Differential Equations [a mere curiosity]

[iCeCuBEz v2]

It is known that the general equation of a homogeneous equation with constant coefficients that has an unrepeated pair of complex conjugate roots a plus or minus bi (with b not equal to 0) is as follows

e^ax(c1cosbx+c2sinbx)=y(x)

an example being

y''-4y'+5y=0

which has the characteristic equation

r^2-4r+5

which equals (by completing the square)

(r-2)^2+1=0

then solve for r in that characteristic equation yielding the complex conjugate roots 2 plus or minus i

then the general solution of the equation is

y(x)=e^(2x)(c1cosx+c2sinx)

I get how to do the procedure I just want to know why the general solution ends up being of the form e^ax(c1cosbx+c2sinbx)