Lecture Notes in Differential Equations - Bruce E. Shapiro

240 LESSON 27. HIGHER ORDER EQUATIONS
An integrating factor is e 2t so that
y = 1 ∫
4 e−2t e 2t (2t + 1)dt (27.103)
= 1 ∫ ∫ ]
[2
4 e−2t te 2t dt + e 2t dt
(27.104)
= 1 ( t
[2
4 e−2t 2 − 1 e
4) 2t + 1 ]
2 e2t (27.105)
= t 4
(27.106)
Therefore y P = t/4.
Alternatively, using the method of undetermined coefficients, we try y P =
ct in the differential equation y (4) − 5y ′′ + 4y = t. Since y ′ = c and y ′′ =
y (4) = 0 we find 4ct = t or c = 1/4, again giving y P = t/4.
Hence the general solution is
y = y P + y H = 1 4 t + C 1e t + C 2 e −t + C 3 e 2t + C 4 e −2t (27.107)
We also have an addition theorem for higher order equations.
Theorem 27.7. If y P,i , i = 1, 2, ..., k are particular solutions of L n y P,i =
f i (t) then
y P = y P,1 + y P,2 + · · · + y P,k (27.108)
is a particular solution of
L n y = f 1 (t) + f 2 (t) + · · · + f k (t) (27.109)
Example 27.7. Find the general solution of y ′′′ − 4y ′ = t + 3 cos t + e −2t .
The general solution is
where
y = y H + y 1 + y 2 + y 3 (27.110)
y H ′′′ − 4y H ′ = 0 (27.111)
y 1 ′′′ − 4y 1 ′ = t (27.112)
y 2 ′′′ − 4y 2 ′ = 3 cos t (27.113)
y 3 ′′′ − 4y 3 ′ = e −2t (27.114)