Lecture Notes in Differential Equations - Bruce E. Shapiro

238 LESSON 27. HIGHER ORDER EQUATIONS
Example 27.5. Find the general solution to y ′′′ + y ′′ − 6y ′ = e t .
The characteristic equation is
0 = r 3 + r 2 − 6r = r(r − 2)(r + 3) (27.80)
which has roots r 1 = 0, r 2 = 2, and r 3 = −3. The general solution to the
homogeneous equation is
y H = C 1 + C 2 e 2t + C 3 e −3t (27.81)
From (27.79), a particular solution is
∫ ∫
∫
y P = e r3t e (r2−r3)s3 e (r1−r2)s2 e −r1s1 f(s 1 )ds 1 ds 2 ds 3 (27.82)
t
s 3 s
∫ ∫ ∫
2
= e −3t e 5s3 e −2s2 e s1 ds 1 ds 2 ds 3 (27.83)
t s 3 s
∫
2
= e −3t e 5s3 e
t
∫s −2s2 e s2 ds 2 ds 3 (27.84)
∫
3
= e −3t e 5s3 e
t
∫s −s2 ds 2 ds 3 (27.85)
∫
3
= −e −3t e 5s3 e −s3 ds 3 (27.86)
t
∫
= −e −3t e 4s3 ds 3 (27.87)
t
= − 1 4 e−3t e 4t (27.88)
= − 1 4 et (27.89)
Hence the general solution is
y = y P + y H = − 1 4 et + C 1 + C 2 e 2t + C 3 e −3t (27.90)
In general it is easier to use undetermined coefficients to determine y P
if a good guess for its form is known, rather than keeping track of the
integrals in (27.79). Failing that the bookkeeping still tends to be easier if
we reproduce the derivation of (27.79) by factoring the equation one root
at a time than it is to use (27.79) directly. In general the best ”guess” for
the form of a particular solution is the same for higher order equations as
it is for second order equations. For example, in the previous example we
would have looked for a solution of the form y P = ce t .