Differential Equations Sanchez

Differential Equations Sanchez Part II -Summary 6 Annihilators

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  c2 sin 2t  c3 cos 2t. Therefore an annihilator of y  5  7 sin 2t  4cos 2t is D D2 1 Note : y  4y  0 is the same as DD2 1 y  0

Problem 1. Find an annihilator for each of the following functions: Function Annihilator 1. y  5  2x  3ex  4sin x 1. D2 D 1(D2 1) 2. y  e2x cosh x  sinh x 2.D-3  ex  ex ex  ex   e2x    e2x ex  e3x    2 2  3. y  xe2x cos3x 3. The characteristic roots are 2  3i double roots. 2 D2  4D 13 4. y  t 2e2t  et  4tet 4. (D 1)2 (D  2)3 5. y  3 4x  sin x 5. D2 (D2 1) 6. y  3ex  5e2x 6. (D-1)(D+2) 7. 2t +5 7. D2 8. t 2et 8. D 13 9. t 2et cos 5t 9. The characteristic roots are 1 5i, triple roots. 3 D2  2D  26 10. t  2e2t  5t 2et  7 sin 2t 10. D2 D  2(D 1)3 D2  4 11. 1 6x  2x3 11. D4

2 x 2 12. 3x  xe cos 2x 12. D3 D2  2D  5

-1- 13. ex  2xex  x2ex 13. (D 1)(D 1)3

2 14. D(D 1)(D  2) 14. 2  ex   4  4ex  e2x 1 cos 2x 1 1 2 15. cos2 x    cos 2x 15. D(D  4) 2 2 2  ex  ex  3 3 2 3 16. x2 sinh x  x2   16. (D 1) (D 1  D 1    2 

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 homogeneous solution (complementary solution) of the differential

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  x x  Yh  c1  c2e  c2xe An annihilator for 5  t is D2  D3(D  1)2 y  0 2 x  x  y  k1  k 2x  k3x  k 4e  k5xe 2  Yp  Ax  Bx

-2- t t 2t b) (D  2)(D 1)y  3e  5  Yh  c1e  c2e An annihilator for 3et  5 is D(D 1)  D(D 1)(D  2)(D 1)y  D(D 1)et  5 0 t t 2t t  y  k1  k 2e  k 3e  k 4e  Yp  A  Be 2 2t t 2t 2t c) D(D 1)(D  2) y  5te  Yh  c1  c2e  c3e  c4xe An annihilator is (D  2)2  D(D 1)(D  2)4 y t 2t 2t 2 2t 3 2t  y  c1  c2e  c3e  c4xe  c5x e  c6x e 2 2t 3 2t  Yp  Ax e  cBx e d) D2  9D2  2D  2y  5cos3t  3et sin t 2  4  4(1)(2) 2  2i a2  2a  2  0  a    1 i 2 2 t  Yh  c1 sin 3t  c2 cos 3t  e c3 sin t  c4 cos t An annihilator for 5cos 3t  3et sin t is D2  9D2  2D  2 2 2  D2  9 D2  2D  2 y  y  c1 sin 3t  c2 cos3t  xc3 sin 3t  c4 cos3t t t  e c5 sin t  c6 cos t xe c7 sin t  c8 cos t t  Yp  xA sin 3t  B cos 3t xe Csin t  Dcos t 2 e) 2D3  3D2  3D  2y  ex  ex   (D 1)(D  2)(2D 1)y  e2x  2  e2x 1 x x 2x 2  Yh  c1e  c2e  c3e An annihilator for e2x  2  e2x is D(D 1)(D 1) 1 x 2 x x 2x 2  D(D 1) (D  2)(2D 1)y  0  y  c1  c2e  c3xe  c4e  c5e x  Yp  A  Bxe

-3- Problem 3. Solve the IVP y -5y  x - 2 2 5x D  5Dy  x  2  D(D  5)y  x  2  Yh  C1  C2e An annihilator for x  2 is D2  D3 (D  5)y  0 2 5x 2  y  c1  c2x  c3x  c4e  Yp  Ax  Bx

Yp  A  2Bx, Yp  2B  2B  5A  2Bx  x  2  1  5A  2B  2 B   9 1   10  Y  x  x2   9 p  10B  1  A  25 10  25 9 1  Y  C  C e5x  x  x2 1 2 25 10

. Problem 4. Solve the given differential equation by un-determinate coefficients y  y 12y  e4x 2 4x 4x 3x 4x D  D 12y  e  D  4)(D  3y  e  Yh  c1e  c2e An annihilator for e4x is (D  4) 2 3x 4x 4x  D  4) (D  3y  0  y  c1e  c2e  c3xe 4x 4x 4x 4x 4x  Yp  Axe , Yp  Ae  4Axe , Yp  8Ae 16xe y  y 12y  e4x  8Ae4x 16xe4x  Ae4x  4Axe4x 12Axe4x  e4x 1 1 1  7Ae4x  e4x  A   Y  xe4x  y  c e3x  c e4x  xe4x 7 p 7 1 2 7

Problem 5. Solve the DE of problem 3 by using the exponential shifting. (D2  D 12)y  e4x  e4x (D  4)(D  3)y  1  (D  4  4)(D  4  3)e4x y 1  D(D  7)e4xy 1  (D  7)e4xy x  c  e7x (D  7)e4xy xe7x  ce7x  (D  7  7)e7xe4xy xe7x  ce7x  De3x y xe7x  ce7x 1 1 c 1  e3xy  xe7x  e7x  e7x  c  e3xy  xe7x  c e7x  c 7 49 7 2 7 1 2 1  y  xe4x  c e4x  c e3x 7 1 2