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DEPARTMENT OF MATHEMATICS

181202- MATHEMATICS – II

PART A (Q & A), PART B QUESTION BANK

| | |

| |UNIT-I ORDINARY DIFFERENTIAL EQUATIONS |

| |Solve[pic] |

|1. | |

| | [pic] |

| |A.E is m2 – 6m + 13= 0 |

| |[pic] |

| |Hence the solution is y = e3x ( A Cos2x + B Sin2x) |

| |Solve (D2 – 2D + 1) y = ex |

|2. | |

| | A.E is m2 – 2m + 1 =0 |

| |(m – 1)2 = 0 => m = 1,1 |

| |C.F => (A + Bx) ex |

| |P.I = [pic] |

| |Solution is y = C.F + P.I = (A + Bx) ex + [pic] |

| |Find the particular integral of ( D2 + 5D + 6) y = 11 e5x |

|3. | |

| | P.I = [pic] |

| |Put D = 5 hence P.I = [pic] |

|4 |Solve (D – 1) 2 y = sinh2x. |

| | A.E is (m – 1)2 = 0 |

| |m = 1,1 |

| |C.F = (A + Bx) ex |

| |P.I = [pic] |

| |= >[pic] |

| |Solution is y = C.F + P.I => (A + Bx) ex + [pic] |

| |Solve [pic] |

| | |

|5. | |

| | (x2 D2 + 4xD + 2) y = 0 |

| |Put x = ez ,z = logx, xD = ( ,x2 D2 = ( (( - 1) |

| |(( (( - 1+ 4 ( + 2) z = 0 => ((2 + 3 ( + 2) z = 0 |

| |A.E is m2 + 3m + 2 = 0 => m = -2, -1. |

| |Z = A e-2z + B e-z |

| |=> y = A e-2(logx) + B e-(logx) => y = [pic] |

|6. |Find the particular integral of ( D2 + 2) y = x2 |

| | P.I = [pic] |

| |= [pic] |

| |P.I = [pic] |

| |Find the particular integral of (D2 – 2 D + 1)y = coshx |

|7. | |

| | [pic] |

| |P.I = [pic] |

| |Solve [pic] |

| | |

|8. | |

| | Dy – x = 0……….(1) |

| |y – Dx = 0………..(2) |

| |Eliminate x from these two eqns => (D2 – 1)x = 0 |

| |A.E is m2 – 1 = 0 => m = 1, -1. |

| |=> x = Aet + B e-t |

| |=> [pic]= y |

| |Hence x = Aet + B e-t , y = Aet - B e-t |

| |Solve [pic] |

| | |

|9. | |

| | (D2 - 4D + 3) y = 0 |

| |A.E is m2 – 4m + 3 = 0 => m= 3, m = 1. |

| |C.F = Ae3x + Bex |

| |Sin3x cos2x = [pic] |

| |P.I = [pic] |

| |Put D2 = -52 |

| |P.I = [pic] |

| |y = Ae3x + Bex + [pic] |

| |Transform the equation ((2x + 3)2 D2 – (2x + 3) D – 12) y = 6x into linear differential equation with constant coefficients. |

| | |

|10. | |

| | Put 2x + 3 = ez , z = log ( 2x + 3) |

| |( 2x + 3)2 D2 = 4 ((2 - ( ) |

| |( 2x + 3)D = 2( |

| |Hence the D.E is (4(2 - 6( -12) y = z ez - 9 |

| |Find the Particular integral of (D – 3)2 y = e3x cosx |

|11. | |

| | P.I = [pic] = [pic]= -e3x cosx |

| |Find the Particular integral of (D2 + 1)y = sin2 x |

|12. | |

| | [pic] |

| |P.I = [pic] |

| |P.I = [pic] |

| |Find the Particular integral of (D2 + 4D + 4)y = e-2x x |

|13. | |

| | P.I = [pic]=[pic] |

| |Transform the equation [pic] into linear differential equation with constant coefficients. |

| | |

|14. | |

| | Put x = ez ,z = logx, xD = ( ,x2 D2 = ( ((-1) |

| |the given eqn => ((((-1) +6( + 2)y = ez z |

| |((2 + 5( + 2) z = z ez |

|15. |Write Cauchy’s homogeneous linear equation. |

| | [pic] |

| |where ai’s are constants and X is a function of x. |

|16. |Write Legendre’s linear equation. |

| | [pic] |

| |A1 , A2,…….are constants. |

|17. |Find the Particular integral of (D + 2) (D – 1)2 = e-2x |

| | P.I = [pic]=[pic] |

| |Transform the equation [pic] into linear differential equation with constant coefficients. |

| | |

|18. | |

| | Put x = ez ,z = logx, xD = ( ,x2 D2 = ( ((-1) |

| |((2 - 2( + 1) z = (z e-z)2 |

|19. |Find the Particular integral of (D2 + 4) y = sin2x |

| | P.I = [pic] |

| |Put D2 = -22 , P.I = [pic] |

| |Transform the equation ((3x + 5)2 D2 – 6(3x + 5) D + 8) y = 0 into linear differential equation with constant coefficients. |

|20. | |

| | Put 3x + 5 = ez , z = log ( 3x + 5) |

| |( 3x + 5)2 D2 = 9 ((2 - ( ) |

| |( 3x + 5)D = 3( |

| |Hence the given eqn becomes ( 9 ((2 - ( ) - 18( + 8 )z = 0 => (9(2 -27( + 8)z = 0. |

| | |

| |PART B |

|1. |Solve ( D2 + 16 ) y = cos3x |

| |Solve by the method of variation of parameters [pic]. |

|2 | |

|3. |Solve (x2 D2 – 3 x D + 4 ) y = x2 cos(logx) |

| |Solve [pic] given x = 1 and y = 0 at t = 0. |

|4. | |

|5. |Solve (x2 D2 – 2 x D - 4 ) y = x2 + 2logx. |

| |Solve [pic]. |

|6. | |

|7. |Solve ( D2 + 4D+3 ) y = e–x sinx. |

|8. |Solve ( D2 + 1 ) y = x sinx by the method of variation of parameters. |

|9. |Solve (3x + 2)2 y’’ + 3 (3x + 2) y’ – 36 y = 3x2 + 4x + 1. |

|10. |Solve (D2 -1) y = x ex sinx. |

| | |

| |UNIT-II VECTOR CALCULUS |

|1. |If [pic], find grad φ at (1, -1, 2) |

| | [pic] |

| |Find [pic]at (2, -2, -1). |

|2. | |

| |[pic] |

| |Find the directional derivative of ( = 3x2+2y-3z at (1, 1, 1) in the direction[pic]. |

|3. | |

| | [pic]= [pic] |

|4 |Find the unit normal vector to the surface x2 + xy + z2 = 4 at the point (1,-1,2). |

| | |

| |( = x2 + xy + z2 , [pic] = [pic] =[pic] |

| |[pic]. |

|5. |Find the angle between the surfaces x logz = y2 – 1 and x2y = 2 –z at the point (1, 1, 1). |

| | |

| |Let (1 = y2 –xlogz -1 |

| |[pic](1 = - log z [pic] , ([pic](1) (1,1,1) = [pic] and |[pic](1| = [pic] |

| |Let (2 = x2y – 2+z |

| |[pic](2 = [pic], ([pic](2)(1,1,1) = [pic] and |[pic](2| = [pic] |

| |[pic], |

| |=[pic] ([pic][pic] |

|6. |In what direction from (-1,1,2) is the directional derivative of (= xy2z3 a maximum. Find also the magnitude of this maximum. |

| | |

| |Given (= xy2z3 |

| |[pic][pic] and [pic] at (1,1,2) = [pic] |

| |[pic] The maximum directional derivative occurs in the direction of |

| |[pic]=[pic] |

| |The magnitude of this max. directional derivative = [pic] |

| |Prove that [pic] |

|7. | |

| |[pic] = [pic] = [pic] = [pic] = [pic]. |

| |If [pic], find div curl [pic] |

|8. | |

| |[pic] |

| |div (curl [pic]) = [pic]= 0 |

| |Find ‘a’, such that (3x-2y +z) [pic]is solenoidal. |

|9. | |

| | Div [pic] = (((3x-2y +z )[pic] = 3 + a + 2 = 5 + a |

| |Div[pic] = 0 ( a = –5. |

| |If [pic] and [pic] are irrotational vectors prove that [pic] is solenoidal. |

|10. | |

| |[pic] is irrotational ( curl [pic] = 0 and [pic] is irrotational ( curl [pic] = 0 |

| |[pic] [pic]( curl [pic]) - [pic]( curl [pic]) = [pic]0 - [pic]0 = 0. |

| |[pic] is solenoidal. |

| |Show that the vector [pic] is irrotational. |

|11. | |

| |[pic] |

| |If [pic][pic], evaluate [pic] where C is the part of the curve y = x3 |

|12. |between x =1 and x=2. |

| | y = x3 ( dy = 3x2 dx |

| |[pic]=[pic] = [pic] = 31 + 63 = 94. |

| |If [pic]=[pic], evaluate the line integral [pic] from (0,0) to (1,1) along the path y=x. |

|13. | |

| |[pic] = [pic] = [pic] |

| |If [pic].Check whether the integral [pic] is independent of the path c. |

|14. | |

| |This integral is independent of the path of integration if |

| |[pic] |

| |[pic] .Hence the line integral is independent of path. |

|15. |State Stoke’s theorem. |

| | If S is an open surface bounded by a simple closed curve C and if a vector function [pic] is continuous and has continuous partial |

| |derivatives in S and on C, then [pic]=[pic] where [pic]is the unit vector normal to the surface. |

|16. |State Green’s Theorem |

| |If M(x ,y) and N(x ,y) are continuous function with continuous partial derivatives in a region R of the xy plane bounded by a simple |

| |closed curve C, then [pic].Where C is the curve described in positive direction |

|17. |State Gauss Divergence Theorem |

| |If V is the volume bounded by a closed surface S and if a vector function [pic]is continuous partial derivative in V and on S , then[pic]|

| |Using Divergence theorem, evaluate [pic] over the surface of the sphere x2 + y2 + z2 = a2. |

|18. | |

| |By Divergence theorem, |

| |[pic] =[pic] =[pic] = [pic] = [pic]. |

|19. |Find the area of a circle of radius ‘a’ using Green’s theorem. |

| |Area = [pic] on x2+ y2 =a2 , We have x = a cos ( y= asin( , (:0(2( |

| |Therefore Area = [pic] = [pic] = (a2. |

| |If S is any closed surface enclosing a volume V and[pic], prove that[pic]. |

|20. | |

| | Div [pic] = a+b+c ; [pic] = (a+b+c) [pic] = (a+b+c)V. |

| | |

| |PART B |

|1. |Verify stoke’s theorem for [pic] in the square region in the XY plane bounded by the lines x = 0, y = 0, x = a and y = a. |

|2. | Find the directional derivative of Ф = 2xy + z2 at the point (1,-1,3) in the direction of [pic] |

|3. |Using Gauss – Divergence theorem, evaluate [pic] where [pic] and |

| |S is the surface bounded by the cube x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. |

|4. |Verify Green’s theorem for [pic] where C is the boundary of the square formed by the lines x = 0, y = 0, a = a, y = a. |

|5. |Show that [pic] is Irrotational and find its scalar potential |

|6. |Find the work done when a force [pic] moves a particle in the xy – |

| |Plane from (0, 0) to (1,1) along the parabola y2 = x. |

|7. |Find the angle between the normals to the surface xy3z2 =4 at the points (-1,-1,2) and (4,1,-1) |

| |Ir [pic] is the position vector of the point (x,y,z) , Prove that [pic] |

|8. | |

| |Evaluate [pic] where C is the square bounded by the line x=0 ,x=1 , y =0 and y=1. |

|9. | |

| | |

|10 |Prove that [pic] |

| | |

| |UNIT-III ANALYTIC FUNCTIONS |

| | |

|1. |Define an analytic function (or) holomorphic function (or) Regular function. |

| |A function is said to be analytic at a point if its derivative exists not only at that point but also in some neighborhood of that point.|

| |Define an entire or an integral function. |

|2. | |

| |A function which is analytic everywhere in the finite plane is called an entire function. An entire function is analytic everywhere |

| |except at z = [pic] |

| |Ex. ez, sinz, cosz, sinhz, coshz |

| |State the necessary condition for f(z) to be analytic [Cauchy – Riemann Equations] |

|3. | |

| |The necessary conditions for a complex function f(z) = u(x,y) + iv(x,y) to be analytic in a region R are |

| |[pic] (i.e) ux = vy and vx = - uy |

|4 |State the sufficient conditions for f(z) to be analytic |

| |If the partial derivatives ux, uy, vx and vy are all continuous in D and ux = vy and uy= - vx. Then the function f(z) is analytic in a |

| |domain D. |

|5. |State the polar form of the C – R equation. |

| |In Cartesian co – ordinates any point z is z = x + iy |

| |In polar co – ordinates it is z = rei[pic] where r is the modulus and [pic] is the argument. Then the C- R equation in polar co – |

| |ordinates is given by |

| |[pic] |

|6. |State the basic difference between the limit of a function of a real variable and that of complex variable. |

| | In the real variable, x →x0 implies that x approaches x0 along the X – axis (or) a line parallel to the X – axis. |

| |In complex variable, z →z0 implies that z approaches z0 along any path joining the points z and z0 that lie in the Z – plane. |

| |Define harmonic function. |

|7. | |

| |A real function of two real variables x and y that possesses continuous second order partial derivatives and that satisfies Laplace |

| |equation is called a harmonic function. |

| |Define conjugate harmonic function. |

|8. | |

| |If u and v are harmonic functions such that u + iv is analytic, then each is called the conjugate harmonic function of the other. |

| |Define conformal transformation. |

|9. | |

| |Consider the transformation w = f(z), where f(z) is a single valued function of z, a point z0 and any two curves C1 and C2 passing |

| |through z0 in the Z plane,will be mapped onto a point w0 and two curves [pic]and [pic] in the W plane. If the angle between C1 and C2 at|

| |z0 is the same as the angle between [pic]and [pic]at w0 both in magnitude and direction, then the transformation w = f(z) is said to be |

| |conformed at the point z0. |

| |Define Isogonal transformation. |

|10. | |

| |A transformation under which angles between every pair of curves through a point are preserved in magnitude but opposite in direction is |

| |said to be isogonal at that point. |

| |Define Bilinear transformation. |

|11. | |

| |The transformation [pic]’ ad – bc [pic] where a, b,c,d are complex numbers is called a bilinear transformation. This is also called as |

| |Mobius or linear fractional transformation. |

| |Define Cross Ratio. |

|12. | |

| |Given four points [pic] in this order , the ratio [pic]is called the cross ratio of the four points. |

| |Show that f(z) = [pic]is differentiable at z = 0 but not analytic at z = 0. |

|13. | |

| |Let z = x + iy and [pic] |

| |[pic] |

| |So the CR equations ux = vy and uy= - vx are not satisfied everywhere except at z = 0. So f(z) may be differentiable only at z = 0. Now |

| |ux = 2y, vy = 0 and uy = 2y, vx = 0 are continuous everywhere and in particular at (0, 0). So f(z) is differentiable at z = 0 only and |

| |not analytic there. |

| |Determine whether the function 2xy + i(x2 – y2) is analytic or not. |

|14. | |

| |Let f(z) = 2xy + i(x2 – y2) |

| |u = 2xy ; v = x2 – y2 |

| |ux = 2y, vy = -2y and uy = 2x , vx = 2x |

| |ux [pic] vy and uy [pic] - vx |

| |CR equations are not satisfied. |

| |Hence f(z) is not an analytic function |

|15. |Prove that an analytic function whose real part is constant must itself be a constant. |

| |Let f(z) = u + iv |

| |Given u = constant. [pic] ux = 0 and uy = 0 |

| |By CR equations |

| |ux = 0 [pic], vy = 0 ; uy = 0 [pic] vx = 0 |

| |f1(z) = ux + ivx = 0 + i0 |

| |f1(z) = 0 [pic] f(z) = c |

| |f(z) is a constant. |

|16. |Show that the function u = 2x – x3 + 3xy2 is harmonic |

| |Given u = 2x – x3 + 3xy2 |

| |ux = 2 – 3x2 + 3y2 and [pic] ; uy = 6xy and [pic] |

| |[pic] +[pic] = -6x + 6x = 0 |

| |Hence u is harmonic |

|17. |Find a function w such that w = u + iv is analytic, if u = exsiny |

| |Given u = exsiny |

| |[pic] |

| |By Milne Thomson’s method |

| |[pic] |

| | |

| |Find the image of the hyperbola x2 – y2 = 10 under the transformation w = z2 |

|18. | |

| |w = z2 [pic] |

| |u = x2 – y2 and v = 2xy ; x2 – y2 = 10 (i.e) u = 10 |

| |Hence the image of the hyperbola x2 – y2 = 10 under the transformation w = z2 is u = 10 which is a straight line in w plane. |

|19. |Obtain the invariant points of the transformation |

| |[pic] , The invariant points are given by [pic] |

| |[pic] |

| |Define a critical point of the bilinear transformation. |

|20. | |

| |The point at which the mapping w = f(z) is not conformal, (i.e) f1(z) = 0 is called a critical point of the mapping. |

| |PART B |

| |Find the fixed points of the transformation [pic] |

|1. | |

|2 |Give an example such that u and v are harmonic but u+iv is not analytic |

|3. |Prove that ex cos y is harmonic function. |

|4. |Define bilinear transformation and what the condition for this to be conformal is. |

|5. |If u + iu is analytic, then prove that v –iu is analytic. |

|6. |Find the image of the circle |z|=2 under the transformation w=3z. |

|7. |Determine the region of the w-plane into which the first quadrant of z-plane mapped by |

| |the transformation w=z2. |

|8. |Construct the analytic function f(z) = u+iv given that 2u+3v = ex(cos y – sin y). |

|9. |Find the bilinear transformation that maps z = (1, i, –1) into w=(2, i, –2). |

|10. |Show that ex( x cos y – y sin y) is harmonic function. Find the analytic function f(z) for which ex (x cos y – y sin y) is the |

| |imaginary part. |

| |UNIT-IV COMPLEX INTEGRATION |

| |What is the value of [pic]where C is [pic]? |

|1. | |

| |[pic]if f (z) is analytic inside and on C. Since ez is analytic on and inside [pic], the given integral is zero. |

| |Evaluate [pic] where C is [pic]. |

| | |

|2. | |

| |Here f(z) =cos[pic], C is[pic] , |

| |By Cauchy’s integral Formula [pic]=2[pic]. |

| |Evaluate [pic]where C is [pic]using Cauchy’s integral formula. |

| | |

|3. | |

| |Z= -2 is out of the given circle [pic], [pic] |

| |Evaluate[pic]where C is [pic]. |

|4. | |

| |We know that Cauchy’s integral formula is [pic] |

| |Given [pic]=[pic]Here f (z) = 3z2+7z+1, a=–1 lies outside[pic]. Therefore by Cauchy’s integral formula [pic]=0. |

| |Obtain the Laurent expansion of the function [pic]in the neighborhood of its |

|5. |Singular point. Hence find the residue at that point. |

| |Z=1 is a pole of order 2 |

| |Put z-1=u .Then [pic]becomes[pic] |

| |[pic] |

| |Residue of f(z) = coefficient of [pic] =e |

| |Expand f(z)=[pic]in the region [pic]. Using this result, expand [pic]and tan-1z in |

|6. |powers of z. |

| |[pic]By using binomial series expansions,[pic] |

| |[pic] |

| |[pic]=[pic]…….(1) |

| |Now [pic] |

| |Changing z by –z2 in(1)[pic] -----(2) |

| |If f(z)=tan-1z, then f ’(z)= [pic] |

| |Hence By integrating (2) w.r.to z , tan-1z=[pic] |

| |Obtain the Taylor’s series expansion of log(1+z) when [pic] |

|7. | |

| |Let f (z) =log(1+z) f (0)=log1=0 |

| | |

| |[pic] [pic] |

| |[pic] [pic] |

| |[pic] [pic] |

| |[pic] [pic] |

| |[pic] |

| |[pic] = [pic] |

| |What is the Nature of the singularity at z=0 of the function[pic]. |

| |f(z) =[pic] The function f(z) is not defined at z=0.But by L’Hospital’s rule, |

|8. |[pic][pic][pic][pic] |

| |Since the limit exists and is finite, the singularity at z=0 is a removable singularity. |

| |Evaluate [pic]where C is [pic] |

| | |

|9. | |

| |Given [pic] [pic] both lies outside c. |

| |Therefore [pic] |

| |Find the residue of the function f(z)= [pic]at a simple pole. |

| | |

|10. | |

| |[pic], z = 2 is a simple pole. Res[f(z)]z=2 = [pic] |

| |State Laurent’s series. |

|11. | |

| |If C1,C2 are two concentric circles with centre at z=a and radii r1 and r2 (r10 . |

| |(ii) f(t) is of exponential order. |

| |Find the Laplace transform of [pic]. |

| | |

|2. | |

| |[pic]= [pic] = [pic]= [pic] |

| |If L[f(t)] = F(s), prove that L{f(t/5)}= 5 F(5s). |

|3. | |

| |[pic] = [pic] |

| |put [pic] ( 5du = dt |

| |([pic]= [pic] = [pic] = 5 F(5s). |

| |Find the Laplace transform of unit step function. |

|4 | |

| |The unit step function is defined as [pic] |

| |and L[ua(t)] = [pic] |

|5. |Find the Laplace transform of f(t) = cos2 3t. |

| | |

| |L[cos2 3t] = [pic] = [pic] = [pic]. |

| |Does [pic] exist? |

|6. | |

| |[pic] |

| |[pic][pic][pic] does not exist. |

|7. |Obtain the Laplace transform of sin2t – 2tcos2t in the simplified form. |

| |[pic] = [pic] |

| |=[pic] |

| |= [pic] |

| |= [pic] |

| |Find [pic]. |

|8. | |

| |[pic] = [pic] |

| |= [pic] |

| |=e-t [pic] |

| |= e-t (cos t + sin t) |

| |What is the Laplace transform of f(t) = f(t +10), 0< t < 10? |

|9. | |

| |Given that f(t) is a periodic function with period 10 |

| |L{f(t)} = [pic] |

| |Put p=10, L{f(t)} = [pic] |

| |If L{f(t)} = [pic], find the value of [pic]. |

| | |

|10. | |

| |[pic]= [pic] = [pic]= [pic]= [pic] |

| |Find [pic] |

| | |

|11. | |

| |[pic]=[pic] |

| |=[pic] - 2 [pic] |

| |= [pic] – [pic]= [pic] |

| |Find the Laplace transform sin32t |

|12. | |

| | |

| |L[sin32t] = ¼ L[3sin2t – sin6t] = ¾ L[sin2t] – ¼ L[sin6t] |

| |= [pic] - [pic] = [pic] - [pic]. |

| |Find [pic] |

| | |

|13. | |

| | Let F(s) = [pic] |

| |F((s) = [pic] = [pic] |

| |([pic]; [pic] |

| |[pic] |

| |Solve using Laplace transform [pic] given that y(0)=0. |

| | |

|14. | |

| |Taking L.T. on both sides, we get L[y(] + L[y] = L[e-t] |

| |(s+1) L[y] =[pic] |

| |L[y] =[pic]; y =[pic] = t e-t. |

|15. |Give an example for a function that do not have Laplace transform. |

| |Consider f(t)=[pic], since [pic], hence [pic]is not of exponential order function. Hence f(t)=[pic], does not have Laplace transform. |

| |Can F(s)= [pic] be the transform of some f(t)? |

|16. | |

| | [pic][pic] ( 0 |

| |Hence F(s) cannot be Laplace transform of f(t). |

| |Evaluate [pic]. |

|17. | |

| |Let [pic] |

| |= L[sint] L[cost] (by convolution theorem) |

| |= [pic] |

| |[pic] =[pic] |

| |Give an example for a function having Laplace transform but not satisfying the continuity condition. |

|18. | |

| | f (t) = [pic]has Laplace transform even though it does not satisfy the continuity condition. i.e. It is not piecewise continuous|

| |in (0,() as [pic] |

|19. |Define a periodic function and give examples |

| |A function f (t) is said to be periodic function if f(t + p) =f(t) for all t. The least value of p> 0 is called the period of p. For |

| |example, sint and cost are periodic functions with period 2( |

| |State the convolution theorem |

| | |

|20. | |

| |The convolution of f(t) and g(t) is defined as [pic] and it is denoted by f(t) * g(t). |

| | |

| |PART B |

|1. |Find the Laplace transform of f(t) = [pic]and f(t+2a) = f(t) for all t. |

|2. |Find the Laplace transform of f(t) = [pic]and f(t+2a) = f(t) for all t. |

|3. |Find the inverse Laplace transform of [pic] using convolution theorem. |

|4. |Find the Laplace transform of [pic] |

|5. |Verify initial and final value theorems for the function f(t) = 1 + e-t(sin t + cos t). |

|6. |Find the inverse Laplace transform of [pic] |

|7. |Using Laplace transform solve the differential equation y((-3y(-4y=2e-t with y(0) = y((0) = 1. |

|8. |Solve the equation y((+ 9y=cos2t with y(0) =1 y ([pic]) = –1. |

|9. |Evaluate [pic] using Laplace transform. |

|10 |Find the Laplace transform of (i) t2 e–tcost (ii) coshat cosat |

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