Lecture Notes in Differential Equations - Bruce E. Shapiro

239
Example 27.6. Find the general solution of y (4) − 5y ′′ + 4y = t
The characteristic equation is 0 = r 4 − 5r 2 + 4 = (r 2 − 1)(r 2 − 4) so the
roots are 1, -1, 2, and –2, and the homogeneous solution is
y H = C 1 e t + C 2 e −t + C 3 e 2t + C 4 e −2t (27.91)
Using the factorization method: we write the differential equation as
(D − 1)(D + 1)(D − 2)(D + 2)y = (D − 1)z = t (27.92)
where z = (D + 1)(D − 2)(D + 2)y. Then z ′ − z = t. An integrating factor
is e −t , so that
z = e t ∫
te −t dt = e t [ −(t + 1)e −t] = −t − 1 (27.93)
where we have ignored the constant of integration because we know that
they will lead to the homogeneous solutions. Therefore
z = (D + 1)(D − 2)(D + 2)y = (D + 1)w = −t − 1 (27.94)
where w = (D − 2)(D + 2)y. Equation (27.94) is equivalent to w ′ + w =
−t − 1, so that
[∫
]
w = −e −t e t (t + 1)dt
(27.95)
[∫ ∫ ]
= −e −t te t dt + e t dt
(27.96)
Therefore
= −e −t [ (t − 1)e t + e t] (27.97)
= −t (27.98)
w = (D − 2)(D + 2)y (27.99)
= (D − 2)u = −t (27.100)
where u = (D + 2)y = y ′ + 2y. Equation (27.99) is u ′ − 2u = −t. An
integrating factor is e −2t and the solution for u is
∫
[
u = −e 2t te −2t dt = −e 2t − 1 2 t − 1 ]
e −2t = 1 4 2 t + 1 (27.101)
4
Therefore
y ′ + 2y = 1 (2t + 1) (27.102)
4