Solve the difference equation

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1 Solve the differece equatio Solutio: y + 3 3y + + y 0 give tat y 0 4, y 0 ad y 8. Let Z{y()} F() Taig Z-trasform o both sides i (), we get y + 3 3y + + y 0 () Z y + 3 3y + + y Z 0 Z y + 3 3Z y + + Z y 0 () By liear property Substitutig (3) ad (4) i (), we get To fid poles: The poles are, Z y F 3 y 0 y y (3) Z y + F y 0 (4) 3 F 3 y 0 y y 3 F y 0 + F 0 3 F 4 3 (0) 8 3 F 4 + F F F F Z{y()} y() Z Let F() F ,, F 4 ( + ) 4 + ( + )

2 F 4 + ( + ) Residue at the pole of order oe R ( ) F ( ) 4 + ( + ) 4 + ( + ) 4() + ( + ) 8 3 Residue at the pole of order oe R + F ( ) + 4( ) 3 Solve the differece equatio Solutio: y Z y ( ) 3 Sum of all residues y + + 6y + + 9y give tat y 0 y 0 Let Z{y()} F() Taig Z-trasform o both sides i (), we get y + + 6y + + 9y () Z y + + 6y + + 9y Z Z y + + 6Z y + + 9Z y () By liear property

3 Substitutig (3) ad (4) i (), we get Z y + F y 0 y (3) Z y + F y 0 (4) F y 0 y + 6 F y 0 + 9F() F + 6 F + 9F() F F Z{y()} y() Z Let F() F To fid poles: , 3, The poles are 3, Residue at the pole 3 of order two R d 3 d + 3 F d 3 d + 3 d 3 d

4 Residue at the pole of order oe R F + 3 Fid Z Solutio: + )( y Z y Sum of all residues + 5 Let F() + )( + F + )( + + )( + To fid poles: + )( + 0, The poles are, Residue at the pole of order oe R + F + + )( + ( + ) + Residue at the pole of order oe R

5 + F + + )( + ( + ) + Z + )( + Sum of all residues Z + )( + Fid Z Solutio: F Let F() )( To fid poles: )( 0, The poles are, Residue at the pole of order oe R F 0 0 )( 0 0 Residue at the pole of order oe R F 0 )(

6 0 0 0 Z Z Sum of all residues Fid Z a)( b Solutio: Let F() a)( b F a)( b + a)( b To fid poles: a)( b 0 a, b The poles are a, b Residue at the pole a of order oe R a a F a a + a)( b + a b a+ a b Residue at the pole of order oe R b a F b b + a)( b b + a b+ b a

7 Z a)( b Sum of all residues Z a)( b a+ a b + b+ b a Fid Z 8 )(4 + usig covolutio teorem. Solutio: Z 8 )(4 + Z Z + 4 Let F ad G + 4 f Z F Z g Z G Z Z 8 )(4 + Z F G f g f g( ) Z 8 )(

8 Fid Z 6 )(3 + usig covolutio teorem. Solutio: Z 6 )(3 + Z Z + 3 Let F ad G + 3 f Z F Z g Z G Z Z 6 )(3 + Z F G f g f g( ) Z 6 )(

9 State ad prove secod shiftig theorem i Z-trasform: Statemet: Let F Z f() the Z f( + ) F f 0 f f f Proof: Z f( + ) 0 f( + )...() Multiplyig ad dividig by i R.H.S of (), we get f( + ) 0 ( +) f + f f f() f 0 f f f 0 Sice 0 f() f 0 + f + f + f + Z f( + ) F f 0 f f f Solve x ( y ) p + y ( x )q (x y ). Solutio: x ( y ) p + y (x )q (y x ). The equatio is of the form Pp + Qq R Here P x ( y ), Q y (x ), R (y x )

10 Itegratig o both sides, we get dx x ( y ) dy y (x ) d (y x ) x dx + y dy y + x d (y x ) x dx + y dy (y x ) d (y x ) x dx + d dy y log x + log y log + log c Itegratig o both sides, we get x + y + c c 3 x +y + The geeral solutio is log x + log y + log log c log xy log c c xy xdx + ydy x x y + x y y d (y x ) xdx + ydy x y d (y x ) xdx + ydy (y x ) d (y x ) xdx + ydy d Φ c, c 3 0 Φ xy, x +y + 0 Solve m y p + (x l)q (ly mx). Solutio:

11 The equatio is of the form Pp + Qq R Here P m y, Q x l, R ly mx dx m y dy x l d ly mx ldx + mdy + d lm ly + mx lm + ly mx ldx + mdy + d 0 ldx + mdy + d 0 Itegratig o both sides, we get lx + my + c xdx + ydy + d mx xy + yx ly + ly mx xdx + ydy + d 0 xdx + ydy + d 0 Itegratig o both sides, we get x + y + c c 3 x +y + The geeral solutio is Φ c, c 3 0 Φ lx + my +, x +y + 0 Solve D 3 7DD 6D 3 si x + y Solutio: To fid complemetary fuctio: Put D m ad D, m 3 7m 6 0 m 3 m 6m 6 0 m m 6 m + 0 m m + m 6 m + 0 m + m m 6 0

12 m,,3 f y x + f y x + f 3 y + 3x To fid Particular Itegral: P. I si x + y D 3 7DD 6D 3 si x + y 4D 7D( ) 6 D si x + y 4D + 7D + 6D si x + y 3D + 6D 3D 6D si x + y (3D + 6D ) 3D 6D 3D si x + y 6D si x + y 9 36D 6 cos x + y 6 si x + y 9( 4) 36( ) 6 cos x + y 6 si x + y x 8D x 8 (6 cos x + y 6 si x + y )dx 6x 8 si x + y + cos x + y P. I x 6 si x + y + cos x + y The geeral solutio is f y x + f y x + f 3 y + 3x + x 6 si x + y + cos x + y Solve D 3 D e x + 3x y Solutio: To fid complemetary fuctio: Put D m ad D, m 3 m 0 m m 0 m 0,0, f y + xf y + f 3 y + x

13 To fid Particular Itegral: The geeral solutio is P. I e x e x D 3 D 3 () (0) ex 3 ex 8 P. I Solve D 3 7DD 6D 3 x y + e x+y Solutio: To fid complemetary fuctio: Put D m ad D, To fid Particular Itegral: P. I 3x y D 3 D 3x y D 3 D D 3 D x D y 3 + D D x y 6 D 3 D 3 3x y + D 3 3 D 3 x y + 6 D 4 x yx5 0 + x6 60 D x f y + xf y + f 3 y + x + ex 8 + yx5 0 + x6 60 m 3 7m 6 0 m 3 m 6m 6 0 m m 6 m + 0 m m + m 6 m + 0 x y D 3 7DD 6D 3 m + m m 6 0 m,,3 f y x + f y x + f 3 y + 3x x y D 3 7D 6D 3 D 3 x y D 3 7D + 6D 3 D 3 7D + 6D 3 D 3 x y D 3 + 7D + 6D 3 D 3 x y D 3 D 3 x y x5 y 60

14 e x+y P. I D 3 7DD 6D 3 e x+y ex+y ex+y The geeral solutio is f y x + f y x + f 3 y + 3x + x5 y 60 ex+y Solve 6DD + 5D xy + e x si y Solutio: The auxiliary equatio is m 6m m,5 The complemetary fuctio CF is φ y + x + φ y + 5x To fid particular itegral: xy P. I 6DD + 5D xy 6 D D + 5 D xy 6 D D 5 D 6 D D 5 D xy + 6 D D 5 D xy x + x + x + xy + 6 D D xy 5 D xy

15 xy + 6 x D xy + 6 D 3 x P. I x3 y 6 + 6x4 4 x3 y 6 + x4 4 D xm m! xm + m +! P. I e x si y 6DD + 5D e y e y ex 6DD + 5D e x+y 6DD + 5D e x y 6DD + 5D e x+y 6 () + 5() e x y 6 ( ) + 5 xe x+y D 6D ex y if deomiator becomes ero te muliply umerator by x ad differetiatig deomiator partially wit respect to D P. I The geeral solutio is Two mars: xe x+y () 6() ex y xex +y 8 ex y CF + P. I + P. I f y + x + f y + 5x + x3 y 6 + x4 4 xex+y 8 Defie uit impulse step fuctio ad its Z trasform. Defie covolutio of two sequeces. ex y If F 5 3 fid f 0 ad lim t f(t) Form a differece equatio by elimiatig the arbitrary costats A from y A 3

16 Defie uit step fuctio ad its Z trasform. State Fial value theorem for Z trasforms. If F fid f 0. Fid P.I of + 4DD e x. Solve 4DD 4D 0. Solve D 3 + D 4DD 4D 3 0

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