Boyce/DiPrima 9 th ed, Ch 3.4:
Repeated Roots; Reduction of Order
Elementary Differential Equations and Boundary Value Problems, 9 th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.
Recall our 2 nd order linear homogeneous ODE a y
  b y
  cy

0 where a , b and c are constants.
Assuming an exponential soln leads to characteristic equation: y ( t )
 e rt  ar
2  br
 c

0
Quadratic formula (or factoring) yields two solutions, r
1 r

 b
 b
2 
4 ac
2 a
& r
2
:
When b 2 – 4 ac = 0, r
1 one solution: y
1
( t )

= r
2
= ce
 b t / 2 a b /2 a , since method only gives

Second Solution: Multiplying Factor v ( t )
We know that y
1
( t ) a solution
 y
2
( t )
 cy
1
( t ) a solution
Since y
1 and y
2 are linearly dependent, we generalize this approach and multiply by a function v , and determine conditions for which y
2 y
1
( t )
 e
 b t / 2 a a solution is a solution:

try y
2
( t )
 v ( t ) e
 b t / 2 a
Then y
2
( t )
 v ( t ) e
 b t / 2 a y

2
( t )
 v

( t ) e
 b t / 2 a  b
2 a v ( t ) e
 b t / 2 a y

2
( t )
 v

( t ) e
 b t / 2 a  b
2 a v

( t ) e
 b t / 2 a  b
2 a v

( t ) e
 b t / 2 a b
2

4 a
2 v ( t ) e
 b t / 2 a

Finding Multiplying Factor v ( t ) a y
  b y
  cy

0
Substituting derivatives into ODE, we seek a formula for v : e
 b t / 2 a


 a

 v

( t )
 b a v

( t )
 b
2
4 a
2 v ( t )


 b

 v

( t )
 b
2 a v ( t )


 cv ( t )




0 a v

( t ) a v

( t )
 b v

( t )


 b
2
4 a
 b b
2

2 a
2 v ( t )
 b v

( t )

4 a
 c

 v ( t )

0 b
2
2 a v ( t )
 cv ( t )

0 a v

( t )


 b
2
4 a

2 b
2
4 a

4 ac
4 a

 v ( t )

0 a v v

( t

( t )
)


 b
2 
4 a
0

4 ac v ( t )

 v ( t )
 k
3 t


0 k
4
 a v

( t )



 b
2
4 a

4 ac
4 a

 v ( t )

0

General Solution
To find our general solution, we have: y ( t )
 k
1 e
 bt / 2 a
 k
1 e
 bt / 2 a
 c
1 e
 bt / 2 a

 k
 k
2
3 t v ( t

) k e
4
 bt
 e
/ 2 a
 bt / 2 a
 c
2 te
 bt / 2 a
Thus the general solution for repeated roots is y ( t )
 c
1 e
 bt / 2 a  c
2 te
 bt / 2 a

Wronskian
The general solution is y ( t )
 c
1 e
 bt / 2 a  c
2 te
 bt / 2 a
Thus every solution is a linear combination of y
1
( t )
 e
 bt / 2 a
, y
2
( t )
 te
 bt / 2 a
The Wronskian of the two solutions is
W ( y
1
, y
2
)( t )

 e
 bt / 2 a b e
 bt / 2 a
2 a

1 te bt
2
 a bt /

 e
2 a
 bt / 2 a
 e
 bt / a 
1 bt
 e
 bt / a
2 a
 e
 bt / a 
0 for all t bt
2 a
Thus y
1 and y
2 form a fundamental solution set for equation.

Example 1
(1 of 2)
Consider the initial value problem y
 
4 y
 
4 y

0
Assuming exponential soln leads to characteristic equation: y ( t )
 e rt  r
2 
4 r

4

0

( r

2 )
2 
0
 r
 
2
So one solution is y
2
( t )
 v ( t ) y
1 e
( t

2
) t
 e

2 t y

2
( y

2
( t t
)
) and a second solution is found:
 v

( t ) e

2 t
 v

( t ) e

2 t

2 v ( t ) e

2 t

4 v

( t ) e

2 t

4 v ( t ) e

2 t
Substituting these into the differential equation and simplifying yields v " ( t )

0 , v ' ( t )
 k
1
, v ( t )
 k
1 t
 k
2 where c
1 and c
2 are arbitrary constants.

Example 1
(2 of 2)
Letting k
1

1 and k
2

0 , v ( t )
 t and y
2
( t )
 te 
2 t
So the general solution is y ( t )
 c
1 e

2 t  c
2 te

2 t
Note that both y
1 and y
2 tend to 0 as values of c
1 and c
2
Using initial conditions y ( 0 )
 c
1
2 c
1

1
 and c
2 y ' ( 0

)


3
1

3

 c
1

1 , c
2

5 t
 
2.0
y t
1.5
1.0
regardless of the y ( t )

( 1

5 t ) e

2 t
Therefore the solution to the
IVP is y ( t )
 e

2 t 
5 te

2 t
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
t

Example 2
(1 of 2)
Consider the initial value problem y
  y
 
0 .
25 y

0 , y
 

2 , y
  

1 / 3
Assuming exponential soln leads to characteristic equation: y ( t )
 e rt  r
2  r

0 .
25

0

( r

1 / 2 )
2 
0
 r

1 / 2
Thus the general solution is y ( t )
 c
1 e t / 2  c
2 te t / 2
Using the initial conditions:
1
2 c
1 c
1
 c
2


2

1

3


 c
1

2 , c
2
 
2
3
Thus y ( t )

2 e t / 2 
2
3 te t / 2
4 y t
3
2
1
1
2
1 y ( t )
 e t / 2
( 2

2 / 3 t )
2 3 4 t

Example 2
(2 of 2)
Suppose that the initial slope in the previous problem was increased y
 

2 , y
  

2
The solution of this modified problem is y ( t )

2 e t / 2  te t / 2
Notice that the coefficient of the second
6 y t term is now positive. This makes a big difference in the graph, since the
2
4 exponential function is raised to a positive power:
 
1 / 2

0
2
1 red : y ( t )
 e t / 2
( 2
 t ) blue : y ( t )
 e t / 2
( 2

2 / 3 t )
2 3 4 t

Reduction of Order
The method used so far in this section also works for equations with nonconstant coefficients: y
  p ( t ) y
  q ( t ) y

0
That is, given that y
1 y
2
( t )
 v ( t ) y
1
( t ) y

2
( t )
 v

( t ) y
1
( t )
 is solution, try y
2 v ( t ) y
1

( t ) y
2

( t )
 v

( t ) y
1
( t )

2 v

( t ) y
1

( t )

= v v ( t ) y
1

( t )
( t ) y
1
:
Substituting these into ODE and collecting terms, y
1 v
 

2 y

1
 py
1
  y
1
  p y

1
 qy
1
 v

0
Since y
1 is a solution to the differential equation, this last equation reduces to a first order equation in v y
1 v
 

2 y

1
 py
1
 v
 
0

:

Example 3: Reduction of Order
(1 of 3)
Given the variable coefficient equation and solution y
1
2 t
2 y
 
3 t y
  y

0 , t

0 ; y
1
( t )
 t

1
,
, use reduction of order method to find a second solution: y
2
( t )
 v ( t ) t

1 y

2
( t ) y

2
( t )
 v

( t ) t

1
 v

( t ) t

1
 v ( t ) t

2

2 v

( t ) t

2 
2 v ( t ) t

3
Substituting these into the ODE and collecting terms,
2 t
2
 v
 t

1 
2 v
 t

2 
2 vt

3
  v
 t

1  vt

2

 vt

1 
0

2 v
 t

4 v
 
4 vt

1 
3 v
 
3 vt

1  vt

1 
0


2 t v
  v
 
2 t u
  u

0
0 , where u ( t )
 v

( t )

Example 3: Finding v ( t )
(2 of 3)
To solve
2 t u
  u

0 , u ( t )
 v

( t ) for u , we can use the separation of variables method:
2 t du dt
 u

0
 u

 t
1 / 2 e
C
 du u

  1
2 t dt u
 ct
1 / 2
,
 ln u since t

0 .

1 / 2 ln t

C
Thus v
  ct 1 / 2 and hence v ( t )

2 / 3 c t 3 / 2
 k

Example 3: General Solution
(3 of 3)
Since
Recall v ( t ) y
2
( t )



2 / 3
2 / 3 c c t 3 / t 3 /
2
2

 k k y
1
( t )
 t

1
 t

1 
2 / 3 c t 1 / 2
 k t

1
So we can neglect the second term of y
2 y
2
( t )
 t 1 / 2 to obtain
The Wronskian of
W ( y
1
, y
1
( t ) and y
2
)( t )

3 / 2 t

3 / 2 y
2
( t )

0 , can be computed t

0
Hence the general solution to the differential equation is y ( t )
 c
1 t

1  c
2 t 1 / 2