Transforms and Boundary Value Problems

Transcription

1 Transforms and Boundary Vaue Probems (For B.Tech Students (Third/Fourth/Fifth Semester Soved University Questions Papers Prepared by Dr. V. SUVITHA Department of Mathematics, SRMIST Kattankuathur 6.

2 CONTENTS Nov. 6 Q&A May 7 Q&A Nov. 7 Q&A May 8 Q&A

3 Soved University Question Papers-5MA, SRMIST B.Tech. Degree Examination, November 6 Third/Fourth/Fifth Semester 5MA-Transforms and Boundary Vaue Probems Time: Three hours Max. Marks: Part - A ( Marks Answer ALL Questions. The compete integra of p + q is (A ax + by (B a(x + y + b (C ax + by + c (D ax by + a So: Given p + q. This is of the form F (p, q. Hence the compete integra is ax + by + c where a + b. i.e., b a Therefore the compete integra is ax + ( a y + c.. The compete integra of p qx is (A ax ay+c (B ax +ay+c (C ay +ax+c (D x y+c So: Given p qx. This is of the form F (x, p, q. Let q a then p ax. We know that d pdx + qdy d axdx + ady Integrating, we get ax + ay + c.. The compementary function of (D + DD D x y is (A φ (y x + φ (y x (B φ (y + x + φ (y + x (C φ (y + x + φ (y x (D φ (y + x + φ (y x So: The auxiiary equation is m + m where D m, D. (m (m + m,. C.F. φ (y + x + φ (y x. Ans. D 4. Find the particuar integra of (D + DD + D e x+y (A ex+y 9 So: P.I Ans. A (B ex y (C ex+y 9 9 D e x+y ex+y + DD + D 9 (D ex y 9 where D, D. u 5. The P.D.E. x + u x y + u y of the form (A Eiptic (B Paraboic (C Hyperboic (D None of these So: Here A, B, C. Hence B 4AC 4 4. Therefore the equation is paraboic. 6. sin x is periodic function with period (A (B (C (D 4 Ans. C

4 V. Suvitha, Department of Mathematics, SRMIST 7. a a f(xdx if f(x is (A Odd (B Even (C Periodic (D Neither even nor odd Ans. A 8. For haf range cosine series of f(x cos x in (, the vaue of a is (A 4 (B (C 4 (D So: a f(xdx cos xdx [sin x]. Ans. D 9. The one dimensiona wave equation is (A u t α u x (B u t a u x (C y t α y x (D u x y a t. How many initia and boundary condition are required to sove u t a u x (A Two (B Three (C Five (D Four Ans. D. The one dimensiona heat equation is (A u x + u y u (B t α u x (C u t a u x (D u x α u t. One dimensiona heat equation is used to find (A Density (B Temperature Distribution (C Time (D Dispacement. The Fourier transform of f(x is (A f(xe ist dt (B (C f(te isx dx (D f(xe isx dx f(se isx dx 4. Under Fourier cosine transform f(x x is (A Compex (B Inverse function (C Cosine function (D Sef-reciproca function So: We know that F c (x n Γ(n s n cos n. Taking n F c(/ x / s. Ans. D 5. F [e iax f(x] is (A F (s + a (B F (s a (C F (as (D F So: F [e iax f(x] Ans. A e iax f(xe isx dx ( s a e i(s+ax f(xdx F (s + a.

5 Soved University Question Papers-5MA, SRMIST 6. F (f(x g(x is (A F (s + G(s (B F (s G(s (C F (sg(s (D F (s/g(s Ans. C 7. Z(5 is (A (B 5. So: We know that Z(k (C 5. (D k Z(5 5. ( 8. Z n ( ( (A og if > (B og ( ( + + (C og (D og if > ( So: Z n n n n + + ( + og ( ( ( ( Z og og og. n Ans. A ( 9. Find Z ( (A n + (B n (C n (D n So: Z {n} d Z( by property d Z {n} d ( [ ] (.. d ( (. ( Taking inverse Z -transform, we get n Z (. n. Poes of φ( ( ( are (A, (B, (C, (D So: The poes are ( (,. Part - B (5 4 Marks Answer ANY FIVE Questions if <.. Form the PDE by eiminating the arbitrary function from f(x + y. So: Given f(x + y ( Equation ( partiay differentiating w.r.to x and y, we get x f (x + y.x and y f (x + y.y Therefore p x f (x + y where p ( x

6 4 V. Suvitha, Department of Mathematics, SRMIST and q y f (x + y where q y. ( From ( and (, we get p x q. Hence py qx. y. Find the compete integra of p + q. So: Given p + q. ( This is of the form F (p, q,. Let f(u. Assume that u x + ay then p d du, q a d du. Substituting the vaues of p, q in (, we get ( d ( + a d du du d + a ( /+ Integrating, we get ( / + + a x + ay + b. u + a + b du + a. Find the RMS vaue of f(x x x, in < x <. b (f(x dx a So: RMS vaue of f(x b a (x x dx (x + x 4 x dx (x + x 4 dx since x is odd State any two assumptions in deriving one dimensiona wave equation and write its a possibe soutions. So: The motion takes pace entirey in one pane. This pane is chosen as the xy pane Gravitationa force may be negected The effect of friction is negigibe The string is perfecty fexibe, etc. (A e λx + B e λx (C e λat + D e λat The possibe soutions are y(x, t (A cos λx + B sin λx(c cos λat + D sin λat (A x + B (C t + D

7 Soved University Question Papers-5MA, SRMIST 5 5. Find the Fourier cosine transform of f(x e ax, a >. So: F c (e ax e ax cos sxdx a a + s. [ ] e ax a ( a cos sx + s sin sx + s sin θ 6. Prove that Z(sin nθ if >. cos θ + Proof: We know that Z {a n } if > a. Taking a eiθ a Z { e inθ} e iθ (cos θ + i sin θ ( cos θ i sin θ [( cos θ + i sin θ] Z(cos nθ + i sin nθ [( cos θ i sin θ][( cos θ + i sin θ] [( cos θ + i sin θ] [( cos θ + sin θ]. Equating imaginary parts, we get (sin nθ sin θ cos θ + if >. 7. Find Z[{n(n }]. So: Z[{n(n }] Z[ { n n } ] Z { n } Z {n} Now Z {n} [See Q. No. 9, Page No. ] and ( Z { n } Z {n.n} d ( [ ( ]..( d ( ( 4 ( + Hence Z[{n(n }] ( ( (. Part - C (5 6 Marks Answer ALL Questions ( + (. 8. a.i. Find the singuar soution of px + qy + p + q. So: Given px + qy + p + q. This is Cairaut s form. The compete soution is ax + by + a + b. ( Partiay differentiating w.r.to a and b, we get x + a and y + b. a x and b y Substituting ( in (, we get x y + x 4 + y 4 4 x y Hence 4 + x + y. ii. Find the genera soution of x( y p + y(x q (y x. dx So: The auxiiary equations are x( y dy y(x d (y x. Taking the Lagrange s mutipiers x, y,, we get xdx x ( y ydy y (x d (y x xdx + ydy + d xdx + ydy + d Each is equa to x ( y (

8 6 V. Suvitha, Department of Mathematics, SRMIST Hence xdx + ydy + d. Integrating, we get x + y + a x + y + a a Aso, taking the Lagrang s mutipiers x, y,, we get dx x y dy y x d y x dx x + dy y + d dx x + dy y + d Each is equa to ( y Hence dx x + dy y + d Integrating, we get og x + og y + og og b xy b. Therefore, the genera soution is φ(x + y +, xy. (OR b. Sove (D D D sin(x + y + x y. So: The auxiiary equation is m m where D m, D. m (m m,,. The Compementary function (C.F. is φ (y + xφ (y + φ (y + x Particuar integra D D [sin(x + y + x y] D D D sin(x + y + D D D x y D P.I + P.I Now P.I D D sin(x + y D sin(x + y D D( repace by D, DD sin(x + y D cos(x + y and [ ] P.I D D x y D D D x y D ( D + D D + D + x y D [ ] D + D x y [x D y + 6D ] D x x5 y + x6 6 The compete soution is C.F. + P.I + P.I φ (y + xφ (y + φ (y + x cos(x + y + x5 y + x6 6.

9 Soved University Question Papers-5MA, SRMIST 7 9. a. Find the Fourier series of f(x x + x in (, of periodicity. Hence deduce that n 6. So: Given the function f(x is neither even nor odd. Let f(x a + a n cos nx + b n sin nx ( n n where a f(xdx, a n f(x cos nxdx and b n f(x sin nxdx. To find a, a n b n : a + a n (x + x dx xdx + x dx x dx, since x is odd and x is even. (x + x cos nxdx + [ ( sin nx x n [ 4( n n b n x cos nxdx + x cos nxdx, since x cos nx is odd cos n n x ( ( ] cos nx sin nx + n n ], since sin sin n (x + x sin nxdx x sin nxdx + x cos nxdx x sin nxdx x sin nxdx +, since x sin nx is odd [ ( ( ] cos nx sin nx x [ ( ] cos n n n n ( n n Substituting the vaues of a, a n, b n in (, we get x + x + [ 4 ( n n n cos nx ] n sin nx Deduction: x is an end point in the range. Hence the vaue of the Fourier series at x is equa to [f( + f( ] [( + + ( + ]. Hence + n 4( n cos n 4 n n n. Therefore n n 6.

10 8 V. Suvitha, Department of Mathematics, SRMIST (OR b. Find the Fourier series upto second harmonic from the foowing data: 4 5 x y So: Let f(x a + a n cos nx + b n sin nx where a f(x, n n m a n f(x cos nx and bn f(x sin nx. m m x f(x cos x sin x cos x sin x Now a [ ].9 a f(x cos x [ ] a f(x cos x [ ]. 6 b f(x sin x [ ].7 6 b f(x sin x [ ] Hence f(x cos x. cos x +.7 sin x.577 sin x.. a. A tighty string of ength has its end fastened at x, x. At t, the string is in the form f(x k(x x and then reeased. Find the dispacement at any point on the string at a distance x from one end and at any time t >. So: The dispacement of the string y(x, t is governed by u t a u x. The boundary conditions are (i y(, t, t (ii y(, t, t. The initia ( conditions are y (iii, x (iv y(x, kx( x, x. t t The proper soution is y(x, t (A cos λx + B sin λx(c cos λat + D sin λat. ( Using boundary condition (i in (, A(C cos λat + D sin λat A. A in (, we get y(x, t B sin λx(c cos λat + D sin λat. ( Appying the boundary condition (ii in (, B sin λ(c cos λat + D sin λat. B and sin n λ n λ n.

11 Soved University Question Papers-5MA, SRMIST 9 λ n in (, we get y(x, t B sin nx Using the initia[ condition (iii, we get y nx B sin C sin nat. na + D cos nat. na ] t ( y B sin nx [ + D. na ] B, D t t D in equation (, y(x, t B sin nx C cos nat The most genera soution is y(x, t n ( C cos nat + D sin nat. ( B n sin nx Using initia condition (iv, we get y(x, This is haf-range Fourier sine series. Therefore B n k k k(x x sin nx dx (x x nx cos n [ cos n. n + n n ( x k. n [ ( n + ] 8k 4k n [ ( n ] n if n is odd if n is even cos nat B n sin nx sin nx n K(x x. ], since sin sin n Substituting the vaue of B n in (4, we get y(x, t (OR nodd + ( 8k nx n sin cos nx n cos nat. b. A rod of ength cm has its end A and B kept at C and 8 C respectivey unti steady state conditions prevai. The temperature at each end is suddeny reduced to C and kept so. Find the resuting temperature function u(x, t. So: The P.D.E. of one dimensiona heat fow is u t α u x. ( In steady state, the P.D.E. becomes d u. ( dx In steady state, the soution is u(x ax + b. ( The initia conditions are u( and u( 8. Using these conditions in (, we obtain u( + b b and u( a + b 8 a + a. Therefore u(x x +. When the temperatures at each ends are reduced to ero. In transient state, the boundary conditions are (4

12 V. Suvitha, Department of Mathematics, SRMIST (i u(, t for a t (ii u(, t for a t. The initia condition is (iii u(x, x + for < x <. In transient state, the proper soution is u(x, t (A cos λx + B sin λxe α λ t. (4 Using (i in (4, we get u(, t Ae α λ t A. A in (4, u(x, t B sin λxe α λ t. (5 Using (ii in (5, we get u(, t B sin xe α λ t. Since B, sin λ λ n λ n. λ n nx in (5, we get u(x, t B sin e α n t/9. The most genera soution is u(x, t n Using (iii in (6, we get u(x, B n sin nx x +. n This is a haf range sine series. Therefore B n 5 (x + sin nx dx (x + nx cos n [ 8.( n n n 4 n [ 4( n ]. Substituting the vaue of B n in (6, we get u(x, t 4 n [ 4( n ] sin nx e α n t/9. n B n sin nx e α n t/9. (6 sin nx n ], since sin sin n. Find the Fourier transform of f(x given by f(x find the vaue of So: F {f(x} ( x cos x sin x x ( x cos dx. f(xe isx dx ( x e isx dx ( x (cos sx + i sin sxdx { x if x < if x > ( x cos sxdx + i ( x sin sxdx and hence

13 Soved University Question Papers-5MA, SRMIST F {f(x} ( x cos sxdx + i., since ( x sin sx is odd [( x s ( cos s [ s + 4 s [s cos s sin s] Using inversion formua f(x F {f(x} e isx ds ( ( ( ] sin sx cos sx sin sx ( x s + ( s ( ] sin s s x [s cos s sin s] (cos sx i sin sxds. s ( s cos s sin s Equating rea parts, we get s Putting x ( s cos s sin s s cos s ds 8 ( s cos s sin s s cos s ds 6. ( x cos x sin x Hence x cos x dx 6. (OR 4 s [s cos s sin s] e isx ds cos sxds ( x. dx b. i. Evauate (a + x (b + x using transform method. So: Consider f(x e ax and g(x e bx. F c (s. a a + s and G c(s. b b [See Q. No. 5, Page No. 5]. + s Using Parseva s identity F c (sg c (sds f(xg(xdx ab (a + s (b + s ds e (a+bx dx dx (a + x (b + x ab(a + b. ii. Find Fourier sine transform of So: ( F s x. x. sin sx x dx.. a. i. Find the inverse Z -transform of f( So: Given f( ab (a + s (b + s ds a + b sin θ dθ, putting sx θ θ ( using ong division method.

14 V. Suvitha, Department of Mathematics, SRMIST By ong division Hence n f(n n Therefore f(, f(, f( 4, f( 8, f(4, etc. ii. Find the inverse Z -transform of using residues. ( ( So: Given f( ( (. n 5 f( n has simpe poes at and. ( ( Therefore f(n R where R is the sum of the residue of f( n. R {Residue} im( and ( ( R {Residue} im( ( ( n. Therefore f(n R + R n. n n (OR b. Sove the equation y n+ + 6y n+ + 9y n n given y y using Z -transform. So: Taking Z -transform on both sides of the equation, we get Z(y [ n+ + 6Z(y n+ + 9Z(y n Z( n Y ( y y ] + 6 [(Y ( y ] + 9Y ( ( Y ( Y ( ( ( Y ( ( ( +. Y ( has simpe poe at and poe of order at. Therefore y(n R where R is the sum of the residue of Y ( n. n R {Residue} im( ( ( + n 5 and d n R {Residue} im ( + [ d ( ( + (.n n n ] [ ]. im ( ( n 5n 5. [ ] Hence y(n n 5 + ( n 5n 5.

15 Soved University Question Papers-5MA, SRMIST B.Tech. Degree Examination, May 7 Third/Fourth/Fifth Semester 5MA-Transforms and Boundary Vaue Probems Time: Three hours Max. Marks: Part - A ( Marks Answer ALL Questions. The PDE formed by eiminating the arbitrary function from f(x + y is (A px qy (B py qx (C p qy (D px q [So: See Q.No., Page No..]. Sove (D D D (A f (y x + f (y x + f (y + x (B f (y + f (y + f (y + x (C f (y + xf (y + f (y + x (D f (y + xf (y + f (y x So: The auxiiary equation is m m where D m, D. m (m m,, C.F. is f (y + xf (y + f (y + x. Ans. C. The particuar integra of (D x y is (A x5 y (B x y (C x 4 y (D x y So: P.I. D x y y. x D x4 y 4. The compete integra of px + qy + pq is (A px+qy+ab (B ax+by+ab (C ax+by+pq (D px+qy+pq So: Given px + qy + pq. This is cairaut s form. Hence the compete integra is ax + by + ab. 5. The constant a of the Fourier series for the function f(x x in x is (A (B (C (D So: a f(xdx xdx ( x. 6. The root mean square vaue of f(x in a x b is b (f(x dx b (f(x dx a a (A (B (C (D b a b + a b a (f(xdx b a 7. A function f(x is periodic with period T, if (A f(x+t f(t (B f(x+t f(x (C f(x+t f(x (D f(x+t.f(x

16 4 V. Suvitha, Department of Mathematics, SRMIST 8. The sum of Fourier series of f(x in x at x is f( + f( (A f( (B f( (C (D Ans. C 9. In wave equation y t a y x, a (A T m Ans. A (B k c (C m T (D stands for k m. The one dimensiona heat equation in steady state is (A u t (B u t (C u x u t Ans. D (D u x. The suitabe soution of u t α u xx is (A u (Ax + BC (B u (Ae λx + Be λx e α λ t (C u (A cos λx + B sin λxe α λ t (D u A t + B Ans. C. The steady state temperature of a rod of ength cm, whose ends are kept respectivey at C and C is (A x (B x + (C x + (D x + 5 So: In steady state, the P.D.E. becomes d u dx Therefore the soution is u(x ax + b ( The initia conditions are u( and u(. Using these conditions in (, we obtain u( + b b and u( a + a + a. Therefore u(x x +.. F [F (s.g(s] (A f(x.g(x (B f(x + g(x (C f(x g(x (D f(x g(x So: By definition F {f(x g(x} F (s.g(s f(x g(x F [F (s.g(s] Ans. C 4. F [f(x a] (A e ias F (a (B e iax F (a (C e ias F (x (D e ias F (s So: By definition F (s F {f(x} f(xe isx dx F {f(x a} F {f(t} Ans. D 5. If f(s F [f(x] then (A f(x dx (B f(x ae isx dx, Putting x a t dx dt f(te i(a+ts dt e ias f(x dx f(s ds (C f(te its dt e ias F (s. f(s dx (D f(s ds

17 Soved University Question Papers-5MA, SRMIST 5 6. Under Fourier cosine transform f(x x is (A Sef-reciproca function (B Inverse function (C Cosine function (D Compex function Ans. A [So: See Q.No. 4, Page No..] 7. Z[na n ] a (A ( a (B ( a (C So: We know that Z[nf(t] Ans. A [ 8. Z cos n ] (A + a ( a (D ( a Z[na n ] d [ ] d a df ( d (B (C + (D ( cos θ So: We know that Z(cos nθ cos θ +. When θ ( Z cos n Z [ 4 ] ( a (A a n (B na n+ [ (C na n ] (D a n+ [ So: We know that Z a ( a na n Z Ans. C. Poes of φ( n ( + ( are (A (B (C (D So: The poes are given by. Ans. A ( a Part - B (5 4 Marks Answer ANY FIVE Questions ] na n. a ( a.. Form a partia differentia equation by eiminating arbitrary constants a and b from (x a + (y b + c. So: Given (x a + (y b + c ( Differentiating partiay w.r.to x and y, we get (x a + and (y b + x y. Therefore x a p, y b q where p x, q y. Putting in (, p + q + c (p + q + c.. Define Root Mean Square (RMS vaue and find the RMS vaue of f(x x in < x <.

18 6 V. Suvitha, Department of Mathematics, SRMIST So: RMS vaue of f(x b a. (f(x dx b a ( x dx (( x. State various possibe soutions of one dimensiona heat equation. (A e λx + B e λx C e α λ t u(x, t (A cos λx + B sin λxc e α λ t (A x + B C 4. State convoution theorem and Parseva s identity for Fourier transform. So: Convoution theorem: The Fourier transform of the convoution of f(x and g(x is the product of their Fourier transforms. i.e., F {f(x g(x} F (s.g(s. Parseva s identity: If F (s is the Fourier transform of f(x then f(x dx F (s ds. 5. Find Z[. n + 5.( n ]. So: Z[. n + 5.( n ] Z[ n ] + 5Z[( n ] by property We know that Z[a n ] if > a. a Therefore Z[ n ] + 5Z[( n ] Cassify the PDE: xu xx + u yy. So: Here A x, B, C, < ; if x >, the equation is eiptic B 4AC 4x > ; if x <, the equation is hyperboic ; if x, the equation is paraboic 7. Sove: p qx So: Given p qx. This is of the form F (x, p, q. Let q a then p ax. we know that d pdx + qdy d axdx + ady Integrating, ax + ay + c. ( which is the compete integra. Differentiating partiay w.r.to c and then, we get. x Hence there is no singuar integra. Put c φ(a in (, ax + ay + φ(a ( Differentiating ( with respect to a, x + y + φ (a ( Eiminating a between ( and (, we get the genera integra.

19 Soved University Question Papers-5MA, SRMIST 7 Part - C (5 6 Marks Answer ALL Questions 8. a. i. Find the genera soution of x( y p + y(x q (y x. Ans. See Q.No: 8. a. ii., Page No. 5. ii. Sove (D 4D cos x cos y. So: The auxiiary equation is m 4 where D m, D. m ± The Compementary function (C.F. is φ (y + x + φ (y x. Particuar integra D cos x cos y 4D (D [cos(x + y + cos(x y] 4D (D 4D cos(x + y + (D cos(x y 4D P.I + P.I P.I (D cos(x + y 4D cos(x + y ( repace by D 4, DD 9 cos(x + y and 64 P.I (D cos(x y 4D cos(x y ( repace by D 4, DD 9 cos(x y 64 Therefore P.I + P.I 64 [cos(x + y + cos(x y] cos x cos y. Hence C.F.+P.I. φ (y + x + φ (y x + cos x cos y. (OR b. Sove (D 7DD 6D x y + sin(x + y. So: The auxiiary equation is m 7m 6 where D m, D (m + (m m 6 (m + (m + (m m,,. The Compementary function (C.F. is φ (y x + φ (y x + φ (y + x

20 8 V. Suvitha, Department of Mathematics, SRMIST Particuar integra D [x y + sin(x + y] 7DD 6D D x y + 7DD 6D D sin(x + y 7DD 6D P.I + P.I [ P.I D x y 7DD 6D D ( ( 7D 7D ( 7D ] D + 6D D x y D + D + 6D D + D + 6D D x y D x y y. [ ] x D y. [ ] x 4 x5 y D 6 and P.I D sin(x + y 7DD 6D sin(x + y repace by D, D 4 D + 8D + 4D sin(x + y + y.sin(x 7D + 4D 9D + 8D. 9D 8D 8D 64D sin(x + y.(9d 8D sin(x + y [8( 64( 4] [9 cos(x + y 8 cos(x + y.] 55 7 cos(x + y cos(x + y C.F. + P.I + P.I φ (y x + φ (y x + φ (y + x + x5 y 6 cos(x + y a. Find the Fourier series expansion of f(x x + x in (,. Hence find the sum of the series So: Given the function f(x is neither even nor odd. Let f(x a + a n cos nx + b n sin nx ( n where a f(xdx, a n Here. To find a, a n, b n : a + (x + x dx n f(x cos nx dx and b n xdx + x dx x dx, since x is odd and x is even f(x sin nx dx.

21 Soved University Question Papers-5MA, SRMIST 9 a 8 a n + x (x + x cos nx dx x cos nx nx dx, since x cos sin nx n x cos nx n x cos nx dx + is odd + 4 cos n. n, since sin sin n 6( n n b n (x + x sin nx dx x sin nx dx +, since x sin nx nx cos x n sin nx n x sin nx dx + is odd sin nx n x cos nx dx cos n n x sin nx dx 4( n n Substituting the vaues of a, a n, b n in (, we get x + x ( n n n cos nx 4 ( n sin nx n n Deduction: x is an end point in the range. Hence the vaue of the Fourier series at x is equa to [f( + f( ] [( + + ( + ( ] 4. Hence ( n n cos n 4 6 n 8. Therefore n 6. n n n (OR b. Find the Fourier series upto second harmonic from the foowing data: 4 5 x : y f(x : Ans. See Q.No. 9. b., Page No. 8.

22 V. Suvitha, Department of Mathematics, SRMIST. a. A string is stretched and fastened to two points apart. Motion is started by dispacing the string in the form y a sin x from which it is reeased at time t. Find the dispacement y(x, t. So: The dispacement of the string y(x, t is governed by u t a u x. The boundary conditions are (i y(, t, t (ii y(, t, t The initia ( conditions are y (iii, x (iv y(x, a sin x t t, x. The proper soution is y(x, t (A cos λx + B sin λx(c cos λat + D sin λat ( Using boundary condition (i in (, A(C cos λat + D sin λat A A in (, y(x, t B sin λx(c cos λat + D sin λat ( Appying the boundary condition (ii in (, B sin λ(c cos λat + D sin λat B and sin n λ n λ n λ n in (, y(x, t B sin nx ( C cos nat + D sin nat ( Again using the [ initia condition (iii, we have y nx B sin C sin nat. na + D cos nat. na ] ( t y B sin nx [ + D. na ] B, D t t D in equation (, y(x, t B sin nx C cos nat The most genera soution is y(x, t n Using initia condition (iv, y(x, n B n sin nx B n sin nx cos nat a sin x. B sin x + B sin x + B sin x + a sin x Comparing both sides, B a, B n, n. Substituting the vaue of B n in (4, we get y(x, t a sin x cos at (OR b. A rod of ength cm has its end A and B kept at C and 9 C respectivey unti steady state conditions prevai. If the temperature at each end is then suddeny reduced to C and maintained so, find the temperature u(x, t at a distance x from A at time t. So: The P.D.E. of one dimensiona heat fow is u t α u x ( In steady state, the P.D.E. becomes d u dx ( Therefore the soution is u(x ax + b ( The initia conditions are u( and u( 9 Using these conditions in (, we obtain u( + b b and u( a + 9 a + a. Thereforre u(x x +. When the temperatures at each ends are reduced to ero. (4

23 Soved University Question Papers-5MA, SRMIST In transient state, the boundary conditions are (i u(, t for a t (ii u(, t for a t The initia condition is (iii u(x, x + for < x < In transient state, the proper soution is u(x, t (A cos λx + B sin λxe α λ t (4 Using (i in (4, we get u(, t Ae α λ t A. A in (4, u(x, t B sin λxe α λ t (5 Using (ii in (5, we get u(, t B sin xe α λ t Since B, sin λ λ n λ n λ n nx in (5, we get u(x, t B sin e α n t/4 The most genera soution is u(x, t Using (iii in (6, we get u(x, n This is a haf range sine series. Hence B n (x + sin nx dx (x + n nx cos n [ 9.( n. +. n n 6 n [ ( n ] B n sin nx e α n t/4 (6 B n sin nx Substituting the vaue of B n in (6, we get u(x, t 6 n [ ( n ] sin nx e α n t/4 n x +. sin nx n ], since sin sin n. a. Find the Fourier transform of f(x given by f(x evauate So: sin x x dx. F {f(x} f(xe isx dx ] a a a e isx dx [ e isx. is a is [eias e ias ] sin as by sin x eix e ix s i Using inversion formua, we get f(x F {f(x} e isx ds f(x ( { if x < a if x > a > sin as e isx ds s and hence

24 V. Suvitha, Department of Mathematics, SRMIST ( sin as s { if x < a (cos sx i sin sxds f(x if x > a > ( sin as Equating rea parts, cos sxds s ( sin s Putting x and a ds s ( sin s Hence ds ( sin x s dx x. (OR b. Find the Fourier sine and cosine transform of f(x e ax, a > and hence deduce their inversion formua. F s (e ax e ax sin sxdx [ ] e ax a ( a sin sx s cos sx + s. s a + s ( F c (e ax. a a [See Q. No. 5, Page No. 5] ( + s By inversion formua of (, we have f(x F s (s sin sxds e ax s sin sx a + s ds e ax, a >. x sin αx Changing the variabes, a + x dx e αa, a >. Again, by inversion formua of (, we have ( f(x F c (s cos sxds e ax cos sx a + s ds a e ax, a >. cos αx Changing the variabes, a + x dx a e αa, a >. ( s a + s sin sxds a a + s cos sxds. a. i. Find (A Z(n (B Z(na n. Ans. (A See Q.No. 7, Page No. 5.; (B Z(na [ n See Q.No. 7, Page No. 5. ii. Using partia fraction method, evauate Z ]. ( ( So: Let X( ( ( X( ( ( Now ( ( A + B A( + B(

25 Soved University Question Papers-5MA, SRMIST put A and B Therefore X( + X( [ + ] Taking inverse transform on both sides, we get Z [X(] [ Z + ] x(n [ + n+ ]. (OR b. Sove the difference equation y n+ 5y n+ + 6y n with y and y using transform method. So: Taking Z -transform on both sides of the equation, we get Z(y n+ 5Z(y n+ + 6Z(y n Z( [ Y ( y y ] 5 [(Y ( y ] + 6Y ( ( 5 + 6Y ( + Y ( ( ( Y ( ( ( (. Y ( has poe at,,. Therefore y(n R where R is the sum of the residue of Y ( n n+ R {Residue} im( ( ( ( n+ R {Residue} im( ( ( ( n+ n+ R {Residue} im( ( ( ( n+ Hence y(n n+ + n+.

26 4 V. Suvitha, Department of Mathematics, SRMIST B.Tech. Degree Examination, November 7 Third/Fourth/Fifth Semester 5MA-Transforms and Boundary Vaue Probems Time: Three hours Max. Marks: Part - A ( Marks Answer ALL Questions. The partia differentia equation formed by eiminating the arbitrary function from the reation f(x + y is (A px qy (B py qx (C p qy (D px q [See Q.No., Page No..]. The compete soution of px + qy + p q (A ax+by +a b (B bx+ay +ab (C ax+by + ab (D ax+by +ab So: Given px + qy + p q. This is Cairaut s form. Hence the compete integra is ax + by + a b. Ans. A. The compementary function of (D DD + D (A φ(y + x φ (y x (B φ(y + x φ (y + x (C φ (y x + φ (y + x (D φ (y + x φ (y + x So: The auxiiary equation is m m + where D m, D. (m (m m,. C.F. is φ (y + x + φ (y + x. 4. The particuar integra of (D DD e x (A e x So: (B e x 4 (C ex 4 (D ex 4 P.I. D e x ex DD 4 4 where D, D x. e x xex D D 4 4 x e x. 5. The constant a of the Fourier series for the function f(x x in x (A (B (C (D [See Q.No. 5, Page No..] 6. The RMS vaue of f(x x in x is (A (B (C (D - b (f(x dx a So: RMS vaue of f(x b a Ans. C x dx.

27 Soved University Question Papers-5MA, SRMIST 5 7. If f(x x in (, then the vaue of b n is (A (B (C - (D So: If f(x is an even function of x in (, then b n. 8. Haf range cosine series for f(x in (, is (A a + a n cos nx (B a n 4 + (a n + b n (C n a Ans: + a n cos nx n 9. The one dimensiona wave equation is (A u t α u x (B u t a u x b n sin nx n (C y t α y x (D a n cos nx n (D u x y a t. How many initia and boundary condition are required to sove u t a u x (A Two (B Three (C Five (D Four Ans. D. One dimensiona heat equation is used to find (A Density (B Temperature Distribution (C Time (D Dispacement. The proper soution in steady state heat fow probem is (A u (Ae λx + Be λx e α λ t (B u Ax + B (C u (A cos λx + B sin λxe α λ t (D u (Ae λx + Be λx (Ce λat + De λat. The Fourier transform of f(x is (A f(xe ist dt (B (C f(xdx (D f(xe isx dx f(xe isx dx 4. If F [f(x] F (s then ( F [e iax f(x] is s (A F (as (B F (C F (s a (D F (s + a a Ans. D [See Q.No. 5, Page No..] 5. F (f(x g(x is (A F (s + G(s (B F (s G(s (C F (sg(s (D F (s/g(s Ans. C 6. If F [f(x] F (s then f(x dx (A Ans. A F (s ds (B F (x dx (C F (x dx (D F (s ds

28 6 V. Suvitha, Department of Mathematics, SRMIST 7. What is Z[( n ] (A (B (C + + So: We know that Z(a n Ans. A (D a Z[( n ] 8. Z[na n ] a a (A ( a (B ( a (C ( a (D Ans. C [See Q.No. 7, Page No. 5.] [ 9. Z cos n ] (A (B + (C (D 4 Ans. D [See Q.No. 8, Page No. 5.] +. + n. Poes of φ( ( ( are (A, (B, (C, (D So: The poes are ( (,.. Sove p tan x + q tan y tan. dx So: The auxiiary equations are tan x Taking the first two ratios, we get og sin x og sin y + og a a sin x sin y. ( a Part - B (5 4 Marks Answer ANY FIVE Questions dx tan y dx tan Simiary taking the ast two ratios, we get b sin y ( sin. sin x Therefore the genera soution is φ sin y, sin y. sin. Find the compete integra of p + q x + y. So: Given equation is separabe type. Therefore p x y q K (say p x + k and q y k. We know that d pdx + qdy. d x + kdx + y kdy Integrating, we have (x + k/ + (y k/ + c.. Find the haf range Fourier sine series f(x x in < x <. So: Let f(x b n sin nx where b n f(x sin nxdx. n Now b n x sin nxdx [ ( ( ] cos nx sin nx x n n Therefore x n ( n sin nx. n [ ] n ( n.

29 Soved University Question Papers-5MA, SRMIST 7 4. State any two assumptions in deriving one dimensiona wave equation and write its a possibe soutions. See Q.No. 4, Page No If F [f(x] F (s then prove that F [f(ax] a F Proof: By definition F {f(ax} Put ax t dx dt a. case i: a > F {f(ax} ( s e i a ( s e i a case ii: a < F {f(ax} From ( and (, we get F {f(ax} a F ( s a e isx f(axdx. t f(t dt t f(t dt ( s. a a ( s a F a a a F where a. ( cos θ 6. Prove that Z(cos nθ if >. cos θ + Ans. See Q.No. 6, Page No. 5. and equating rea parts, we get ( cos θ Z(cos nθ if >. cos θ + [ ] 7. Find Z. ( ( So: Let X( ( (. X( n n has simpe poes at and. ( ( ( n R {Residue} im and ( ( ( n R {Residue} im ( ( n. Hence x(n R where R is the sum of the residues of X( n x(n n. Part - C (5 6 Marks Answer ALL Questions. ( ( s a. ( 8. a.i. Find the singuar soution of px + qy + p q. So: Given px + qy + p q. This is Cairaut s form. The compete soution is ax + by + a b. ( Partiay differentiating w.r.to a and b, we get x + ab and y + ba. Therefore x ab and y ba ( x a y b ab k (say. Hence a ky and b kx. Substituting the vaues of a and b in ( and (, we get

30 8 V. Suvitha, Department of Mathematics, SRMIST k xy and kxy + kxy + k4 x y ( kxy + kx y xy kxy. Hence 7 8 k x y 6 + 7x y. ii. Sove x(y p + y( xq (x y. dx So: The auxiiary equations are x(y Each is equa to dx + dy + d x(y d(x + y +. Integrating, we have x + y + a. dx + dy + d dy y( x Aso, taking the Lagrangian mutipiers x, y,, we have dx x y dy y x d x y Each ratio is equa to dx x + dy y + d (y dx x + dy y + d Hence dx x + dy y + d. Integrating og x + og y + og og b xy b. Hence, the genera soution is φ(x + y +, xy. (OR d (x y. b. Sove (i (D + DD 6D x y + e x+y (ii (p + q x + y. So: i. The auxiiary equation is m + m 6 where D m, D. (m (m + m,. The compementary function (C.F. is φ (y x + φ (y + x. Particuar integra D (x y + e x+y + DD 6D D x y + + DD 6D D e x+y + DD 6D P.I + P.I [ ( ] P.I D x y DD + 6D + DD 6D D + D x y ( ( DD + 6D DD + 6D D + + x y D [ D D x4 y x5 6 D D ] x y D [x y D x ]

31 Soved University Question Papers-5MA, SRMIST 9 P.I D + DD 6D e x+y ex+y 6 where D, D C.F. + P.I + P.I φ (y x + φ (y + x + x4 y ii. Given (p + q x + y. (p + (q x + y. This is of the form F ( k p, k q. Put Z k+ then P Z x5 6 + ex+y. 6 x Z. p where p x x. Simiary Q q. Therefore the equation reduces to P + Q 4(x + y. P 4x 4y Q 4a (say Hence P 4a + 4x and Q 4y 4a We know that dz P dx + Qdy dz a + x dx + [ y ady x Integrating, Z a + x + a x sinh + y y a a a y ] cosh + b a x a + x + a sinh x a + y y a a cosh y a + b. 9. a. Find the Fourier series to represent (x x in the interva (,. Deduce the vaue of So: Given the function f(x is neither even nor odd. Let f(x a + a n cos nx + b n sin nx ( n n where a f(xdx, a n f(x cos nxdx and b n f(x sin nxdx. To find a, a n, b n : a a n (x x dx xdx x dx x dx, since x is odd and x is even (x x cos nxdx x cos nxdx x cos nxdx x cos nxdx, since x cos nx is odd [ ( ( ( ] sin nx cos nx sin nx x x + n n n

32 V. Suvitha, Department of Mathematics, SRMIST [ cos n ], since sin sin n n 4( n b n n (x x sin nxdx x sin nxdx x sin nxdx x sin nxdx, since x sin nx is odd [ ( ( ] cos nx sin nx x [ ( ] cos n n n n ( n n Substituting the vaues of a, a n, b n in (, we get x x [ 4 ( n n n cos nx + ] n sin nx Deduction: x which is a point of continuity. Hence the sum of the Fourier series equas the vaue [ of the function at x. Therefore f( ] +. (OR b. Find the Fourier series upto second harmonic from the foowing data: 4 5 x y So: Let f(x a + a n cos nx + b n sin nx where a f(x, n n m a n f(x cos nx and bn f(x sin nx. m m x f(x cos x sin x cos x sin x Now a [ ] a f(x cos x [ ].67 6

33 Soved University Question Papers-5MA, SRMIST a f(x cos x [ ].48 6 b f(x sin x [ ] b f(x sin x [ ].44 6 Hence f(x.5.67 cos x +.48 cos x sin x.44 sin x.. a. A tighty string of ength has its end fastened at x, x. At t, the string is in the form f(x kx( x and then reeased. Find the dispacement at any point on the string at a distance x from one end and at any time t >. Ans. See Q.No.. a., Page No. 8. (OR b. A rod of ength cm has its end A and B kept at C and 9 C respectivey unti steady state conditions prevai. If the temperature at each end is then suddeny reduced to C and maintained so, find the temperature u(x, t at a distance x from A at time t. Ans. See Q.No.. b., Page No.. { a x if x < a. a. Find the Fourier transform of f(x given by f(x if x > a > hence prove that So: F {f(x} sin x x cos x x dx 4. a a a a a f(xe isx dx a a (a x (cos sx + i sin sxdx (a x cos sxdx + i (a x e isx dx a a (a x sin sxdx (a x cos sxdx + i. since (a x sin sx is odd ( ( sin sx cos sx [(a x ( x s s ( cos as ( ] sin as [ a s + s 4 s [sin as as cos as] Using inversion formua f(x F {f(x} e isx ds + ( ( ] sin sx a s

34 V. Suvitha, Department of Mathematics, SRMIST f(x 4 s [sin as as cos as] e isx ds s [sin as as cos as] (cos sx i sin sxds a x ( sin as as cos as Equating rea parts, s cos sxds (a x ( sin s s cos s Putting x and a s ds ( sin s s cos s Hence ds ( sin x x cos x 4 dx 4. s (OR b. i. Find the Fourier cosine and sine transform of f(x e ax, a >. Ans. See Q.No.. b., Page No.. dx ii. Evauate (a + x (b + x using transform method. Ans. See Q.No.. b. i., Page No... a. i. Find Z [F (] if F ( Ans. See Q.No..a.i, Page No.. ii. Find the inverse Z -transform of f( using residues + So: Given f( ( + (. f( n n ( + ( + ( x has simpe poes at and. + ( + ( f(. Therefore x(n R where R is the sum of the residue of f( n R {Residue} im ( + n ( + ( + ( R {Residue} im( n ( + ( + ( n.5 Therefore x(n R + n.5 (5.n. (OR b. Sove the equation y n+ 7y n+ +y n n given y y using transform. So: Taking Z -transform on both sides of the equation, we get Z(y n+ 7Z(y n+ + Z(y n Z( n [ Y ( y y ] 7 [(Y ( y ] + Y ( ( 7 + Y ( Y ( Y ( ( ( ( 4. Y ( has poe at,, 4. ( ( 7 + Therefore y(n R where R is the sum of the residue of Y ( n R {Residue} im( ( ( ( 4 n n n

35 Soved University Question Papers-5MA, SRMIST R {Residue} im( ( ( ( 4 n R {Residue} 4 im( 4 4 ( ( ( 4 4n n Hence y(n n n + n. n n

36 4 V. Suvitha, Department of Mathematics, SRMIST B.Tech. Degree Examination, November 8 Third/Fourth/Fifth Semester 5MA-Transforms and Boundary Vaue Probems Time: Three hours Max. Marks: Part - A ( Marks Answer ALL Questions. The partia differentia equation formed by eiminating the arbitrary function from the reation f(x + y is (A px qy (B py qx (C p qy (D px q [So: See Q.No., Page No. ]. The compete integra of q py is (A ax + ay + b (B ax ay + b (C ax + by (D xy So: Given q py. This is of the form F (y, p, q. Let p a then q ay. We know that d pdx + qdy d adx + aydy Integrating, ax + ay + b. Ans. A. The soution which has equa number in arbitrary constants and independent variabes is known as (A Genera integra (B Compete integra (C Particuar integra (D Singuar integra 4. The particuar integra of D x y is (A x5 y (B x y (C x 4 y (D x y So: P.I. D x y y. x 4 D 4 x5 y. Ans. A 5. The period of the periodic function tan x is (A (B (C (D Ans. A 6. If f(x is discontinuous at x a, then the sum of the Fourier series x a is (A f(a f(a + (B f(a f(a + (C f(a f(a + (D f(a + f(a + Ans. D 7. In the haf range cosine series of f(x cos x in (, the vaue of a is (A 4 (B (C 4 (D Ans. D [So: See Q.No. 8, Page No. ]

37 Soved University Question Papers-5MA, SRMIST 5 8. If y is the RMS vaue of f(x in (,, then the vaue of (A y Ans. D (B y (C y (D y a 4 + (a n + b n is 9. How many initia and boundary condition are required to sove u t a u x (A Two (B Three (C Four (D Five Ans. C. The most genera soution for the dispacement y(x, t of the string of the ength vibrating between fixed end points with non-ero initia veocity is ( nx ( nat ( nx ( nat (A B n sin cos (B B n sin sin n n ( nx ( nat ( nx ( nat (C B n cos sin (D B n cos cos n n So: The dispacement y(x, t is governed by u t a u x. The boundary conditions are (i y(, t, t (ii y(, t, t The initia conditions are ( y (iii y(x,, x (iv, x. t t The proper soution is y(x, t (A cos λx + B sin λx(c cos λat + D sin λat ( Using boundary condition (i in (, A(C cos λat + D sin λat A A in (, y(x, t B sin λx(c cos λat + D sin λat ( Appying the boundary condition (ii in (, B sin λ(c cos λat + D sin λat B and sin n λ n λ n λ n in (, y(x, t B sin nx ( C cos nat + D sin nat ( Again using the initia condition (iii, B sin nx.c B, C. C in equation (, y(x, t B sin nx D sin nat The most genera soution is y(x, t B n sin nx sin nat. n. The one dimensiona heat equation u t α u x, α (A k (B T δ m (C k (D k ρc c Ans. C stands for. The steady state temperature of a rod of ength whose ends are kept respectivey at C and 4 C is (A u(x x (C u(x x + (B u(x x + (D u(x x n

38 6 V. Suvitha, Department of Mathematics, SRMIST So: In steady state, the P.D.E. becomes d u dx Therefore the soution is u(x ax + b ( The initia conditions are u( and u( 4. Using these conditions in (, we obtain u( + b b and u( a + 4 a + a. Therefore u(x x +. Ans. A or C. If F [f(x] F (s then F [f(ax] ( s (A af (s (B F (as (C a F a Ans. C [So: See Q.No. 5, Page No. 7] (D af ( s a 4. The Fourier cosine transform of f(x e ax, (where a > ( ( s a (A s + a (B s + a ( ( s a (C s a (D s a [So: See Q.No. 5, Page No. 5] 5. The vaue of F c [x.f(x] is d (A ds [F s(s] (B d ds [F s(s] (C i d ds [F s(s] (D ( i d ds [F s(s] So: By definition F s (s f(x sin sxdx Differentiating w.r.to s, we get d ds [F s(s] xf(x cos sxdx d ds [F s(s] F c {x.f(x}. Ans. A 6. If F [f(x] F (s then f(x dx (A F (s ds (B 7. The vaue of Z [ ] is n! ( (B e F (s ds (C F (s ds (D (A e( (C e (D e [ ] So: Z [ ] n! n n! n Z + n! +! + Ans. A [ ] 8. The inverse Z -transform of ( is (A n (B n (C n + (D n [So: See Q.No. 9, Page No. 5 and repace by a ] F (s ds! + e

39 Soved University Question Papers-5MA, SRMIST 7 9. If Z(f(n F ( and Z(g(n G(, then the Z -transform of f(n g(n is (A F ( + G( (B F ( G( (C F (.G( (D F ( G( Ans. C. Poes of φ( n ( + ( are (A - (order (B (order (C (order (D - (order So: The poes are given by. Ans. C Part - B (5 4 Marks Answer ANY FIVE Questions. Compute the compete integra of p + q x + y. Ans. See Q.No., Page No. 6.. Sove (D 4DD + 4D e x+y. So: The auxiiary equation is m 4m + 4 (m m,. The Compementary function (C.F. is φ (y + x + xφ (y + x Particuar integra (P.I. D e x+y 4DD + 4D P.I. D e x+y 4DD + 4D ex+y where D, D ex+y x. e x+y D 4D x. 4 4 ex+y where D, D x. ex+y x. ex+y Hence the soution is C.F. + P.I. φ (y + x + xφ (y + x + x ex+y.. Find the haf range Fourier sine series f(x x( x in < x <. So: Let f(x b n sin nx where b n f(x sin nxdx n

40 8 V. Suvitha, Department of Mathematics, SRMIST Now b n [ (x x sin nxdx [ (x x ( cos nx n cos n n n cos ( x 4 n [( n ] 8 n if n is odd Therefore x( x 8 n sin nx. nodd ] ( sin nx n since sin sin n ( cos nx ] + ( n 4. List out a the possibe and correct soutions for the one dimensiona wave equation with non-ero veocity probem. Ans. Possibe soution: See Q. No. 4, Page No. 4 and the correct soution is y(x, t (A cos λx + B sin λx(c cos λat + D sin λat. 5. A rod of ength has its ends A and B kept at C and C respectivey unti steady state conditions prevai. Find the steady state temperature of the rod. So: In steady state, the P.D.E. becomes d u dx Therefore the soution is u(x ax + b ( The initia conditions are u( and u(. Using these conditions in (, we obtain u( + b b and u( a a a. Therefore u(x x. 6. Determine the Fourier transform of e x. F { e x } f(xe isx dx e x (cos sx + i sin sxdx e x cos sx + i e x e isx dx e x sin sxdx e x cos sxdx +, since e x sin sx is odd [ e x s ( cos sx + s sin sx + ( s + ]

41 Soved University Question Papers-5MA, SRMIST 9 7. Using convoution theorem, cacuate the inverse Z -transform of So: [ Z ] [ ] ( a Z a. a ( ( Z.Z a a n a n a n a m.a n m m n a n a n (n + m ( a. Part - C (5 6 Marks Answer ALL Questions 8. a. i. Find the singuar integra of px + qy + + p + q. So: The given equation is Cairaut s form. Therefore the compete integra is ax + by + + a + b. ( Differentiating ( w. r. to a and b, we get x + ( + a + b / a.a x ( + a + b and y + ( + a + b /.b y Now x y a + b + a + b x y + a + b x y ( a x + a + b x x y ( b y + a + b y x y b + a + b ( + a + b Substituting in (, we get x x y y x y + x y x y x y x y x y. Hence x + y +. ii. Sove (D D D sin(x + y + x y. Ans. See Q.No. 8. b., Page No. 6. (OR b. i. Form the PDE by eiminating φ from φ(x + y +, x + my + n. So: Rewriting the given equation as x + y + f(x + my + n. Differentiating partiay w.r.to x and y, we get x + ( x f (x + my + n y + y f (x + my + n + n x ( + m y and (. (

42 4 V. Suvitha, Department of Mathematics, SRMIST Divide ( and (, we get x + p y + q + np where p, m + nq x y q. (x + p(m + nq (y + q( + np xm + xnq + pm + pnq y + ynp + q + qnp. (m ynp + (xn q y xm. ii. Find the genera soution of (x y x p y q. So: The auxiiary equation are dx x dy y dx Taking the first two ratios, x dy y. Integrating, x y + a x + y a. dx + dy Now x y d d(x + y d (x y x + y. Integrating, og(x + y og + og b x + y The genera soution is φ ( x + y, x + y. d (x y. 9. a. Obtain the Fourier series expansion of f(x x in < x < and hence find the sum of the series So: Given the function is even function. Therefore b n. The Fourier series is f(x a + a n cos nx ( n where a f(xdx and a n f(x cos nxdx. To find a, a n : a x dx ( x a n x cos nxdx [ ( sin nx x n [ cos n n x b. ( ( ] cos nx sin nx + n n ], since sin sin n 4( n n Substituting the vaues of a, a n in (, we get x + ( n 4 cos nx. n n By Parseva s identity f(x dx a 4 + x 4 dx n 4 n (a n + b n. n

43 Soved University Question Papers-5MA, SRMIST 4. x 4 dx ( x n n n 4 n 4 n n (OR b. Compute the first three harmonics of the haf range cosine series of y f(x from So: Let f(x a + Put θ x x 4 5 y f(x a n cos nx where a f(x, an nx f(x cos, n m m and 6. Therefore the vaues are θ are, 6, x 6, x 6, 4x 6, 5x 6. x f(x cos θ cos θ cos θ Now a [ ] 4 6 a f(x cos θ [ ] a f(x cos θ [ ].8 6 a f(x cos θ [ ] Hence f(x cos θ.8 cos θ.6667 cos θ.. a. A tighty stretched string with( fixed end point x and x is initiay in a x position given by y(x, y sin. If it is reeased from rest in this position, find the dispacement y(x, t of the string at any point. So: The dispacement of the string y(x, t is governed by u t a u x. The boundary conditions are (i y(, t, t (ii y(, t, t The initia ( conditions are y ( x (iii, x (iv y(x, y sin, x. t t The proper soution is y(x, t (A cos λx + B sin λx(c cos λat + D sin λat ( Using boundary condition (i in (, A(C cos λat + D sin λat A

44 4 V. Suvitha, Department of Mathematics, SRMIST A in (, y(x, t B sin λx(c cos λat + D sin λat ( Appying the boundary condition (ii in (, B sin λ(c cos λat + D sin λat B and sin n λ n λ n λ n in (, y(x, t B sin nx ( C cos nat Again using the [ initia condition (iii, y nx B sin C sin nat. na + D cos nat. na ] t ( y B sin nx [ + D. na ] B, D t t D in equation (, y(x, t B sin nx C cos nat The most genera soution is y(x, t n Using initia condition (iv, y(x, ( x Since sin 4 Therefore B sin x [ n sin x sin x + B sin x B n sin nx B n sin nx ]. + B sin x + D sin nat + y 4 cos nat y sin ( x. [ sin x sin x ]. Equating ike coefficients, we get B y 4, B, B y 4, B 4 B 5. Substituting these vaue in (4, we get y(x, t y 4 (OR sin x cos at y x sin 4 ( (4 cos at. b. A rod of ength cm has its end A and B kept at C and 8 C respectivey unti steady state conditions prevai. The temperature at each end is suddeny reduced to C and kept so. Estimate the resuting temperature function u(x, t of the rod. Ans. See Q.No.. b., Page No. 9. { x, in x <. Find the Fourier transform of f(x and hence deduce the vaue, in x > ( sin t 4 of dt. t So: F {f(x} f(xe isx dx ( x e isx dx ( x (cos sx + i sin sxdx ( x cos sxdx + i., since ( x sin sx is odd

45 Soved University Question Papers-5MA, SRMIST 4 [ ( sin sx F {f(x} ( x s [ ( cos s ( ] s s ( cos s s By Parseva s identity F (s ds f(x dx.4 sin4 (s/ s 4 ds 8. sin 4 (s/ s 4 ds ( x dx ( x dx (. sin (s/ s Put t s t s. Therefore dt ds and t to t 8 sin 4 [ ] t ( x (t 4.dt sin 4 t t 4 dt. b. Using Parseva s identity, evauate (OR dx (x + a and ( ] cos sx s x dx (x + a. So: Consider f(x e ax, a >. The Fourier cosine transform F c (e ax a a + s [See Q. No. 5, Page No. 5] and The Fourier sine transform F s (e ax s a [See Q. No..b., Page No. ]. + s By Parseva s identity for cosine transform F c (s ds f(x dx a (s + a ds e ax dx a [ ] e ax (s + a ds a ds Therefore (s + a 4a a. Changing the variabe s into x, we get By Parseva s identity for sine transform s (s + a ds e ax dx s [ ] e ax (s + a ds a s ds Therefore (s + a 4a a. Changing the variabe s into x, we get dx (x + a 4a. F s (s ds f(x dx x dx (x + a 4a.

46 44 V. Suvitha, Department of Mathematics, SRMIST. a. i. Using ong division method, find the inverse Z -transform of So: Let F ( By ong division + Hence n ( ( f(n n ( ( Therefore f(, f(, f(, f( 7, f(4 5, etc. ii. Sove y(n + 4y(n + + 4y(n, given y( and y(. So: Taking Z -transform on both sides of the equation, we get Z[y(n + ] 4Z[y(n + ] + 4Z[y(n] [ Y ( y( y( ] 4 [(Y ( y(] + 4Y ( ( 4 + 4Y ( + 4 Y ( 4 ( Y ( ( 4 (. Y ( has poe of order at. Therefore y(n R where R is the sum of the residue of Y ( n d R im d ( n ( 4 ( im [ n. + n n ( 4 ] n ( n. Hence y(n n ( n. (OR b. i. Using residues method, find the inverse Z -transform of ( + ( 5 So: Let f( ( + ( 5 f(n n ( ( + ( 5 The poes are given by ( + ( 5, 5 which are simpe poes. R {Residue} im ( + n ( ( + ( ( n n ( R {Residue} 5 im( 5 5 ( + ( 5 7 (5n Therefore x(n R 5 7 ( n + 7 (5n. ii. Sove y(n + + 6y(n + + 9y(n n, given y( y(. Ans. See Q.No.. b., Page No..