What Is the Form of G When k Is Complex?

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The discussion focuses on the form of the function G when the constant k is complex in the ordinary differential equation G'' = -kG. It establishes that for k > 0, the solution takes the form G = Acos(sqrt(k)x) + Bsin(sqrt(k)x). When k is a complex number, the roots are expressed as r = +/- sqrt(-ki), leading to a different representation of G. The implications of complex k on the solution's form are crucial for understanding the behavior of the system described by the ODE.

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madeinmsia
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G'' = -kG
k is a constant
solving this ODE, r = +/- sqrt(-k)
if k > 0, then r = +/- sqrt(k)i
so G is in the form Acos(sqrt(k)x) + Bsin(sqrt(k)x)

so, what if k is a complex number, then
r = +/- sqrt(-ki)

then what is the form of G?
 
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madeinmsia said:
so, what if k is a complex number, then
r = +/- sqrt(-ki)

then what is the form of G?

Don't you know how to calculate http://www.mathpropress.com/stan/bibliography/complexSquareRoot.pdf?
 

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