Lecture Notes in Differential Equations - Bruce E. Shapiro

279
Case 2: Two equal real roots. Suppose r 1 = r 2 = r. Then
r = 1 − α
2
(29.26)
because the square root must be zero (ie., (1 − α) 2 − 4β = 0) to give a
repeated root.
We have one solution is given by y 1 = t r = t (1−α)/2 . To get a second
solution we can use reduction of order.
From Abel’s formula the Wronskian is
( ∫ ) αtdt
W = C exp −
t 2 = C exp
(
−α
∫ ) dt
= Ce −α ln t = C t
t α (29.27)
By the definition of the Wronskian a second formula is given by
W = y 1 y 2 ′ − y 2 y 1 ′ (29.28)
(
= t (1−α)/2) ( ) 1 − α (
y 2 ′ − y 2
′ t (−1−α)/2) (29.29)
2
Equating the two expressions for W ,
(
t (1−α)/2) ( ) 1 − α (
y 2 ′ − y 2
′ t (−1−α)/2) = C 2
t α (29.30)
( ) 1 − α t
y 2 ′ (−1−α)/2
C
−
2 t (1−α)/2 y′ 2 =
(29.31)
t α (t (1−α)/2 )
( ) α − 1
y 2 ′ + y ′ C
2 =
(29.32)
2t t (1+α)/2
This is a first order linear equation; an integrating factor is
(∫ ) ( )
α − 1
α − 1
µ(t) = exp dt = exp ln t = t (α−1)/2 (29.33)
2t
2
If we denote q(t) = Ct −(1+α)/2 then the general solution of (29.32) is
y = 1 [∫
]
µ(t)q(t)dt + C 1 (29.34)
µt
[∫
]
= t (1−α)/2 t (α−1)/2 Ct −(1+α)/2 dt + C 1 (29.35)
∫ dt
= Cy 1
t + C 1y 1 (29.36)
= Cy 1 ln |t| + C 1 y 1 (29.37)