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Problem 15 - Entrance Test
Find the solution to the initial value problem: y'' - 4y' + 4y = 0, y(0) = 1, y'(0) = 0.
A. y = e^(2x) - 2xe^(2x)
B. y = e^(2x) + 2xe^(2x)
C. y = e^(2x)
D. y = xe^(2x)
Check Answer
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Correct: A
Characteristic equation: r^2 - 4r + 4 = 0 => (r - 2)^2 = 0 => r = 2 (repeated root). General solution: y = c1 e^(2x) + c2 x e^(2x). y' = 2c1 e^(2x) + c2 e^(2x) + 2c2 x e^(2x). y(0) = 1: c1 = 1. y'(0) = 0: 2c1 + c2 = 0 => 2 + c2 = 0 => c2 = -2. Thus, y = e^(2x) - 2xe^(2x).