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Thread: Differential equation theory

  1. #1
    Senior Member
    Oct 2009

    Differential equation theory

    This page is about solving a second-order linear differential equation with real and coincident roots. The equation being solved was given earlier but it is almost there halfway down following the word "thus". It is homogeneous, i.e. it equals 0. I have followed the writing down to the line following "So". Can someone explain the rest?
    Attached Thumbnails Attached Thumbnails Differential equation theory-scan-1.jpg  
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  2. #2
    MHF Contributor
    Oct 2008

    Re: Differential equation theory

    Quote Originally Posted by Stuck Man View Post
    I have followed the writing down to the line following "So".
    ... so, down to but not including, "giving $\displaystyle y = Bxe^{mx}$ as a solution" ...?

    If so, the point to get is that

    $\displaystyle y = Bxe^{mx}$ (where m happens to be the repeated root of $\displaystyle az^2 + bz + c = 0 $ so that $\displaystyle m = \tfrac{-b}{2a} $ and also $\displaystyle c = \tfrac{b^2}{4a})$

    has been shown to be a solution to the differential equation because it has been shown to produce the equation. I.e. what has been shown is that setting

    $\displaystyle y = Bxe^{mx}$ (where... )


    $\displaystyle ay'' + by' + c $

    come to zero.

    Maybe the 'Now we know' line would be slightly clearer as:

    "Now we were assuming that the m in the exponent is equal to $\displaystyle \tfrac{-b}{2a}$ which, with the help of $\displaystyle c = \tfrac{b^2}{4a} $, is going to make both of $\displaystyle am^2 +bm + c $ and $\displaystyle 2am + b$ come to zero.

    Or maybe not! But you could also dispense with m in the first place and set,

    $\displaystyle y = Bxe^{\frac{-b}{2a}x}$

    ... as here: Pauls Online Notes : Differential Equations - Repeated Roots

    Otherwise, if the issue is just the putting together of this solution with the previous one at the top of the page, notice that (A + B) in the original version of the first solution has been replaced by A alone, and B in the second solution has nothing to do with B at the top of the page.
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  3. #3
    MHF Contributor

    Apr 2005

    Re: Differential equation theory

    Another way of seeing that is to use the "reduction of order" method. If we have an nth order d.e, and know one solution, u(x), then looking for a solution of the form y(x)= xu(x).

    For example, if a second order differential equation has characteristice equation with at double root, that is, of the form $\displaystyle (r- a)^2= r^2- 2ar+ a^2= 0$, then the differential equation is of the form $\displaystyle y''- 2ay'+ a^2y$. Knowing that [tex]e^{ax}[/itex] is a solution ($\displaystyle (e^{ax})''- 2a(e^{ax})'+ a^2e^{ax}= a^2e^{ax}- 2a^2e^{ax}+ a^2e^{ax}= (a^2- 2a^2+ a^2)e^{ax}= 0$), we try a solution of the form y= $\displaystyle e^{ax}u(x)$. Then $\displaystyle y'= ae^{ax}u+ e^{ax}u'$ and $\displaystyle a^2e^{ax}u+ 2ae^"{ax}u'+ e^{ax}u''$. Putting those into the equation, [tex]a^2e^{ax}u+ 2ae^{ax}u'+ e^{ax}u''- 2a(ae^{ax}u+ e^{ax}u')+ a^2(e^{ax}u)= e^{ax}(u''+ (2a- 2a)u'+ (a^2- 2a^2+ a^2)u)= e^{ax}u''= 0[tex]. The terms involving u only have canceled because $\displaystyle e^{ax}$ is a solution to the differential equation. The terms involving u' only have canceled because a is a double root. Since $\displaystyle e^{ax}$ is never 0, we must have u''= 0. Integrating twice, we have u(x)= C+ Dx so that $\displaystyle y= (C+ Dx)e^{ax}= Ce^{ax}+ Dxe^{ax}$. The "$\displaystyle Ce^{ax}$" we already had but now we know that $\displaystyle xe^{ax}$ is also a solution.
    Last edited by HallsofIvy; Jun 23rd 2012 at 04:31 PM.
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