3.5 Driven Linear ODEs: Undetermined Coefficients I
Driven, constant-coefficient, second-order, linear ODE:
𝑦 ′′ + 𝑎𝑦 ′ + 𝑏𝑦 = 𝑓(𝑡)
Finding the general solution
The general solution 𝑦 of the driven ODE is obtained by adding any particular solution 𝑦𝑑 of the
driven ODE to the general solution 𝑦𝑢 of the undriven ODE (found using the methods of section 3.2):
𝑦 = 𝑦𝑑 + 𝑦𝑢
Finding a particular solution: polynomial driving term
The ODE
𝑦 ′′ + 𝑐𝑦 ′ + 𝑑𝑦 = 𝑎𝑛 𝑡 𝑛 + ⋯ + 𝑎1 𝑡 + 𝑎0
has a particular solution 𝑦𝑑 of one of the following forms:
Case 1: If 𝑑 ≠ 0, take
𝑦𝑑 = 𝐴𝑛 𝑡 𝑛 + ⋯ + 𝐴1 𝑡 + 𝐴0
Case2: If 𝑑 = 0, 𝑐 ≠ 0, take
𝑦𝑑 = 𝐴𝑛+1 𝑡 𝑛+1 + ⋯ + 𝐴2 𝑡 2 + 𝐴1 𝑡
Case 3: If 𝑑 = 0, 𝑐 = 0, take
𝑦𝑑 = 𝐴𝑛+2 𝑡 𝑛+2 + ⋯ + 𝐴3 𝑡 3 + 𝐴2 𝑡 2
Finding a particular solution: polynomial-exponential driving term
The ODE
𝑦 ′′ + 𝑎𝑦 ′ + 𝑏𝑦 = 𝑞 𝑡 𝑒 𝑠𝑡 ,
with 𝑞 𝑡 a polynomial of degree 𝑛, has a particular solution of the form 𝑦𝑑 = ℎ 𝑡 𝑒 𝑠𝑡 , where ℎ(𝑡) is
a polynomial.
Let the polynomial 𝑝 be such that 𝑝 𝐷 [𝑦] = 𝑦 ′′ + 𝑎𝑦 ′ + 𝑏𝑦. Then we substitute our guess for 𝑦𝑑 in
our ODE 𝑝 𝐷 𝑦 = 𝑞 𝑡 𝑒 𝑠𝑡 to obtain:
𝑝 𝐷 ℎ 𝑡 𝑒 𝑠𝑡 = 𝑞 𝑡 𝑒 𝑠𝑡
𝑒 𝑠𝑡 𝑝 𝐷 + 𝑠 ℎ = 𝑞 𝑡 𝑒 𝑠𝑡
𝑝 𝐷+𝑠 ℎ = 𝑞 𝑡
Note that 𝑝 𝐷 + 𝑠 ℎ = ℎ′′ + 𝑎ℎ′ + 𝑏ℎ, for some 𝑎 and 𝑏, so the polynomial ℎ(𝑡) can be
determined using the method above for a polynomial driving term, applied to the ODE
ℎ′′ + 𝑎ℎ′ + 𝑏ℎ = 𝑞 𝑡 .