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Looks same as distinct root solution

  1. Apr 18, 2012 #1
    ay'' + by' + cy = 0

    1) for 2 distinct roots,the solution should be y = Ae^m1x + Be^m2x

    2) for Real and equal roots should be
    y = e^mx(A + Bx) that is it.

    So if same roots as the 2nd one,one solution is y1(x)=Ae^mx, what about the second solution?
    there is a second solution given by y2(x)=Bxe^mx

    is this should be y2(x)=Be^mx also? because looks same as distinct root solution, and we can just put the 2 roots in the same just act like there have 2 distinct roots but are the same.
    Last edited: Apr 18, 2012
  2. jcsd
  3. Apr 18, 2012 #2


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    Re: help

    Hey doey and welcome to the forums.

    I can't remember the derivation (but I have seen it before), but what I recommend for you to show that it is true is take that expression, calculate the first and second derivatives of it which generate the coefficients for your quadratic and then show that you get two equal real roots for the solution of the quadratic.

    It's not the same as going from DE to solution proof, but it is a proof never the less since the function is unique and can be written in DE form, meaning that the DE has the unique solution as the function and thus will have a DE form that you can show to have the properties for the roots.
  4. Apr 18, 2012 #3


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    Re: help

    One way of deriving that solution is to use the "reduction of order" method.

    If [itex]\alpha[/itex] is a double root of the characteristic equation then the characteristic equation must be of the form [itex](r- \alpha)^2= r^2- 2\alpha r+ \alpha^2= 0[/itex] which means that the differential equation must be of the form
    [tex]\frac{d^2y}{dx^2}- 2\alpha\frac{dy}{dx}+ \alpha^2y= 0[/tex]

    And, as you say, one solution is [itex]e^{\alpha x}[/itex]. So look for a solution of the form [itex]y(x)= u(x)e^{\alpha x}[/itex] where u(x) is a function of x. [itex]y'= u'e^{\alpha x}+ \alpha ue^{\alpha x}[/itex] and [itex]y''= u''e^{\alpha x}+ 2\alpha u'e^{\alpha x}+ \alpha^2 u^{\alpha x}[/itex].

    Putting those into the equation gives
    [tex]u''e^{\alpha x}+ 2\alpha u'e^{\alpha x}+ \alpha^2 u e^{\alpha x}- 2\alpha(u'e^{\alpha x}+ \alpha ue^{\alpha x})+ u(x)e^{\alpha x}= 0[/tex]
    which reduces to
    [tex]u''e^{\alpha x}= 0[/tex]
    That is, the terms involving u, without a derivative, cancel because without differentiating, we are treasting u as a constant and [itex]e^{\alpha x}[/itex] satisfies the differential equation (i.e. because [itex]\alpha[/itex] is a root of the characteristic equation). And all terms involving u' cancel because [itex]\alpha[/itex] is a double root of that equation.

    Of course, we can divide both sides by [itex]e^{\alpha x}[/itex] (which is never 0) to get [itex]u''= 0[/itex]. Since the second derivative of u is always 0, the first derivative must be a constant- u'= C for some number C. Integrating again, u=Cx+ D. That means our solution [itex]y= u(x)e^{\alpha x}= (Cx+ D)e^{\alpha x}= Cxe^{\alpha x}+ De^{\alpha x}[/itex]. The "[itex]De^{\alpha x}[/itex]" is the original solution. What is new is the [itex]xe^{-\alpha x}[/itex] term.

    Of course, once you realize that, you don't need to do that analysis every time!
  5. Apr 18, 2012 #4
    Re: help

    i guess i know what you means,but only a way there i not really sure,is there we can just put the front of e^mx with an function ? by making it into u(x)e^mx.
    Last edited: Apr 18, 2012
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