Ordinary differential equation

set x_1 = y and x_2 = y' so that

x_1 ' = y' and

x_2 ' = y '' = y ' - 4y = x_2 - 4x_1

hence the given equation is equivalent to the system

X ' =

(0 1)
(-4 1) X

where X = [x_1 x_2] is a column vector.

The eigenvalues of the matrix A on the right side are

(1/2)(1 +/- i.sqrt(15))

and an eigenvector corresponding to the " + " eigenvalue is

[ 1 - i.sqrt(15), 8]

so the solution we get from this eigenvalue/eigenvector pair is

X(t) = e^( (1/2 + i.(sqrt(15)/2)) t } [1 - i.sqrt(15) 8]

= e^{(1/2)t} (cos ((sqrt(15)/2) t) + i. sin ((sqrt(15)/2) t)) [ 1 - i.sqrt(15) 8]

which can be written in the form U(t) + i. V(t) where

U(t) = [e^{(1/2) t} (cos (sqrt(15)/2) t) + sqrt(15). sin (sqrt(15)/2) t); 8e^{(1/2) t} cos( sqrt(15)/2) t)]

V(t) = [e^{(1/2)t} (sin (sqrt(15)/2) t) -- sqrt(15). cos (sqrt(15)/2) t); 8e^{(1/2)t} sin (sqrt(15)/2) t)]

Finally, since U and V form a linearly independent solution set, the general solution to the system X ' = AX is

X(t) = c_1 U(t) + c_2 V(t).

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