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Homework Help: Don't understand this ODE

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm currently taking a course on ordinary differential equations. I am now reading through the lecture slides but I'm not really sure about the " factorising the equation " part onwards:


    2. Relevant equations

    3. The attempt at a solution

    I'm not sure what is the point of trying to factorize the differential equation and how the CF is made up of exponentials..

    Secondly, how did they simply decide that x*emx gives an independent solution? Why not x2emx then?
  2. jcsd
  3. Dec 13, 2012 #2


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    Science Advisor

    First, the "factorizing" is an "operator" method using that fact that we can treat repeated derivatives (as long as the coefficients are constants) as if they were products and so get a symbolic way of reducing the equation to a series of first order equations.

    Second, If we have a differential equation of the form [itex]d^2y/dx- 2m dy/dx+ m^2y= 0[/itex] so that "trying" a solution of the form [itex]y= e^{rx}[/itex] reduces the equation to [itex]r^2 e^{rx}- 2mr e^{rx}+ m^2e^rx= e^{rx}(r^2- 2mr+ m^2)= e^{rx}(r- m)^2= 0[/itex] and, since [itex]e^{mx}[/itex] is never 0, we must have [itex](r- m)^2= 0[/itex] which, of course, has the single root r= m. That tells us immediately that [itex]y= e^{mx}[/itex] is a solution to the differential equation. Of course, because this is a second order linear, homogeneous solution, we need a second independent solution to be able to write the general solution.

    It is easy to see, by checking, that [itex]y= xe^{mx}[/itex] is a solution: if [itex]y= xe^{mx}[/itex] then [itex]dy/dx= e^{mx}+ mxe^{mx}[/itex] and [itex]d^2y/dx^2= 2me^{mx}+ m^2xe^{mx}[/itex] so that [itex]d^2y/dx^2- 2m dy/dx+ m^2y= (2me^{mx}+ m^2xe^{mx})- 2m(e^{mx}+ mxe^{mx})+ xe^{mx}= (m^2xe^{mx}- 2m^2xe^{mx}+ m^2e^{mx})+ (2me^{mx}- 2me^{mx})= 0[/itex].

    We can then appeal to the theory, that the set of all solutions to a second order linear homogeneous differential equation form a vector space of dimension 2 and so can be spanned by two independent solutions, to see that there cannot be other independent solutions.

    And, of course, we see, by direct substitution, that is [itex]y= x^2e^{mx}[/itex] then [itex]dy/dx= 2xe^{mx}+ mx^2e^{mx}[/itex] and [itex]d^2y/dx^2= 2e^{mx}+ 4mxe^{mx}+ m^2x^2e^{mx}[/itex] and the equation becomes [itex](2e^{mx}+ 4mxe^{mx}+ m^2x^2e^{mx})- 2m(2xe^{mx}+ mx^2e^{mx})+ m^2x^2e^{mx}= (m^2x^2e^{mx}- 2^2x^2e^{mx}+ m^2x^2e^{mx})+ (4mxe^{mx}- 4mxe^{mx})+ (2e^{mx}= 2e^{mx}[/itex], not 0.

    How would we know in advance that worked? Experience- and they are trying to save you the time required to get that experience.
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