Calculus

for what values of r does the function y=e^rx satisfy the differential equation y''-4y'+y=0

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  1. y = 1 e^(rx)

    y' = r e^(rx)

    y" = r^2 e^(rx)

    r^2 - 4 r + 1 = 0

    r = [4 +/- sqrt (16 -4) ] /2

    r = [ 4 +/- 2 sqrt 3 ]2

    r = 2 +/- sqrt 3

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