, D2 = 0

    and pluging them back in :

    [math] x(t) = cos(t)[/math]


    [math]x_{nonhom}(t) = A\cdot e^x - \dfrac{x^2}{2}[/math]


    [math] x'_{nonhom}(t) = A\cdot e^x - x[/math]


    [math]x''_{nonhom}(t) = A\cdot e^x - 1 [/math]


    [math]A\cdot e^x - 1 + 1 = e^x \Rightarrow A = 1[/math]


    [math]x_{nonhom}(t) = e^x - \dfrac{x^2}{2}[/math]

    [math] x = x_{nonhom}(t) + x_{hom}(t)[/math]

    [math] x = e^{x(t)} - \dfrac{x(t)^2}{2} + cos(t) [/math]


    correct me if I'm wrong.

  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.