Lecture Notes in Differential Equations - Bruce E. Shapiro

158 LESSON 18. COMPLEX ROOTS
where
A = C 1 + C 2 (18.49)
B = i(C 1 − C 2 ) (18.50)
Proof.
y H = C 1 e r1t + C 2 e r2t (18.51)
= C 1 e (µ+iω)t + C 2 e (µ−iω)t (18.52)
= e µt (C 1 e iωt + C 2 e −iωt ) (18.53)
= e µt [C 1 (cos ωt + i sin ωt) + C 2 (cos ωt − i sin ωt)] (18.54)
= e µt [(C 1 + C 2 ) cos ωt + i(C 1 − C 2 ) sin ωt] (18.55)
= e µt (A cos ωt + B sin ωt) (18.56)
Example 18.4. Find the general solution of
The characteristic polynomial is
and the roots are
r = −6 ± √ −64
2
Hence the general solution is
for any numbers A and B.
y ′′ + 6y ′ + 25y = 0 (18.57)
r 2 + 6r + 25 = 0 (18.58)
=
−6 ± 8i
2
= −3 ± 4i (18.59)
y = e −3t A cos 4t + B sin 4t]. (18.60)
Example 18.5. Solve the initial value problem
⎫
y ′′ + 2y ′ + 2y = 0⎪⎬
y (0) = 2
⎪⎭
y ′ (0) = 4
(18.61)
The characteristic equation is
r 2 + 2y + 2 = 0 (18.62)