Lecture Notes in Differential Equations - Bruce E. Shapiro

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(b) If λ 1 ≠ λ 2 ∈ R, i.e., T 2 ≥ 4∆,
y i = C i1 e λ1t + C i2 e λ2t (31.32)
(c) It T 2 < 4∆ then λ 1,2 = µ ± iσ, where µ, σ ∈ R, and
y i = e µt (C i1 cos σt + C i2 sin σt) (31.33)
Example 31.1. Solve the system
(
y ′ 6
= Ay, y(0) =
5)
(31.34)
where
by elimination of variables.
A =
( )
1 1
4 −2
(31.35)
To eliminate variables means we try to reduce the system to either two
separate first order equations or a single second order equation that we can
solve. Multiplying out the matrix system gives us
u ′ = u + v, u 0 = 6 (31.36)
v ′ = 4u − 2v, v 0 = 5 (31.37)
Solving the first equation for v and substituting into the second,
v ′ = 4u − 2v = 4u − 2(u ′ − u) = 6u − 2u ′ (31.38)
Differentiating the first equations and substituting
Rearranging,
u ′′ = u ′ + v ′ = u ′ + 6u − 2u ′ = −u ′ + 6u (31.39)
u ′′ + u ′ − 6u = 0 (31.40)
The characteristic equation is r 2 + r − 6 = (r − 2)(r + 3) = 0 so that
u = C 1 e 2t + C 2 e −3t (31.41)
From the initial conditions, u ′ 0 = u 0 + v 0 = 11. Hence
C 1 + C 2 = 6
2C 1 − 3C 2 = 11
(31.42)
Multiplying the first of (31.42) by 3 and adding to the second gives 5C 1 = 29
or C 1 = 29/5, and therefore C 2 = 6 − 29/5 = 1/5. Thus