Transcript Math 260 - National University of Singapore

Ch 4.4: Variation of Parameters
The variation of parameters method can be used to find a
particular solution of the nonhomogeneous nth order linear
differential equation
Ly  y ( n)  p1 (t ) y (n1)   pn1 (t ) y  pn (t ) y  g (t ),
provided g is continuous.
As with 2nd order equations, begin by assuming y1, y2 …, yn
are fundamental solutions to homogeneous equation.
Next, assume the particular solution Y has the form
Y (t )  u1 (t ) y1 (t )  u2 (t ) y2 (t )   un (t ) yn (t )
where u1, u2,… un are functions to be solved for.
In order to find these n functions, we need n equations.
Variation of Parameters Derivation (2 of 5)
First, consider the derivatives of Y:
Y   u1 y1  u2 y2   un yn   u1 y1  u2 y2   un yn 
If we require
then
u1 y1  u2 y2   un yn  0
Y   u1 y1  u2 y2   un yn   u1 y1  u2 y2   un yn 
Thus we next require
u1 y1  u2 y2   un yn  0
Continuing in this way, we require
u1 y1( k 1)  u2 y(2k 1)   un y(nk 1)  0, k  1,, n 1
and hence
Y ( k )  u1 y1(k )   un yn(k ) , k  0,1,, n 1
Variation of Parameters Derivation (3 of 5)
From the previous slide,
Y ( k )  u1 y1(k )   un yn(k ) , k  0,1,, n 1
Finally,

 
Y ( n)  u1 y1(n1)   un yn(n1)  u1 y1(n)   un yn(n)

Next, substitute these derivatives into our equation
y( n)  p1 (t ) y (n1)   pn1 (t ) y  pn (t ) y  g (t )
Recalling that y1, y2 …, yn are solutions to homogeneous
equation, and after rearranging terms, we obtain
u1 y1(n1)   un yn(n1)  g
Variation of Parameters Derivation (4 of 5)
The n equations needed in order to find the n functions
u1, u2,… un are
u1 y1    un y1  0
u1 y1    un yn  0

u1 y1( n 1)    un yn( n 1)  g
Using Cramer’s Rule, for each k = 1, …, n,
uk (t ) 
g (t )Wk (t )
, where W (t )  W ( y1 ,, yn )(t )
W (t )
and Wk is determinant obtained by replacing kth column of
W with (0, 0, …, 1).
Variation of Parameters Derivation (5 of 5)
From the previous slide,
g (t )Wk (t )
, k  1,, n
W (t )
uk (t ) 
Integrate to obtain u1, u2,… un:
uk (t )  
t
t0
g ( s)Wk ( s)
ds, k  1,, n
W (s)
Thus, a particular solution Y is given by
 t g (s)Wk (s) 
Y (t )   
ds yk (t )
t0
W ( s)
k 1 

n
where t0 is arbitrary.
Example (1 of 3)
Consider the equation below, along with the given solutions
of corresponding homogeneous solutions y1, y2, y3:
y  y  y  y  e2t , y1 (t )  et , y2 (t )  tet , y3 (t )  et
Then a particular solution of this ODE is given by
 t e 2 sWk ( s) 
Y (t )   
ds yk (t )
t0 W ( s )
k 1 

3
It can be shown that
et
W (t )  et
et
tet
t  1et
t  2et
e t
 e t  4et
e t
Example (2 of 3)
Also,
0
W1 (t )  0
1
tet
t  1et
t  2et
e t
 e t  2t  1
e t
et
W2 (t )  et
et
0 e t
0  e t  2
1 e t
et
W3 (t )  et
tet
t  1et
et
t  2et
0
0  et
1
Example (3 of 3)
Thus a particular solution is
 t e 2 sWk ( s) 
Y (t )    
ds yk (t )
t0 W ( s )
k 1 

2s
2s
2s 2s
t e  2 s  1
t 2e
t e e
t
t
 et 
ds

te
ds

e
ds
s
s
s


t0
t
t
0 4e
0
4e
4e
et t s
tet t s
e t t 4 s
   e 2s  1ds 
e ds 
e ds


t
t
t
4 0
2 0
4 0
3
Choosing t0 = 0, we obtain
1
1
1
1
Y (t )   et  tet  e t  e 2t
4
2
12
3
More simply,
1
Y (t )  e 2t
3