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Homework Help: Question concerning 2nd order homogeneous linear diff eqs

  1. Aug 4, 2015 #1
    1. The problem statement, all variables and given/known data

    Regarding the case where the auxillary (characteristic) equation has complex roots, we solve the quadratic in the usual way using [itex]i[/itex] to get the general solution

    [tex]y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + i C_2 \sin{\beta x}\right)[/tex]

    And the textbook shows

    [tex]y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + C_2 \sin{\beta x}\right)[/tex]

    without the imaginary number [itex]i[/itex] in the equation.

    At first I just assumed that the [itex]i[/itex] has been subsumed into the constant [itex]C_2[/itex], but then what is happening when we solve an initial value problem of this form, and find that [itex]C_2[/itex] is actually a real number? Where has the [itex]i[/itex] gone?
  2. jcsd
  3. Aug 4, 2015 #2


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    If the initial conditions were such that ##C_2## was real in the first solution, then if we had used the 2nd solution, the initial condition would imply that ##C_2'## was imaginary. The important fact is that sin and cos are linearly independent solutions, so we can use any linear combination of them.
  4. Aug 4, 2015 #3


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    This link provides as clear an explanation as any:

  5. Aug 5, 2015 #4

    Ray Vickson

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    [tex]y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + i C_2 \sin{\beta x}\right)[/tex]
    we can write
    [tex]y(x) = e^{\alpha x}\left(A_1 \cos{\beta x} + A_2 \sin{\beta x}\right).[/tex]
    where ##A_1 = C_1## and ##A_2 = i C_2##. It is just notation, nothing more.

    If ##y(x) = e^{\alpha x}\left(C_1 \cos{\beta x} + i C_2 \sin{\beta x}\right) ## and ##y(0), y'(0)## are given (for example), then we need
    [tex] C_1 = y(0) \\
    \alpha C_1 + \beta i C_2 = y'(0)[/tex]
    [tex] C_1 = y(0), \; C_2 = \frac{i}{\beta} \left( \alpha y(0) - y'(0) \right) [/tex]
    If course, ##iC_2 = (y'(0) - \alpha y(0))/\beta## is real if ##y(0), y'(0)## are real.
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