- #1

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The general solution to y''+k^2 y = 0 (with k>0) is y=A cos(kx)+B sin(kx).

If we then set k=0 the solution reduces to y=A.

Yet, if we set k=0 in the original ODE we get [1] y''=0, whose solution is [2] y=Bx+C.

Where A, B, and C are constants.

I get analogous results for y''-k^2 y = 0.

On the other hand, if I start with y''+2b y'+k^2 y = 0, then it seems that setting either b or k to 0, but not both, can be done either in the ODE itself or in the general solution.

Hrm, I just had another insight while writing this. The form of solution [2] is due to the characteristic equation having a repeated real root of r=0. Is this the reason why my initial attempt failed? Because I was setting the parameter to the same value as the eigenvalue of the ODE?