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

  1. Mar 4, 2009 #1

    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?

  2. jcsd
  3. Mar 4, 2009 #2


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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" [Broken]

    Using this for your equation you get:


    from which:


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


    The iterative scheme is now:


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







    Hope this helps,

    Last edited by a moderator: May 4, 2017
  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 http://mathworld.wolfram.com/LambertW-Function.html" [Broken].
    Last edited by a moderator: May 4, 2017
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