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Why y=e^mx is taken as trial soln ?

  1. Dec 11, 2005 #1
    why y=e^mx is alwayas taken as trial soln in solving 2nd order diff equations. please explain in details.
     
  2. jcsd
  3. Dec 11, 2005 #2

    LeonhardEuler

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    You would not always take [itex]ce^{mx}[/itex] as a trial solution to any second order differential equation, only homogenius linear ones with constant coefficients. In fact you should take [itex]ce^{mx}[/itex] as a trial solution to any homogeneous linear ordinary differential equation with constant coefficients. Here is why: The fundamental theorem of algebra says that any nth degree polynomial has exactly n complex roots (including multilicity). When you assume a trial solution of [itex]ce^{mx}[/itex] to one of these equations what happens? The equation is of the form:
    [tex]a_0y +a_1y' +...+ a_ny^{(n)}=0[/tex]
    Assuming the trial solution:
    [tex]a_0ce^{mx} + a_1mce^{mx} +...+ a_nm^{n}ce^{mx}=0[/tex]
    Canceling [itex]ce^{mx}[/itex] gives:
    [tex]a_0+ a_1m + a_2m^2 +...+a_nm^n=0[/tex]
    Which is an nth degree polynomial. We are guaranteed n solutions to this polynomial, and we know that an nth degree linear differential equation has n linearly independent solutions. So, as long as all the roots are distinct, we have solved the problem since [itex]e^{mx}[/itex] and [itex]e^{lx}[/itex] are linearly independent when [itex]m \neq l[/itex]. Otherwise we need to try other solutions as well. The other solutions that work in this case are of the form [itex]x^{\alpha}e^{mx}[/itex].
     
    Last edited: Dec 11, 2005
  4. Dec 11, 2005 #3

    matt grime

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    You don't do that for *all* DE's, only linear ones, and only to get the homogeneous solutions. It will not work in general. But we can prove that in this particular case that these are the only solutions, though this proof is somewhat difficult and of no interest here.

    Some clever person noted that this method always gives (when done properly) in this case the two things that we know exist.

    So the simple answer is: because it works.
     
  5. Dec 11, 2005 #4

    Tide

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    In addition to what Leonhard and Matt said, you can always do a power series solution (Frobenius). That can be rather tedious but for the types of equations noted the power series method always gives the equivalent of exponential (possibly complex) functions.
     
  6. Dec 11, 2005 #5
    Just as an addendum. Exponentials are very easy to work with. I mean, they are easy to integrate, derivate, multiply, devide, ...

    For example, in computational physics, one will always try to express the wavefunction of a many body problem in terms of the product of many single body wave functions. Such wavefunctions are expressed in terms of Gaussians (ie exponentials), like the SZ and DZP basis sets, for the above calculatory advantages. So exponentials not only rule math, they also rule QM, QFT, ...err, in short : PHYSICS :wink:

    regards
    marlon
     
  7. Dec 11, 2005 #6

    HallsofIvy

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    Actually only for linear equations with constant coefficients.

    The point is that if we are to cancel different derivatives, multiplying only by constants, the derivatives must be the same "kind" of function. Since the derivative of ekx is kekx, that works nicely. Of course, the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) so sine and cosine would also work nicely. The derivative of any polynomial is also a polynomial so polynomials would work. It happens that ekx is simpler to use than those.
     
  8. Dec 29, 2005 #7
    Code (Text):
    The fundamental theorem of algebra says that any nth degree polynomial has exactly n complex roots (including multilicity).
    What do u exactly mean by the above? Say a third degree polynomial does not have exactly 3 complex roots. Complex roots always occur in conjugate pairs ie. in any polynomial complex roots must occur in even nos.
     
  9. Dec 29, 2005 #8
    What do YOU mean? It's a statement of the fundamental theorem of algebra. It's not his statement. And remember that the real numbers are a subset of the complex numbers, all real numbers are complex numbers of the form a + bi, only that b = 0.
     
  10. Dec 29, 2005 #9

    LeonhardEuler

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    What d_leet said pretty much answers the question, but I also just wanted to point out that complex roots do not always come in conjugate pairs. This is the case for polynomials with real coefficients, but not true when complex coefficients are allowed. For example the polynomial you get by expanding (x-(1+i))(x-1)2 would clearly have 1+i as a root but not 1-i. That said, I agree with you that my wording was somewhat misleading. A better way to phrase it would be "has exactly n roots in the field of complex numbers".
     
  11. Dec 30, 2005 #10

    arildno

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    L. Euler:
    Perhaps using a polynomial like x-i gets the message across better than the one you chose..
    Not that there was anything wrong in your answer, though..
     
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