Lecture Notes in Differential Equations - Bruce E. Shapiro

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a closed form expression for a particular solution to L n y = f(t). Such an
expression can be found by factoring the differential equation as
L n y = a n (D − r 1 )(D − r 2 ) · · · (D − r n )y (27.71)
where z = (D − r 2 ) · · · (D − r n )y. Then
An integrating factor is µ = e −r1t , so that
= a n (D − r 1 )z = f(t) (27.72)
z ′ − r 1 z = (1/a n )f(t) (27.73)
(D − r 2 ) · · · (D − r n )y = z (27.74)
= 1 e r1t e
a n
∫t
−r1s1 f(s 1 )ds 1 = f 1 (t) (27.75)
where the last expression on the right hand side of (27.74) is taken as the
definition of f 1 (t). We have ignored the constants of integration because
they will give us the homogeneous solutions.
Defining a new z = (D − r 3 ) · · · (D − r n )y gives
z ′ − r 2 z = f 1 (t) (27.76)
An integrating factor for (27.76) is µ = e −r2t , so that
(D − r 3 ) · · · (D − r n )y = z = e r2t ∫
t
e −r2s2 f 1 (s 2 )ds 2 = f 2 (t) (27.77)
where the expression on the right hand side of (27.77) is taken as the
definition of f 2 (t). Substituting for f 1 (s 2 ) from (27.74) into (27.77) gives
∫
(D − r 3 ) · · · (D − r n )y = e r2t e 1 ∫
−r2s2 e r1s2 e −r1s1 f(s 1 )ds 1 ds 2
a n s 2
t
(27.78)
Repeating this procedure n times until we have exhausted all of the roots,
y P = ernt
a n
∫t
e (rn−1−rn)sn ∫s n
e (rn−2−rn−1)sn−1 · · ·
(27.79)
· · · e
∫s (r1−r2)s2 e
3
∫s −r1s1 f(s 1 )ds 1 · · · ds n
2