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Second Order Linear Homogeneous DE

  1. Feb 10, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]3\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + y = 0[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex]3y'' + 2y' +y = 0[/tex]

    I know the solution is going to be in the form of [tex]y=Ce^{mx}+De^{nx}+...[/tex]
    (Unless there is a multiplicity, in which case I understand that too)

    So I'll just skip to the Aux Equation:
    [tex]3m^{2} + 2m + 1 = 0[/tex]
    [tex]m = \frac{-1+\sqrt{2}i}{3} or \frac{-1-\sqrt{2}i}{3}[/tex]

    Thus, a general form of the solution is:
    [tex]y(x) = C_{1}e^{\frac{-1}{3}}e^{\frac{\sqrt{2}ix}{3}} + C_{2}e^{\frac{-1}{3}}e^{\frac{-\sqrt{2}ix}{3}}[/tex]

    So now I just rename the constants..

    [tex]Ae^{\frac{\sqrt{2}ix}{3}} + Be^{\frac{-\sqrt{2}ix}{3}}[/tex]

    and since...

    [tex]e^{ix} = cos(x) + isin(x)[/tex]

    [tex]A(cos(\frac{\sqrt{2}x}{3})+isin(\frac{\sqrt{2}x}{3})) + B(cos(\frac{-\sqrt{2}x}{3})+isin(\frac{-\sqrt{2}x}{3}))[/tex]

    Since cosine is even and sine is odd...

    [tex]Acos(\frac{\sqrt{2}x}{3}) + iAsin(\frac{\sqrt{2}x}{3}) + Bcos(\frac{\sqrt{2}x}{3}) -iBsin(\frac{\sqrt{2}x}{3})[/tex]

    Usually here is where the imaginary stuff cancels out, but it's just not happening this time and I think I am making a mistake somewhere but I can't see it. What do I do??
     
  2. jcsd
  3. Feb 10, 2012 #2

    micromass

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    Samll mistake, but you also want [itex]e^{-\frac{1}{3}x}[/itex] instead of [itex]e^{-\frac{1}{3}}[/itex]. Indeed, your general solution requires you to multiply entire m with x. This will not save you from the trouble below however.

    Now you found all complex solutions, which is good. But you want all real solutions. Do you see choices of A and B that will get you real solutions?? (you can choose A and B complex if you want to).

    The hardest part is proving that you got all real solutions.
     
  4. Feb 14, 2012 #3
    Thanks Micromass. I see what I did wrong. I'll try to re-derive it. I just don't want to blindly use the formula to solve this type of DE without understanding where it's coming from.
     
  5. Feb 14, 2012 #4

    Mark44

    Staff: Mentor

    For diff. equations of this type, if the roots of the characteristic equation are a [itex]\pm[/itex] bi, a pair of solutions is y1 = eatsin(bt) and y2 = eatcos(bt).
     
  6. Feb 15, 2012 #5
    I figured it out.

    What I was doing wrong was this:

    Once the equation turned into the form with (or without) the exponential and then the sum of sines and cosines, I needed to group the two functions with the complex variable in front and factor them out into the constant. So it's essentially possible that the constant contains a complex variable.

    I don't know what happened in the above work, I lost some things in translation to Latex and now looking over it, it doesn't make much sense! I fully understand how an equation of that form with an aux equation containing both complex and real part solutions works now, and simplifies into exp(x)(sin+cos) (for second order at least).

    Today we went over how the discriminant determines which parts of the solution you get, the three cases (d>0, d<0, d=0) and what that means exactly. Now it all makes sense.
     
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