How to solve 2nd order ODE solution eg. te^t+e^t, for t?

  1. Mar 4, 2009 #1
    Hi,

    I have a second order differential equation with a solution in the form:

    [tex]f(t) = Ae^{t}+Bte^{t}[/tex]

    I want to solve for t, ie. work out for what value of t does the function f(t) have a particular value. But there seems to be no way (that I know of) to do this. Can anyone give me any pointers to what to do here?

    Thanks.
     
  2. jcsd
  3. Mar 4, 2009 #2

    lurflurf

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    Homework Helper

    You will need a function such as the product-log to do that.
     
  4. Mar 5, 2009 #3
    Another way would be to use the Newton-Raphson iterative scheme. Here is a link:

    http://nl.wikipedia.org/wiki/Newton-Raphson

    Using this for your equation you get:

    [tex]f=\alpha=(A+Bt)e^t[/tex]

    from which:

    [tex]g=\alpha-(A+Bt)e^t=0[/tex]

    The function to be solved. The derivative is found to be:

    [tex]g'=-(A+B+Bt)e^t[/tex]

    The iterative scheme is now:

    [tex]t_{n+1}=t_n+\frac{\alpha-(A+Bt)e^t}{(A+B+Bt)e^t}[/tex]

    Start with [itex]t_0=0[/itex], giving for the example [itex]A=B=1[/itex], [itex]\alpha=3[/itex]:

    [tex]0[/tex]

    [tex]1[/tex]

    [tex]0.701213[/tex]

    [tex]0.622262[/tex]

    [tex]0.617657[/tex]

    [tex]0.617642[/tex]

    Hope this helps,

    coomast
     
  5. Mar 5, 2009 #4
    Maple 12 suggests [tex]t = \text{LambertW}\left( \frac{f\cdot \exp{\frac AB}}B\right) - \frac AB[/tex]

    see here.
     
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