Lecture Notes in Differential Equations - Bruce E. Shapiro

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The characteristic equation is
0 = r 3 − 4r = r(r 2 − 4) = r(r − 2)(r + 2) (27.115)
Since the roots are r = 0, ±2, the solution of the homogeneous equation is
y H = C 1 + C 2 e 2t + C 3 e −2t (27.116)
To find y 1 , our first inclination would be to try y = At + B. But e 0t is
a solution of the homogeneous equation so we try y + 1 = t k (at + b) with
k = 1.
y 1 = t k (a + bt)e 0t = at + bt 2 (27.117)
Differentiating three times
Substituting into the differential equation gives
y 1 ′ = a + 2bt (27.118)
y 1 ′′ = 2b (27.119)
y 1 ′′′ = 0 (27.120)
t = y ′′′
1 − 4y ′ 1 = 0 − 4(a + 2bt) = −4a − 8bt (27.121)
This must hold for all t, so equating like coefficients of t gives us a = 0 and
b = −1/8 so that the first particular solution is
y 1 = − 1 8 t2 (27.122)
For y 2 , since r = ±i are not roots of the characteristic equation we try
Differentiating three times,
Substituting into the differential equation for y 2
y 2 = a cos t + b sin t (27.123)
y 2 ′ = −a sin t + b cos t (27.124)
y 2 ′′ = −a cos t − b sin t (27.125)
y 2 ′′′ = a sin t − b cos t (27.126)
3 cos t = a sin t − b cos t − 4(−a sin t + b cos t) (27.127)
= a sin t − b cos t + 4a sin t − 4b cos t (27.128)
= 5a sin t − 5b cos t (27.129)