Lecture Notes in Differential Equations - Bruce E. Shapiro

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The characteristic equation is
r 2 − 9 = (r − 3)(r + 3) = 0 =⇒ r = ±3 (22.99)
Thus the solution of the homogeneous equation is
y H = Ae 3t + Be −3t (22.100)
From equation (22.96) with a = 1, r 1 = 3, r 2 = −3, and f(t) = e 2t , a
particular solution is
y P = 1 ∫ (∫
)
a er2t e (r1−r2)t f(t)e −r1t dt dt (22.101)
∫ (∫ )
= e −3t e 6t e 2t e −3t dt dt (22.102)
∫ (∫ )
= e −3t e 6t e −t dt dt (22.103)
∫
= −e −3t e 6t e −t dt (22.104)
∫
= −e −3t e 5t dt (22.105)
Hence the general solution is
= − 1 5 e−3t e 5t (22.106)
= − 1 5 e2t (22.107)
The first initial condition gives
y = Ae 3t + Be −3t − 1 5 e2t (22.108)
1 = A + B − 1 5 =⇒ B = 6 5 − A (22.109)
To apply the second initial condition we must differentiate (22.108):
y ′ = 3Ae 3t − 3Be −3t − 2 5 e2t (22.110)
hence
2 = 3A − 3B − 2 5 =⇒ 12 5
= 3A − 3B (22.111)