Lecture Notes in Differential Equations - Bruce E. Shapiro

242 LESSON 27. HIGHER ORDER EQUATIONS
Equating coefficients gives a
= 0 and b = -3/5, and hence
y 2 = − 3 sin t (27.130)
5
For y 3 , Since r = −2 is already a root of the characteristic equation with
multiplicity one, we will try
Differentiating three times gives
Substituting for y 3 into its ODE gives
y 3 = ate −2t (27.131)
y ′ 3 = a ( e −2t − 2te −2t) (27.132)
= ae −2t (1 − 2t) (27.133)
y 3 ′′ = a [ −2e −2t (1 − 2t) + e −2t (−2) ] (27.134)
= ae −2t (−4 + 4t) (27.135)
y 3 ′′′ = a [ −2e −2t (−4 + 4t) + e −2t (4) ] (27.136)
= ae −2t (12 − 8t) (27.137)
e −2t = [ ae −2t (12 − 8t) ] − 4 [ ae −2t (1 − 2t) ] = 8ae −2t (27.138)
and therefore a = 1/8, so that
y 3 = 1 8 te−2t (27.139)
Combining all of the particular solutions with the homogeneous solution
gives us the general solution to the differential equation, which is given by
y = C 1 + C 2 e 2t + C 3 e −2t − 1 8 t2 − 3 5 sin t + 1 8 te−2t . (27.140)