y″+2y′+2y=0
↓ (write characteristic equation)
λ2+2λ+2=0
↓ (solve characteristic equation)
λ=−1±i
↓ (write general solution)
y=Ae(−1−i)t+Be(−1+i)t
↓ (apply Euler's formula)
y=Ae−t(cos(−t)+isin(−t))+Be−t(cos(t)+isin(t))
↓ (perform minor algebra/trig rearrangement)
y=(A+B)e−tcos(t)+(B−A)e−tsin(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 C1 and C2 to get the general solution
y=C1e−tcos(t)+C2e−tsin(t)
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