If y=e^kx is a solution of the equation f ''(x) -4f '(x) + 3y = 0 determine the values of k. The back of my text book says k=1 or 3 but im not sure how to work this out. The working out of this question would be greatly appreciated.
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f(x) = e^(kx)
f '(x) = ke^(kx)
f "(x) = k^2e^(kx)
k^2e^(kx) - 4ke^(kx) + 3e^(kx) = 0
e^(kx)[k^2 - 4k + 3] = 0
e^(kx) = 0 => reject no roots exists
(k - 1)(k - 3) = 0
k = 1 , 3
f '(x) = ke^(kx)
f "(x) = k^2e^(kx)
k^2e^(kx) - 4ke^(kx) + 3e^(kx) = 0
e^(kx)[k^2 - 4k + 3] = 0
e^(kx) = 0 => reject no roots exists
(k - 1)(k - 3) = 0
k = 1 , 3
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You are welcome mate, thanks and best regards.
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