Calculus- Early Transcendentals, 2021a

396 Differential Equations
C cos(4t)+Dsin(4t) is also, so it cannot be a solution to the non-homogeneous equation. Instead, we
guess Ct cos(4t)+Dt sin(4t). Then substituting:
(−16Ct cos(4t) − 16Dsin(4t)+8Dcos(4t) − 8C sin(4t))) + 16(Ct cos(4t)+Dt sin(4t))
= 8Dcos(4t) − 8C sin(4t).
Thus C = 1/8, D = 0, and the solution is C cos(4t)+Dsin(4t)+(1/8)t cos(4t).
♣
In general, if f (t)=acos(ωt)+bsin(ωt), and±ωi are the roots of the characteristic equation, then
instead of C cos(ωt)+Dsin(ωt) we guess Ct cos(ωt)+Dt sin(ωt).
Exercises for 10.6
Find the general solution to the differential equation.
Exercise 10.6.1 y ′′ − 10y ′ + 25y = cost
Exercise 10.6.2 y ′′ + 2 √ 2y ′ + 2y = 10
Exercise 10.6.3 y ′′ + 16y = 8t 2 + 3t − 4
Exercise 10.6.4 y ′′ + 2y = cos(5t)+sin(5t)
Exercise 10.6.5 y ′′ − 2y ′ + 2y = e 2t
Exercise 10.6.6 y ′′ − 6y + 13 = 1 + 2t + e −t
Exercise 10.6.7 y ′′ + y ′ − 6y = e −3t
Exercise 10.6.8 y ′′ − 4y ′ + 3y = e 3t
Exercise 10.6.9 y ′′ + 16y = cos(4t)
Exercise 10.6.10 y ′′ + 9y = 3sin(3t)
Exercise 10.6.11 y ′′ + 12y ′ + 36y = 6e −6t
Exercise 10.6.12 y ′′ − 8y ′ + 16y = −2e 4t
Exercise 10.6.13 y ′′ + 6y ′ + 5y = 4
Exercise 10.6.14 y ′′ − y ′ − 12y = t
Exercise 10.6.15 y ′′ + 5y = 8sin(2t)
Exercise 10.6.16 y ′′ − 4y = 4e 2t