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Differential Equations
Part II -Summary 6
Annihilators
Sanchez
Annihilators. An annihilator of a function y=f(t) is a linear differential P(D) that satisfies
the condition P(D)[f(t)] =0. This is the same as saying that f(t) is a solution of the
homogenous differential equation P(D)f(t)=0.
Example: the homogeneous solution of the linear differential equation
y   4y  0 is y  c1  c 2 sin 2t  c 3 cos 2t . Therefore an annihilato r of


Note : y   4y  0 is the same as DD 2  1 y  0
y  5  7 sin 2t  4 cos 2t is D D 2  1
Problem 1. Find an annihilator for each of the following functions:
Function
Annihilator
1. D 2 D  1(D 2  1)
1. y  5  2x  3e x  4 sin x
2. y  e 2x cosh x  sinh x 
 e x  e x e x  e x
 e 2x 


2
2

 2x x
 e e  e 3x


2.D-3
3. The characteri stic roots are 2  3i
3. y  xe 2x cos 3x
double roots .
D 2  4D  132
4. y  t 2 e 2t  e t  4te t
5. y  3  4x  sin x
4. ( D  1) 2 ( D  2) 3
5. D 2 ( D 2  1)
6. (D-1)(D+2)
6. y  3e x  5e 2x
7. 2t +5
7. D 2
8. D  13
9. The characteri stic roots are
8. t 2 e  t
9. t 2 e  t cos 5t
 1  5i, triple roots .
D 2  2D  263

10. t  2e 2t  5t 2 e t  7 sin 2t
10. D 2 D  2( D  1) 3 D 2  4
11. 1  6x  2x 3
11. D 4
12. 3x 2  xe x cos 2x
12. D 3 D 2  2D  5

-1-
2

13. e  x  2xe x  x 2 e x

14. 2  e x
13. ( D  1)( D  1) 3
2  4  4e x  e 2x
15. cos 2 x 
14. D( D  1)( D  2)
1  cos 2x 1 1
  cos 2x
2
2 2
 e x  e x
16. x 2 sinh x  x 2 

2

15. D( D 2  4)





3
16. ( D  1) 3 ( D  13  D 2  1
Finding the general form of the particular solution of a non-homogeneous linear
differential equation by using annihilators.
Step 1. Express the DE in linear differential form, that is, L(y)=g(x)
Step 2. Find the homogeneous solution (complementary solution) of the differential
equation, that is find the general solution of L(y)=0
Step 3.. Find an annihilator L1for g(x), that is L1(g(x)=0
Step 4. Operate on both sides of the non-homogeneous equation with the annihilator L1,
that is,
L1 L(y)  L1 (g(x)  0
Step 5. Find the homogeneou s solution (complemen tary solution) of the differenti al
equation L1 L( y )  0
Step 6. The general for the particular solution of L( y )  g( x ) is given by
e lim inating from the solution of L1 L( y )  0, the terms which belong to the
solution of L( y )  0
Problem 2. Find the general form of a particular solution of the following differential
equation.
a) D( D  1) 2 y  5  x
 Yh  c1  c 2 e  x  c 2 xe  x
An annihilato r for 5  t is D 2  D 3 ( D  1) 2 y  0
 y  k 1  k 2 x  k 3 x 2  k 4 e  x  k 5 xe  x
 Yp  Ax  Bx 2
-2-
b ) ( D  2)( D  1)y  3e t  5  Yh  c1e  t  c 2 e 2t
An annihilato r for 3e t  5 is D( D  1)


 D( D  1)( D  2)( D  1)y  D( D  1) e t  5  0
 y  k 1  k 2 e  t  k 3 e t  k 4 e 2t
 Yp  A  Be t
c) D( D  1)( D  2) 2 y  5te 2t  Yh  c1  c 2 e t  c 3 e 2t  c 4 xe 2t
An annihilato r is ( D  2) 2  D( D  1)( D  2) 4 y
 y  c1  c 2 e t  c 3 e  2t  c 4 xe  2t  c 5 x 2 e  2t  c 6 x 3 e  2t
 Yp  Ax 2 e  2t  cBx 3 e  2t



d ) D 2  9 D 2  2D  2 y  5 cos 3t  3e t sin t
a 2  2a  2  0  a 
2  4  4(1)( 2)
2
2  2i
 1 i
2

 Yh  c1 sin 3t  c 2 cos 3t  e t c 3 sin t  c 4 cos t 


An annihilato r for 5 cos 3t  3e t sin t is D 2  9 D 2  2D  2

 D2  9
2 D 2  2D  22 y

 y  c1 sin 3t  c 2 cos 3t  xc 3 sin 3t  c 4 cos 3t 
 e t c 5 sin t  c 6 cos t   xe t c 7 sin t  c 8 cos t 
 Yp  xA sin 3t  B cos 3t   xe t C sin t  D cos t 

 
e) 2D 3  3D 2  3D  2 y  e x  e  x
 Yh  c1e
x
 c 2e
2x
 c3
2  (D  1)(D  2)(2D  1)y  e 2x  2  e 2x
1
x
e2
An annihilato r for e 2x  2  e  2x is D( D  1)( D  1)
2
 D( D  1) ( D  2)( 2D  1)y  0  y  c1  c 2 e
 Yp  A  Bxe  x
-3-
x
 c 3 xe
x
 c 4e
2x
 c5
1
x
e2
Problem 3. Solve the IVP y  - 5y   x - 2
D 2  5Dy  x  2  D(D  5)y  x  2  Yh  C1  C 2e 5x
An annihilato r for x  2 is D 2  D 3 ( D  5)y  0
 y  c1  c 2 x  c 3 x 2  c 4 e 5x  Yp  Ax  Bx 2
Yp  A  2Bx, Yp  2B  2B  5A  2Bx  x  2
1

  5A  2B  2 B   10
9
1


 Yp 
x  x2
9
25
10
  10B  1
 A
25

9
1
 Y  C1  C 2 e 5x 
x  x2
25
10
.
Problem 4.
Solve the given differential equation by un-determinate coefficients
y   y   12y  e 4x
D 2  D  12y  e 4x  D  4)(D  3y  e 4x  Yh  c1e 3x  c 2e 4x
An annihilato r for e 4x is ( D  4)


 D  4) 2 ( D  3 y  0  y  c1e 3x  c 2 e 4x  c 3 xe 4x
 Yp  Axe 4x , Yp  Ae 4x  4Axe 4x , Yp  8Ae 4x  16xe 4x


y   y   12y  e 4x  8Ae 4x  16xe 4x  Ae 4x  4Axe 4x  12Axe 4x  e 4x
 7 Ae 4x  e 4x  A 
1
1
1
 Yp  xe 4x  y  c1e 3x  c 2 e 4x  xe 4x
7
7
7
Problem 5.
Solve the DE of problem 3 by using the exponential shifting.
( D 2  D  12)y  e 4x  e 4x ( D  4)( D  3)y  1






 ( D  4  4)( D  4  3) e  4x y  1  D( D  7 ) e  4x y  1  ( D  7 ) e  4x y  x  c




 e 7 x ( D  7 ) e  4x y  xe 7 x  ce 7 x  ( D  7  7 ) e 7 x e  4x y  xe 7 x  ce 7 x
 
 D e 3x y  xe 7 x  ce 7 x
 e 3x y 
y
1 7x 1 7x c 7x
1
xe  e  e  c 2  e 3x y  xe 7 x  c1e 7 x  c 2
7
49
7
7
1 4x
xe  c1e 4x  c 2 e  3x
7
-4-