Transcript Math 260

Boyce/DiPrima 9th 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 2nd order linear homogeneous ODE
ay  by  cy  0
where a, b and c are constants.
Assuming an exponential soln leads to characteristic equation:
y(t )  ert  ar2  br  c  0
Quadratic formula (or factoring) yields two solutions, r1 & r2:
 b  b 2  4ac
r
2a
When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives
one solution:
y1 (t )  cebt / 2a
Second Solution: Multiplying Factor v(t)
We know that
y1 (t ) a solution y2 (t )  cy1 (t ) a solution
Since y1 and y2 are linearly dependent, we generalize this
approach and multiply by a function v, and determine
conditions for which y2 is a solution:
y1 (t )  ebt / 2a a solution try y2 (t )  v(t )ebt / 2a
Then
y2 (t )  v(t )e b t / 2 a
b
v(t )e b t / 2 a
2a
2
b
b
b
y2(t )  v(t )e b t / 2 a  v(t )e b t / 2 a  v(t )e b t / 2 a  2 v(t )e b t / 2 a
2a
2a
4a
y2 (t )  v(t )e b t / 2 a 
ay  by  cy  0
Finding Multiplying Factor v(t)
Substituting derivatives into ODE, we seek a formula for v:
e
b t / 2 a
 

 
b
b2
b

a v(t )  v(t )  2 v(t )  b v(t )  v(t )  cv(t )  0
a
4a
2a

 
 

b2
b2
av(t )  bv(t ) 
v(t )  bv(t )  v(t )  cv(t )  0
4a
2a
 b2 b2

av(t )   
 c v(t )  0
 4a 2a

 b 2 2b 2 4ac 
  b 2 4ac 
v(t )  0  av(t )  
v(t )  0
av(t )   


4a 
4a 
 4a 4a
 4a
 b 2  4ac 
v(t )  0
av(t )  
 4a 
v(t )  0  v(t )  k3t  k 4
General Solution
To find our general solution, we have:
y (t )  k1e bt / 2 a  k 2 v(t )e  bt / 2 a
 k1e bt / 2 a  k3t  k 4 e bt / 2 a
 c1e bt / 2 a  c2te bt / 2 a
Thus the general solution for repeated roots is
y(t )  c1ebt / 2a  c2tebt / 2a
Wronskian
The general solution is
y(t )  c1ebt / 2a  c2tebt / 2a
Thus every solution is a linear combination of
y1 (t )  ebt / 2a , y2 (t )  tebt / 2a
The Wronskian of the two solutions is
e bt / 2 a
W ( y1 , y2 )(t )  b bt / 2 a
 e
2a
te bt / 2 a
 bt  bt / 2 a
1   e
 2a 
 bt 
 bt 
 e bt / a 1    e bt / a  
 2a 
 2a 
 e bt / a  0 for all t
Thus y1 and y2 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 )  ert  r 2  4r  4  0  (r  2)2  0  r  2
So one solution is y1(t )  e2t and a second solution is found:
y2 (t )  v(t )e 2t
y2 (t )  v(t )e  2 t  2v(t )e  2 t
y2(t )  v(t )e  2 t  4v(t )e  2 t  4v(t )e  2 t
Substituting these into the differential equation and
simplifying yields v" (t )  0, v' (t )  k1, v(t )  k1t  k 2
where c1 and c 2 are arbitrary constants.
Example 1
(2 of 2)
Letting k1  1 and k 2  0, v(t )  t and y 2(t )  te  2t
So the general solution is
y(t )  c1e2t  c2te2t
Note that both y1 and y 2 tend to 0 as t   regardless of the
values of c1 and c2
Using initial conditions
y (0)  1 and y ' (0)  3
c1
 1
  c1  1, c2  5
 2c1  c2  3
Therefore the solution to the
IVP is y(t )  e2t  5te2t
yt
2.0
y(t )  (1  5t )e2t
1.5
1.0
0.5
t
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Example 2
(1 of 2)
Consider the initial value problem
y  y  0.25y  0, y0  2, y0  1/ 3
Assuming exponential soln leads to characteristic equation:
y(t )  ert  r 2  r  0.25  0  (r 1/ 2)2  0  r  1/ 2
Thus the general solution is
y(t )  c1et / 2  c2tet / 2
Using the initial conditions:

c1
1
c1
2
Thus
 c2
2
2

1   c1  2, c2  

3
3

y (t )  2e
t/2
2 t/2
 te
3
yt
4
y(t )  et / 2 (2  2/3 t )
3
2
1
t
1
1
2
2
3
4
Example 2
(2 of 2)
Suppose that the initial slope in the previous problem was
increased
y0  2, y0  2
The solution of this modified problem is
y(t )  2et / 2  tet / 2
yt
6
Notice that the coefficient of the second
term is now positive. This makes a big
difference in the graph, since the
exponential function is raised to a
positive power:   1/2  0
red : y (t )  et / 2 (2  t )
4
blue : y (t )  et / 2 (2  2/3 t )
2
t
1
2
2
3
4
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 y1 is solution, try y2 = v(t)y1:
y2 (t )  v(t ) y1 (t )
y2 (t )  v(t ) y1 (t )  v(t ) y1 (t )
y2(t )  v(t ) y1 (t )  2v(t ) y1 (t )  v(t ) y1(t )
Substituting these into ODE and collecting terms,
y1v  2 y1  py1 v   y1  py1  qy1 v  0
Since y1 is a solution to the differential equation, this last
equation reduces to a first order equation in v :
y1v  2 y1  py1 v  0
Example 3: Reduction of Order
(1 of 3)
Given the variable coefficient equation and solution y1,
2t 2 y  3ty  y  0, t  0; y1 (t )  t 1 ,
use reduction of order method to find a second solution:
y2 (t )  v(t ) t 1
y2 (t )  v(t ) t 1  v(t ) t  2
y2(t )  v(t ) t 1  2v(t ) t  2  2v(t ) t 3
Substituting these into the ODE and collecting terms,

 

2t 2 vt 1  2vt 2  2vt 3  3t vt 1  vt 2  vt 1  0
 2vt  4v  4vt 1  3v  3vt 1  vt 1  0
 2tv  v  0
 2tu   u  0, where u (t )  v(t )
Example 3: Finding v(t)
(2 of 3)
To solve
2tu  u  0, u(t )  v(t )
for u, we can use the separation of variables method:
2t
du
u  0 
dt
 u t
1/ 2 C
e
du
1

 u  2t dt  ln u  1/2 ln t  C
 u  ct1/ 2 , since t  0.
Thus
v  ct 1/2
and hence
v(t )  2/3 c t 3/ 2  k
Example 3: General Solution
Since
(3 of 3)
v(t )  2/3 c t 3/ 2  k
y2 (t )  2/3 c t 3/ 2  k t 1  2/3 c t1/ 2  k t 1
y1 (t )  t 1
Recall
So we can neglect the second term of y2 to obtain
y2 (t )  t1/ 2
The Wronskian of y1 (t ) and y2 (t ) can be computed
W ( y1 , y 2)(t )  3/2 t  3 / 2  0, t  0
Hence the general solution to the differential equation is
y(t )  c1t 1  c2t1/ 2