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Thread: y''+k^2*y=0, k>0

  1. #1
    Dec 2009

    y''+k^2*y=0, k>0

    I am having some trouble regarding this second order homogeneous differential equation, the one which appears from the title. I am supposed to finde the complete solution of (1)

    (1) y''+k2y=0, where k>0 is a constant

    I am aware of the fact that if i use the Wronskian-determinant i get the complete solution with the arbitary constants c1 and c2


    as i have W(sin(kx), cos(kx)) =-ksin2kx-kcos2kx=-k≠0

    but where the hell do sinkx and coskx come from? I'm so damn confused, and very pissed!

    My book claims that by executing a sample we can conclude that (1) has the solutions coskx and sinkx, where do these come from and what is up with all this sample thing?
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  2. #2
    Senior Member
    Mar 2009
    I don't know about using the Wronskian for a homogeneous DEQ, but this is how I would solve the problem.
    We assume the solution is of the form
    $\displaystyle y(x) = e^{\lambda x}$.
    Now, we substitute this into the original DEQ
    $\displaystyle (\lambda^2 + k^2) e^{\lambda x} = 0$.
    We solve the characteristic equation
    $\displaystyle \lambda^2 + k^2 = 0 \Rightarrow \lambda = \pm k i$.
    Therefore, the solution is of the form
    $\displaystyle y(x) = C_1 \cos k x + C_2 \sin k x$.
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