Lecture Notes in Differential Equations - Bruce E. Shapiro

159
and its roots are given by
The solution is then
r = −2 ± √ (2) 2 − 4(1)(2)
2
The first initial condition gives
and thus the solution becomes
Differentiating this solution
=
−2 ± 2i
2
= −1 ± i (18.63)
y = e −t (A cos t + B sin t) (18.64)
2 = A (18.65)
y = e −t (2 cos t + B sin t) (18.66)
y ′ = −e −t (2 cos t + B sin t) + e −t (−2 sin t + B cos t) (18.67)
The second initial condition gives us
4 = −2 + B =⇒ B = 6 (18.68)
Hence
y = e −t (2 cos t − 6 sin t) (18.69)
Summary. General solution to the homogeneous linear equation with
constant coefficients.
The general solution to
ay ′′ + by ′ + cy = 0 (18.70)
is given by
⎧
⎪⎨ Ae r1t + Be r2t
r 1 ≠ r 2 (distinct real roots)
y = (A + Bt)e
⎪⎩
rt r = r 1 = r 2 (repeated real root) (18.71)
e µt (A cos ωt + B sin ωt) r = µ ± iω (complex roots)
where r 1 and r 2 are roots of the characteristic equation ar 2 + br + c = 0.