y'' + 2y' + 2y = 0
\downarrow (write characteristic equation)
\lambda^2 +2\lambda + 2 = 0
\downarrow (solve characteristic equation)
\lambda = -1 \pm i
\downarrow (write general solution)
y = Ae^{(-1 - i)t} + Be^{(-1 + i)t}
\downarrow (apply Euler's formula)
y = Ae^{-t}(\cos(-t) + i\sin(-t)) + Be^{-t}(\cos(t) + i\sin(t))
\downarrow (perform minor algebra/trig rearrangement)
y = (A + B)e^{-t}\cos(t) + (B - A)e^{-t}\sin(t)i
Where do I go from here to eliminate the i? Plugging in the exponential formulas for sine and cosine leads back to the original general solution, with i's remaining.
Answer
Both (A+B) and (B-A)i are arbitrary constants. Rename them C_1 and C_2 to get the general solution
y=C_1e^{-t}\cos{(t)}+C_2e^{-t}\sin{(t)}
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