Differential equations Matrices

What is the general solution of the system

x' = 2.7x - y
y' = 4.2x + 3.5y

r' = Mr

r = [x]
.....[y]

M = [2.7 ..-1]
.......[4.2 3.5]

We can diagonalize M by finding it's eigenvalues λ1 and λ2 using the characteristic equation |M - Iλ| = 0:

(2.7 - λ)(3.5 - λ) + 4.2 = 0

Which gives (approximately):

=> λ1 = 3.1 + 2i, λ2 = 3.1 - 2i

Hence the diagonal form of M is D:

D = [λ1 .0] = [3.1 + 2i ........0]
......[0 .λ2]....[0..........3.1 - 2i]

Mr = Dr

=> r' = Dr

=> x' = (3.1 + 2i)x
.....y' = (3.1 - 2i)y

=> x = Ae^(λ1t)
.....y = Be^(λ2t)

x' = Aλ1e^(λ1t) = 2.7Ae^(λ1t) - Be^(λ2t)
y' = Bλ2e^(λ2t) = 4.2Ae^(λ1t) + 3.5Be^(λ2t)

put t=0

Aλ1 = 2.7A - B
Bλ2 = 4.2A + 3.5B

(2.7 - λ1)A - B = 0
4.2A + (3.5 - λ2)B = 0

These two equations represent a single constraint, hence A and B are related by:

A = B/(2.7 - λ1)

or:

A = B(λ2 - 3.5)/4.2

Which gives (approximately):

A = (-0.1 - 0.5i)B

so (again, approximately):

x = -(0.1 + 0.5i)Be^(λ1t)
y = Be^(λ2t)

I think.

Before you go. Everything is correct up to the equations: x = Ae^(λ1t), y = Be^(λ2t). Please forget the rest!!

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Either that or double check it for yourself. I sort of did this ad hoc so can't promise that it is the *correct* way. Thank you though.

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