Chapter 0 Ordinary Differential Equations 9
Now collect terms in the derivatives ofv. The preceding equation becomes
u 1 v′′+(
2 u′ 1 +k(t)u 1)
v′+(
u′′ 1 +k(t)u′ 1 +p(t)u 1)
v= 0.However,u 1 is a solution of Eq. (24), so the coefficient ofvis 0. This leaves
u 1 v′′+(
2 u′ 1 +k(t)u 1)
v′= 0 , (25)which is a first-order linear equation forv′. Thus, a nonconstantvcan be
found, at least in terms of some integrals.
Example.
Consider the equation
(
1 −t^2
)
u′′− 2 tu′+ 2 u= 0 , − 1 <t< 1 ,which hasu 1 (t)=tas a solution. By assuming thatu 2 =v·tand substituting,
we obtain
(
1 −t^2
)
(v′′t+ 2 v′)− 2 t(v′t+v)+ 2 vt= 0.After collecting terms, we have
(
1 −t^2
)
tv′′+( 2 − 4 t^2 )v′= 0.From here, it is fairly easy to find
v′′
v′=4 t^2 − 2
t( 1 −t^2 )=− 2
t +1
1 −t−1
1 +t(using partial fractions), and then
lnv′=−2ln(t)−ln( 1 −t)−ln( 1 +t).Finally, each side is exponentiated to obtain
v′=^1
t^2 ( 1 −t^2 )=^1
t^2+^1 /^2
1 −t+^1 /^2
1 +t,
v=−1
t+1
2 ln∣∣
∣∣^1 +t
1 −t∣∣
∣∣.
D. Higher-Order Equations
Linear homogeneous equations of order higher than 2 — especially order 4 —
occur frequently in elasticity and fluid mechanics. A general,nth-order homo-
geneous linear equation may be written
u(n)+k 1 (t)u(n−^1 )+···+kn− 1 (t)u(^1 )+kn(t)u= 0 , (26)