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Section 5.2 Problem 17:

Solve the initial value problem.

4y'' - 12y' + 9y = 0

y(0) = 3

y'(0) = (5)/(2)


User Mick F
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1 Answer

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This DE has characteristic equation


4r^2 - 12r + 9r = (2r - 3)^2 = 0

with a repeated root at r = 3/2. Then the characteristic solution is


y_c = C_1 e^(\frac32 x) + C_2 x e^(\frac32 x)

which has derivative


{y_c}' = \frac{3C_1}2 e^(\frac32 x) + \frac{3C_2}2 x e^(\frac32x) + C_2 e^(\frac32 x)

Use the given initial conditions to solve for the constants:


y(0) = 3 \implies 3 = C_1


y'(0) = \frac52 \implies \frac52 = \frac{3C_1}2 + C_2 \implies C_2 = -2

and so the particular solution to the IVP is


\boxed{y(x) = 3 e^(\frac32 x) - 2 x e^(\frac32 x)}

User River Satya
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