solve differential equations with complex eigenvalues and eigenvectors?

Question

i can come up with the general formula for the DE but since there are complex eigenvalues and eigenvectors i get lost when converting it to sine and cosine, just wandering if there is a simpler way to this without expanding out the whole equation and then simplifying.

an example is:

x' = 3x -5y
y' = 4x -5y

Answer 1

Assuming your prime is dx/dt = 3x - 5y then differentiating wrt t, we have

x'' = 3x' - 5y'

x'' = 3(3x - 5y) - 5(4x - 5y)

x'' = -11x + 10y

x'' = -11x + 2(3x - x')

x'' + 2x' + 5x = 0

auxiliary equation m² + 2m + 5 = 0

m = -1 ± 2i

so the solution for x is ..... x = e^(-t)[Acos2t + Bsin2t]

similarly eliminating x you will have

y'' + 2y' + 5y = 0

and so the solution for y is ...... y = e^(-t)[Ccos2t + Dsin2t]

Answer 2

As I understand
example
y´+5y=4x
y´+5y=0 y= C e^-5x
Particular solution
y=ax+b
a+5ax+5b=4x so 5a=4 and a = 4/5
a+5b=0 and b= -4/25
y=Ce^-5x+4/5 x-2/25