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Help on Diff Equation!

  1. Sep 17, 2008 #1
    Question:
    I have to find the constant a such that [tex] y(x) = e^{ax}[/tex] is a solution for x" = x

    My answer:
    I tried 2 ways of reasoning and they both led to my answer that a can be any number (but zero).
    Can anyone see if they are correct?? If not, pls give me some hint!

    1. x= ln(y)/a but x" is a 2nd derivative with respect to x. so x" = 0 => x=0 =>[tex] a\in R [/tex] \0

    2. x has to be a constant for the solution to work, and it has to be specifically 0, so a is any real number but 0.
     
  2. jcsd
  3. Sep 17, 2008 #2

    Defennder

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    I don't see what does y = e^(ax) got to do with x'' = x. The latter is actually a 2nd order derivative with respect to some variable like t (or y) is it not? Because it is evident that d^2/dx^2 x = 0.
     
  4. Sep 17, 2008 #3
    Thanks for replying!

    Actually this is one part of a big question where similar DE are given with the same solution(where others have y) so the solution y=exp(ax) seems irrelevant in this case.

    The question didn't state x" is being differentiated w.r.t. what variable, so I just assumed it's dx...

    This may sound dumb... but.. :confused::
    So you mean as the whole equation is based on x, I could simply jump ahead to conclude that a is unbounded? even without the "not equal to zero" condition? I added the "not equal 0" condition as x = ln(y)/a (is this unnecessary?)

    Also does it mean that my #1 or 2 reasoning are totally unnecessary?
     
  5. Sep 17, 2008 #4
    I'm confused. Can't you just make a characteristic equation and make the equation
    r^2 - 1 = 0, and then solve for r? Yielding simply 1 and -1? I haven't done this in a while, sorry.
     
  6. Sep 17, 2008 #5

    Defennder

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    No I still don't follow you at all. You mentioned that it was part of a larger question, so perhaps you could post that part of the question as it is? Because I don't know what you're talking about.
     
  7. Sep 17, 2008 #6
    sr6622: sorry, I still haven't learned characteristic equation. so I can't use the technique yet.. but thanks!

    Defennder:
    Ohhhh. I mean it's a big question but each part has nothing to do with each other. They are just unrelated, like part i) is y' + y =0 and we are asked to do the same thing (finding the constant a with y(x)=exp(ax) is the solution).

    so I guess as the "solution" y(x)=exp(ax) is irrelevant I can have a to be any number I want then.,,?
     
    Last edited: Sep 17, 2008
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