Y=Aexp(Bx)+Fexp(Gx) is the solution of (y'') + (-12y') + ( 27y) = 0....?

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Y=Aexp(Bx)+Fexp(Gx) is the solution of (y'') + (-12y') + ( 27y) = 0....?

[From: ] [author: ] [Date: 14-05-07] [Hit: ]

2 = A + F.- 3 = 9A + 3F.A + F = 2,9A + 3F = - 3.......

y'' − 12y' + 27y = 0

First we solve characteristic equation to find B and G
r² − 12r + 27 = 0
(r − 3) (r − 9) = 0
r = 3, r = 9

y = Ae^(3x) + Fe^(9x)
y' = 3Ae^(3x) + 9Fe^(9x)

y(0) = 2 ------> A + F = 2
y'(0) = −3 ----> 3A + 9F = −3

Solving we get: A = 7/2, F = −3/2

y = 7/2 e^(3x) − 3/2 e^(9x)

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The cardinal rule is to solve two unknowns you need two equations. We have one for y. No problem, we can take its derivative y'.
Put the boundary conditions in
y = Aₑ9x + Fₑ3x.
y=2 when x=0 gives
2 = A + F.
y' = 9Aₑ9x + 3Fₑ3x
y'= - 3 when x=0 gives
- 3 = 9A + 3F.
Now solve the pair
A + F = 2,
9A + 3F = - 3.

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