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2nd order homogenous constant coefficient ODE question

  1. Mar 26, 2014 #1


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    If we have a constant coefficient second order homogeneous ODE, the way to solve this is to suppose a solution of the exponential type. This yields a second order polynomial equation (the "characteristic equation") that the exponent must satisfy. In case the solutions of the characteristic equation are complex, we get a solution of the form [itex]c1e^{wx}+c2e^{\overline{w}x}[/itex] (call this form 1). The general solution of the ODE is usually presented in the form e^x(Acos(bx)+Bsin(bx)) (call this form 2) by using Euler's formula for e^{a+ib}.

    To my eyes, forms 1 and 2 are equivalent. So then, how come trying to solve an initial value problem with form 1 does not work?!?

    Consider for instance y''-2y+2=0 with y(0)=1, y'(0)=0. The roots of the char. equ. are 1+i and 1-i. If we try to find c1 and c2 using the initial values given we are lead to the impossible system c1 + c2 = 1, c1 + c2=0, c1 - c2=0. With form 2 however, we find A=1, B=-1.

    Thanks, I'm confused!
  2. jcsd
  3. Mar 26, 2014 #2


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    Surely you meant y'' - 2y' + 2y = 0, based on the characteristic equation you have.
    If you checked your solution against what you wrote -- y'' - 2y + 2 = 0 -- that's a problem right there.

    Either pair of functions e(1 + i)x, e(1 - i)x and excos(x), exsin(x), is a basis for the solution space of your differential equation. Due to your typo, you might have been trying to calculate the constants that go with your initial conditions using the wrong DE.
  4. Mar 26, 2014 #3


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    So there are three things we can do.
    They are all equivalent.
    1)Work freely with complex numbers
    2)Use complex numbers but convert back to real numbers at the end.
    3)Work only with real numbers.

    As far as your example your equations seem to result from a false assumption that c1 and c2 are real.
    The initial conditions give the system (or some variation of it)

    which is easily solved
  5. Mar 26, 2014 #4


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    Regardless of what equation you were trying to solve, I'm confused about how you got 3 equations from 2 initial conditions.
  6. Mar 27, 2014 #5


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    So sorry, the correct DE is y'' - 2y' + 2y = 0.

    But anyhow, I see what I did wrong there. My mistake was indeed that I assumed for a second that c1 and c2 are real! Thanks for the help.
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