IGNOU MPH-001 Solved Question Paper PDF Download

IGNOU MPH-001 Previous Year Solved Question Paper in Hindi

IGNOU MPH-001 Previous Year Solved Question Paper in English

Q1. Consider heat flow in a uniform bar of length L insulated along its length. The temperature of the bar is modelled by the one-dimensional heat diffusion equation: 4+6 ∂T(x,t)/∂t = K ∂²T(x,t)/∂x² (0 < x < L, t > 0) (i) Separate the equation into two ODEs. (ii) If the initial temperature distribution is given by: T(x,0) = (2L – x) (0 < x < L) solve the heat equation for the boundary conditions on T(x,t) given as: T(0,t) = 0 and ∂T(L,t)/∂x = 0, t ≥ 0

Ans. (i) Separation of the Equation The given one-dimensional heat diffusion equation is: ∂T/∂t = K ∂²T/∂x² We use the method of separation of variables and assume a solution of the form: T(x,t) = X(x)Θ(t) where X is a function of x only and Θ is a function of t only. Substituting this into the heat equation, we get: X(x) dΘ(t)/dt = K Θ(t) d²X(x)/dx² Dividing the equation by K X(x)Θ(t) to separate the variables: (1/KΘ) dΘ/dt = (1/X) d²X/dx² Since the left side depends only on t and the right side depends only on x, both sides must be equal to a constant, which we’ll call -λ² (a negative constant leads to physically meaningful, non-trivial solutions). This yields two Ordinary Differential Equations (ODEs): 1. Spatial Equation: d²X/dx² + λ²X = 0 2. Temporal Equation: dΘ/dt + Kλ²Θ = 0 These are the two separated ODEs for the given partial differential equation.

(ii) Solution of the Boundary Value Problem First, we solve the spatial equation for X(x) with the given boundary conditions. The boundary conditions are T(0,t) = 0 and ∂T(L,t)/∂x = 0. In terms of T(x,t) = X(x)Θ(t), this translates to X(0)Θ(t) = 0 and X'(L)Θ(t) = 0. Since Θ(t) cannot be trivially zero, we must have: X(0) = 0 and X'(L) = 0 The general solution to the spatial equation d²X/dx² + λ²X = 0 is: X(x) = A cos(λx) + B sin(λx) Applying the first boundary condition, X(0) = 0: X(0) = A cos(0) + B sin(0) = A = 0 So, X(x) = B sin(λx). Applying the second boundary condition, X'(L) = 0: X'(x) = Bλ cos(λx) ⇒ X'(L) = Bλ cos(λL) = 0 For a non-trivial solution B ≠ 0, so we must have cos(λL) = 0. This gives λL = (2n-1)π/2, where n = 1, 2, 3, … Thus, the eigenvalues are λ_n = (2n-1)π / (2L) And the corresponding eigenfunctions are X_n(x) = sin((2n-1)πx / (2L)) .

Now we solve the temporal equation dΘ/dt + Kλ_n²Θ = 0. Its solution is: Θ_n(t) = C_n e^(-Kλ_n²t)

The general solution of the heat equation is given by the superposition principle: T(x,t) = Σ[n=1 to ∞] X_n(x)Θ_n(t) = Σ[n=1 to ∞] B_n sin((2n-1)πx / (2L)) e^(-K((2n-1)π/2L)²t) where B_n = B * C_n is a new constant.

The coefficients B_n are determined using the initial condition T(x,0) = (2L-x): T(x,0) = Σ[n=1 to ∞] B_n sin((2n-1)πx / (2L)) = (2L-x) This is a Fourier sine series for the function (2L-x). The coefficients B_n are given by: B_n = (2/L) ∫[0 to L] (2L-x) sin((2n-1)πx / (2L)) dx Integrating by parts: B_n = (2/L) [ (2L-x) (-2L/(2n-1)π)cos((2n-1)πx/2L) |[0 to L] – ∫[0 to L] (-1) (-2L/(2n-1)π)cos((2n-1)πx/2L) dx ] B_n = (2/L) [ (L 0) – (2L -2L/(2n-1)π) – (2L/(2n-1)π) * (2L/(2n-1)π)sin((2n-1)πx/2L) |[0 to L] ] B_n = (2/L) [ 4L²/((2n-1)π) – (4L²/((2n-1)²π²)) * sin((2n-1)π/2) ] B_n = 8L/((2n-1)π) – (8L/((2n-1)²π²)) (-1)^(n+1)

The final solution is: T(x,t) = Σ[n=1 to ∞] [8L/((2n-1)π) + (8L(-1)^n)/((2n-1)²π²)] sin((2n-1)πx/2L) e^(-K((2n-1)π/2L)²t)

Q2. (a) The generating function for Bessel’s functions of the first kind and integral order is given by: 5 g(x,t) = e^((x/2)(t – 1/t)) = Σ[n=-∞ to +∞] J_n(x)t^n Show that: J₀(x) + 2J₂(x) + 2J₄(x) + … = 1 (Note: The question paper may have a typo. The standard identity will be proven.) (b) The expression for Bessel’s function of the first kind of order m is given by: 5 J_m(x) = Σ[k=0 to ∞] (-1)^k / (k! Γ(k+m+1)) * (x/2)^(2k+m) Using this relation, derive the recurrence relation: x J_m'(x) = m J_m(x) – x J_(m+1)(x)

Ans. (a) Proof of the Identity for Bessel Functions The generating function for Bessel functions of the first kind is: g(x,t) = e^((x/2)(t – 1/t)) = Σ[n=-∞ to +∞] J_n(x)t^n To prove the identity, we substitute t = e^(iθ). t – 1/t = e^(iθ) – e^(-iθ) = (cosθ + i sinθ) – (cosθ – i sinθ) = 2i sinθ So, the generating function becomes: g(x, e^(iθ)) = e^((x/2)(2i sinθ)) = e^(ix sinθ) Substituting t = e^(iθ) on the series side: Σ[n=-∞ to +∞] J_n(x) (e^(iθ))^n = Σ[n=-∞ to +∞] J_n(x) e^(inθ) Thus, we have the Jacobi-Anger expansion: e^(ix sinθ) = Σ[n=-∞ to +∞] J_n(x) e^(inθ) Taking the real part of this equation: cos(x sinθ) = Re [ Σ J_n(x) (cos(nθ) + i sin(nθ)) ] Using the relation J_(-n)(x) = (-1)^n J_n(x), the sum can be simplified: Σ J_n(x) e^(inθ) = J₀(x) + Σ[n=1 to ∞] (J_n(x)e^(inθ) + J_(-n)(x)e^(-inθ)) = J₀(x) + Σ[n=1 to ∞] (J_n(x)e^(inθ) + (-1)^n J_n(x)e^(-inθ)) The real part of this is: cos(x sinθ) = J₀(x) + Σ[n=1 to ∞] J_n(x) (cos(nθ) + (-1)^n cos(nθ)) = J₀(x) + 2J₂(x)cos(2θ) + 2J₄(x)cos(4θ) + … This is the Fourier cosine series for cos(x sinθ). Now, let’s set θ = 0 in this equation: LHS = cos(x sin(0)) = cos(0) = 1 RHS = J₀(x) + 2J₂(x)cos(0) + 2J₄(x)cos(0) + … RHS = J₀(x) + 2J₂(x) + 2J₄(x) + … Equating the LHS and RHS, we get the desired identity: 1 = J₀(x) + 2J₂(x) + 2J₄(x) + …

(b) Derivation of the Recurrence Relation The series expression for the Bessel function of order m is: J_m(x) = Σ[k=0 to ∞] ((-1)^k / (k! Γ(k+m+1))) * (x/2)^(2k+m) Differentiating with respect to x: J_m'(x) = d/dx [ Σ[k=0 to ∞] ((-1)^k / (k! Γ(k+m+1))) * (x/2)^(2k+m) ] J_m'(x) = Σ[k=0 to ∞] ((-1)^k / (k! Γ(k+m+1))) (2k+m) (x/2)^(2k+m-1) (1/2) J_m'(x) = Σ[k=0 to ∞] ((-1)^k (2k+m)) / (k! Γ(k+m+1) 2) * (x/2)^(2k+m-1) Multiply the equation by x: x J_m'(x) = Σ[k=0 to ∞] ((-1)^k (2k+m)) / (k! Γ(k+m+1)) (x/2) * (x/2)^(2k+m-1) x J_m'(x) = Σ[k=0 to ∞] ((-1)^k (2k+m)) / (k! Γ(k+m+1)) (x/2)^(2k+m) Split the term (2k+m) as m + 2k: x J_m'(x) = Σ ((-1)^k m / (k! Γ(k+m+1))) (x/2)^(2k+m) + Σ ((-1)^k 2k / (k! Γ(k+m+1))) (x/2)^(2k+m) The first term is exactly m J_m(x) . For the second term (the k=0 term is zero), cancel k from k!: Second term = Σ[k=1 to ∞] ((-1)^k 2 / ((k-1)! Γ(k+m+1))) (x/2)^(2k+m) Let k’ = k-1. Then k = k’+1. = Σ[k’=0 to ∞] ((-1)^(k’+1) 2 / (k’! Γ(k’+1+m+1))) (x/2)^(2(k’+1)+m) = – Σ[k’=0 to ∞] ((-1)^k’ 2 / (k’! Γ(k’+m+2))) (x/2)^(2k’+m+2) Factor out an x: = – x Σ[k’=0 to ∞] ((-1)^k’ / (k’! Γ(k’+(m+1)+1))) * (x/2)^(2k’+m+1) The summation is the series for J_(m+1)(x). So, the second term is -x J_(m+1)(x) . Combining the terms, we get the recurrence relation: x J_m'(x) = m J_m(x) – x J_(m+1)(x) .

Q3. (a) Starting with generating function of Legendre polynomials of first kind: 4 g(x,t) = (1 – 2xt + t²)^(-1/2) = Σ[n=0 to ∞] P_n(x)t^n show that: P_n(-x) = (-1)^n P_n(x) (b) Using the generating function of Hermite polynomials: 6 g(x, t) = e^(2xt – t²) = Σ[n=0 to ∞] H_n(x) t^n/n! evaluate the integral: ∫[-∞ to +∞] x e^(-x²) H_m(x) H_n(x) dx

Ans. (a) Proof of the Property for P_n(-x) The generating function for Legendre polynomials is: g(x,t) = (1 – 2xt + t²)^(-1/2) = Σ[n=0 to ∞] P_n(x)t^n In this function, replace x with -x: g(-x,t) = (1 – 2(-x)t + t²)^(-1/2) = (1 + 2xt + t²)^(-1/2) From the series form of the generating function, we have: g(-x,t) = Σ[n=0 to ∞] P_n(-x)t^n Now, in the original generating function, replace t with -t: g(x,-t) = (1 – 2x(-t) + (-t)²)^(-1/2) = (1 + 2xt + t²)^(-1/2) From the series form of the generating function, we have: g(x,-t) = Σ[n=0 to ∞] P_n(x)(-t)^n = Σ[n=0 to ∞] (-1)^n P_n(x)t^n From the expressions above, we see that g(-x,t) = g(x,-t) . Equating their series expansions: Σ[n=0 to ∞] P_n(-x)t^n = Σ[n=0 to ∞] (-1)^n P_n(x)t^n Comparing the coefficients of t^n on both sides, we directly obtain: P_n(-x) = (-1)^n P_n(x) This property shows that P_n(x) is an even function for even n, and an odd function for odd n.

(b) Evaluation of the Integral with Hermite Polynomials We need to evaluate the integral I = ∫[-∞ to +∞] x e^(-x²) H_m(x) H_n(x) dx. We use the recurrence relation for Hermite polynomials: 2x H_n(x) = H_(n+1)(x) + 2n H_(n-1)(x) This gives us: x H_n(x) = (1/2) H_(n+1)(x) + n H_(n-1)(x) Substitute this expression into the integral: I = ∫[-∞ to +∞] e^(-x²) H_m(x) [ (1/2) H_(n+1)(x) + n H_(n-1)(x) ] dx I = (1/2) ∫ e^(-x²) H_m(x) H_(n+1)(x) dx + n ∫ e^(-x²) H_m(x) H_(n-1)(x) dx Now we use the orthogonality relation for Hermite polynomials: ∫[-∞ to +∞] e^(-x²) H_m(x) H_k(x) dx = 2^k k! √π δ_mk where δ_mk is the Kronecker delta. Applying this orthogonality relation to the two parts of our integral: 1. The first part: (1/2) ∫ e^(-x²) H_m(x) H_(n+1)(x) dx This term is non-zero only if m = n+1 . In that case, its value is: (1/2) * [ 2^(n+1) (n+1)! √π ] = 2^n (n+1)! √π 2. The second part: n ∫ e^(-x²) H_m(x) H_(n-1)(x) dx This term is non-zero only if m = n-1 . In that case, its value is: n [ 2^(n-1) (n-1)! √π ] = n 2^(n-1) (n-1)! √π = (1/2) * 2^n n! √π Thus, the value of the integral is non-zero in two cases:

• If m = n+1 , then I = 2^n (n+1)! √π
• If m = n-1 , then I = n * 2^(n-1) (n-1)! √π = (2^n n! √π) / 2

In all other cases (i.e., |m-n| ≠ 1), the integral is zero.

Q4. (a) Show that the vectors (v₁, v₂, v₃); v₁ = (1, 0, 0)ᵀ, v₂ = (1, 1, 0)ᵀ, v₃ = (1, 1, 1)ᵀ form the basis in ℝ³, the vector space of three-dimensional vectors. 4 (b) Let A = [[1, a, b], [0, 1, c], [0, 0, 1]], B = [[1, d, e], [0, 1, f], [0, 0, 1]] Calculate AB and show that it is also in form of upper triangular matrix. Calculate the value of d, e, f, so that AB = I. Find the condition that AB = BA. 6

Ans. (a) Proof of Basis for Vectors To show that the vectors v₁, v₂, and v₃ form a basis in ℝ³, we must prove that they are linearly independent . Since the dimension of ℝ³ is 3, any set of three linearly independent vectors will form a basis. To test for linear independence, we set up the following vector equation: c₁v₁ + c₂v₂ + c₃v₃ = 0 where c₁, c₂, c₃ are scalars, and 0 is the zero vector. c₁[1, 0, 0]ᵀ + c₂[1, 1, 0]ᵀ + c₃[1, 1, 1]ᵀ = [0, 0, 0]ᵀ This leads to the following system of linear equations: 1. c₁ + c₂ + c₃ = 0 2. c₂ + c₃ = 0 3. c₃ = 0 From equation (3), we have c₃ = 0 . Substituting this into equation (2), we get c₂ + 0 = 0, which means c₂ = 0 . Substituting the values of both c₂ and c₃ into equation (1), we get c₁ + 0 + 0 = 0, which means c₁ = 0 . Since the only solution is c₁ = c₂ = c₃ = 0, the vectors v₁, v₂, v₃ are linearly independent. Being three linearly independent vectors in ℝ³, they form a basis for the vector space ℝ³.

(b) Matrix Calculations Given the upper triangular matrices: A = [[1, a, b], [0, 1, c], [0, 0, 1]] and B = [[1, d, e], [0, 1, f], [0, 0, 1]] 1. Calculation of AB: AB = [[1, a, b], [0, 1, c], [0, 0, 1]] * [[1, d, e], [0, 1, f], [0, 0, 1]] AB = [[1 1+a 0+b 0, 1 d+a 1+b 0, 1 e+a f+b*1], [0 1+1 0+c 0, 0 d+1 1+c 0, 0 e+1 f+c*1], [0 1+0 0+1 0, 0 d+0 1+1 0, 0 e+0 f+1*1]] AB = [[1, a+d, b+e+af], [0, 1, c+f], [0, 0, 1]] Since all elements below the main diagonal are zero, the product matrix AB is also an upper triangular matrix . 2. Values of d, e, f for AB = I: We want AB to be equal to the identity matrix I: [[1, a+d, b+e+af], [0, 1, c+f], [0, 0, 1]] = [[1, 0, 0], [0, 1, 0], [0, 0, 1]] Equating the corresponding elements: a + d = 0 ⇒ d = -a c + f = 0 ⇒ f = -c b + e + af = 0 ⇒ b + e + a(-c) = 0 ⇒ e = ac – b So, the values are: d = -a, f = -c, e = ac – b . This shows that B is the inverse of A (A⁻¹). 3. Condition for AB = BA: First, calculate BA: BA = [[1, d, e], [0, 1, f], [0, 0, 1]] * [[1, a, b], [0, 1, c], [0, 0, 1]] BA = [[1, a+d, b+e+dc], [0, 1, c+f], [0, 0, 1]] For AB = BA, we must have: [[1, a+d, b+e+af], [0, 1, c+f], [0, 0, 1]] = [[1, a+d, b+e+dc], [0, 1, c+f], [0, 0, 1]] Most elements are identical. We only need to equate the (1,3) elements: b + e + af = b + e + dc af = dc This is the condition for the matrices A and B to commute. Note that a, c, d, f can be any values as long as this relation holds. For example, if a=c=d=f=0, they commute.

Q5. (a) Obtain the normalized eigen vectors for a real symmetric matrix: 6 [[cosθ, sinθ, 0], [sinθ, -cosθ, 0], [0, 0, -2]] and find its diagonalizing matrix. (b) Show that: u(x, y) = 3x²y + 2x² – y³ – 2y² is harmonic in some domain. Calculate its conjugate harmonic function v(x, y) and construct the complex function f(z). 4

Ans. (a) Eigenvalues, Eigenvectors and Diagonalizing Matrix Let the given matrix be M: M = [[cosθ, sinθ, 0], [sinθ, -cosθ, 0], [0, 0, -2]] 1. Finding the Eigenvalues: The eigenvalues are the solutions to the characteristic equation det(M – λI) = 0. det([[cosθ-λ, sinθ, 0], [sinθ, -cosθ-λ, 0], [0, 0, -2-λ]]) = 0 Expanding the determinant: (-2-λ) * [ (cosθ-λ)(-cosθ-λ) – sin²θ ] = 0 (-2-λ) * [ -cos²θ – λcosθ + λcosθ + λ² – sin²θ ] = 0 (-2-λ) * [ λ² – (cos²θ + sin²θ) ] = 0 (-2-λ) * (λ² – 1) = 0 (-2-λ) (λ – 1) (λ + 1) = 0 Thus, the eigenvalues are λ₁ = 1, λ₂ = -1, λ₃ = -2 . 2. Finding the Eigenvectors: For each eigenvalue, we solve (M – λI)v = 0. For λ₁ = 1: [[cosθ-1, sinθ, 0], [sinθ, -cosθ-1, 0], [0, 0, -3]] [x, y, z]ᵀ = 0 -3z = 0 ⇒ z = 0. (cosθ-1)x + (sinθ)y = 0 (sinθ)x + (-cosθ-1)y = 0 Solving gives an eigenvector v₁’ = [sinθ, 1-cosθ, 0]ᵀ. Using half-angle identities, this is proportional to [2sin(θ/2)cos(θ/2), 2sin²(θ/2), 0]ᵀ, or [cos(θ/2), sin(θ/2), 0]ᵀ. This has a magnitude of 1. So, v₁ = [cos(θ/2), sin(θ/2), 0]ᵀ .

For λ₂ = -1: [[cosθ+1, sinθ, 0], [sinθ, -cosθ+1, 0], [0, 0, -1]] [x, y, z]ᵀ = 0 -z = 0 ⇒ z = 0. (cosθ+1)x + (sinθ)y = 0 Solving gives an eigenvector v₂’ = [sinθ, -(1+cosθ), 0]ᵀ. This is proportional to [sin(θ/2), -cos(θ/2), 0]ᵀ. This has a magnitude of 1. So, v₂ = [sin(θ/2), -cos(θ/2), 0]ᵀ .

For λ₃ = -2: [[cosθ+2, sinθ, 0], [sinθ, -cosθ+2, 0], [0, 0, 0]] [x, y, z]ᵀ = 0 These equations yield x=0, y=0. z can be any value. An eigenvector is v₃’ = [0, 0, 1]ᵀ. This is already normalized. v₃ = [0, 0, 1]ᵀ . 3. Diagonalizing Matrix: The diagonalizing matrix P has the normalized eigenvectors as its columns. P = [[cos(θ/2), sin(θ/2), 0], [sin(θ/2), -cos(θ/2), 0], [0, 0, 1]] This matrix will diagonalize M: P⁻¹MP = D, where D is the diagonal matrix of eigenvalues.

(b) Harmonic Function and Complex Function Given function: u(x, y) = 3x²y + 2x² – y³ – 2y² The question paper appears to have a typo. A common related problem is u(x,y) = 3x²y – y³. Let’s assume u(x, y) = 3x²y – y³ + 2x² – 2y². 1. Verification of Harmonic Property: A function is harmonic if it satisfies Laplace’s equation ∇²u = ∂²u/∂x² + ∂²u/∂y² = 0. ∂u/∂x = 6xy + 4x ⇒ ∂²u/∂x² = 6y + 4 ∂u/∂y = 3x² – 3y² – 4y ⇒ ∂²u/∂y² = -6y – 4 ∇²u = (6y + 4) + (-6y – 4) = 0 Since ∇²u = 0, the function u(x, y) is harmonic . 2. Calculation of Conjugate Harmonic Function v(x, y): Use the Cauchy-Riemann equations: (i) ∂v/∂y = ∂u/∂x = 6xy + 4x (ii) ∂v/∂x = -∂u/∂y = -(3x² – 3y² – 4y) = -3x² + 3y² + 4y Integrate equation (i) with respect to y: v(x, y) = ∫ (6xy + 4x) dy = 3xy² + 4xy + g(x) where g(x) is a function of x. Now, differentiate this v(x,y) with respect to x and compare with equation (ii): ∂v/∂x = 3y² + 4y + g'(x) Comparing to equation (ii): 3y² + 4y + g'(x) = -3x² + 3y² + 4y g'(x) = -3x² g(x) = ∫ -3x² dx = -x³ + C Setting the integration constant C to zero, we get: v(x, y) = 3xy² + 4xy – x³ 3. Construction of Complex Function f(z): f(z) = u(x, y) + i v(x, y) f(z) = (3x²y + 2x² – y³ – 2y²) + i (3xy² + 4xy – x³) Rearranging terms: f(z) = (2x² – 2y² + 4ixy) – ix³ – 3x²y + 3ixy² + iy³ f(z) = 2(x² – y² + 2ixy) – i(x³ + 3x²(iy) + 3x(iy)² + (iy)³) Using z=x+iy, z²=x²-y²+2ixy, and z³=x³+3x²(iy)+3x(iy)²+(iy)³. f(z) = 2z² – i(x+iy)³ = 2z² – iz³ Hence, the complex function is f(z) = 2z² – iz³ .

Q6. (a) Determine the value of Gamma function Γ(1/2). 4 (b) Determine and classify all the singularities of the following complex functions: 4+2 (i) z / (e^z + 1) (ii) 1 / (2 sin(z) – 1)²

Ans. (a) Value of Γ(1/2) The definition of the Gamma function is: Γ(z) = ∫[0 to ∞] t^(z-1) e^(-t) dt Setting z = 1/2, we get: Γ(1/2) = ∫[0 to ∞] t^(1/2 – 1) e^(-t) dt = ∫[0 to ∞] t^(-1/2) e^(-t) dt To evaluate this integral, we use the substitution t = x². Then, dt = 2x dx. When t = 0, x = 0; when t → ∞, x → ∞. Making the substitution: Γ(1/2) = ∫[0 to ∞] (x²)^(-1/2) e^(-x²) (2x dx) Γ(1/2) = ∫[0 to ∞] x⁻¹ e^(-x²) (2x dx) Γ(1/2) = 2 ∫[0 to ∞] e^(-x²) dx This is related to the famous Gaussian integral: ∫[-∞ to ∞] e^(-x²) dx = √π Since e^(-x²) is an even function (i.e., f(x) = f(-x)), we can write: ∫[-∞ to ∞] e^(-x²) dx = 2 ∫[0 to ∞] e^(-x²) dx So, 2 ∫[0 to ∞] e^(-x²) dx = √π Therefore, comparing with our expression for Γ(1/2), we find directly: Γ(1/2) = √π

(b) Determination and Classification of Singularities (i) f(z) = z / (e^z + 1) Singularities occur when the denominator is zero: e^z + 1 = 0 ⇒ e^z = -1 Writing -1 in polar form, -1 = e^(iπ). Since e^z is periodic with period 2πi, the general solution is: e^z = e^(i(π + 2nπ)) = e^(i(2n+1)π), where n is an integer. Taking the logarithm of both sides: z = i(2n+1)π, n ∈ ℤ These are the singularities. To classify them, we consider the numerator g(z) = z and denominator h(z) = e^z + 1. At these singularities, the numerator g(z_n) = i(2n+1)π ≠ 0 (unless n=-1/2, which is not an integer). Check the derivative of the denominator: h'(z) = e^z. Evaluate the derivative at the singularities: h'(i(2n+1)π) = e^(i(2n+1)π) = -1 ≠ 0. Since h(z_n) = 0 but h'(z_n) ≠ 0, these singularities are simple poles .

(ii) f(z) = 1 / (2 sin(z) – 1)² Singularities occur when the denominator is zero: (2 sin(z) – 1)² = 0 ⇒ sin(z) = 1/2 Using the definition of sin z, sin(z) = (e^(iz) – e^(-iz)) / (2i): (e^(iz) – e^(-iz)) / (2i) = 1/2 e^(iz) – e^(-iz) = i Letting w = e^(iz), the equation becomes w – 1/w = i. Multiplying by w: w² – 1 = iw ⇒ w² – iw – 1 = 0 Solving this quadratic equation for w: w = [ -(-i) ± √((-i)² – 4(1)(-1)) ] / 2 = [ i ± √(-1 + 4) ] / 2 = (i ± √3) / 2 There are two values for w = e^(iz): 1. e^(iz) = (√3 + i)/2 = e^(iπ/6). So, iz = iπ/6 + 2nπi ⇒ z = π/6 + 2nπ 2. e^(iz) = (-√3 + i)/2 = e^(i5π/6). So, iz = i5π/6 + 2nπi ⇒ z = 5π/6 + 2nπ (where n ∈ ℤ). These are the singularities. Now, to classify them. The denominator is h(z) = (g(z))², where g(z) = 2 sin(z) – 1. The zeros of g(z) are sin(z) = 1/2, which gives the series of z values above. Check g'(z) = 2 cos(z). At z = π/6 + 2nπ, cos(z) = cos(π/6) = √3/2 ≠ 0. At z = 5π/6 + 2nπ, cos(z) = cos(5π/6) = -√3/2 ≠ 0. Since g(z) has a simple zero at these points (because g=0 but g’≠0), then h(z) = (g(z))² has a zero of order 2 at these points. Therefore, the function f(z) = 1/h(z) has poles of order 2 at: z = π/6 + 2nπ and z = 5π/6 + 2nπ, n ∈ ℤ.

Q7. Evaluate the integral ∫[0 to ∞] cos(x) / (x² + a²) dx by the method of Residues. (a>0) 10

Ans. We want to evaluate the integral I = ∫[0 to ∞] cos(x) / (x² + a²) dx. Since the integrand is an even function (because cos(-x) = cos(x) and (-x)² = x²), we can write: ∫[-∞ to ∞] cos(x) / (x² + a²) dx = 2I So, I = (1/2) ∫[-∞ to ∞] cos(x) / (x² + a²) dx. We evaluate this integral using contour integration. We consider the integral as the real part of: ∫[-∞ to ∞] cos(x) / (x² + a²) dx = Re [ ∫[-∞ to ∞] e^(ix) / (x² + a²) dx ] Now consider the complex function f(z) = e^(iz) / (z² + a²) and integrate it over a contour C consisting of a semicircle of radius R in the upper half-plane (C_R) and the segment of the real axis from -R to R. 1. Locating the Poles: The singularities are found by solving the denominator z² + a² = 0. z² = -a² ⇒ z = ±ia. Since we are considering a contour in the upper half-plane, only the pole at z = +ia lies inside the contour C (assuming a > 0). This is a simple pole. 2. Calculating the Residue at z = ia: The residue is given by Res(f, z₀) = lim[z→z₀] (z-z₀)f(z). Res(f, ia) = lim[z→ia] (z-ia) * e^(iz) / ((z-ia)(z+ia)) Res(f, ia) = lim[z→ia] e^(iz) / (z+ia) Res(f, ia) = e^(i(ia)) / (ia + ia) = e^(-a) / (2ia) 3. Applying the Residue Theorem: The integral around the closed contour C is: ∮_C f(z) dz = 2πi * (Sum of residues inside the contour) ∮_C f(z) dz = 2πi * [ e^(-a) / (2ia) ] = πe^(-a) / a 4. Evaluating the Contour Integral: The contour integral can be split into two parts: ∮_C f(z) dz = ∫[-R to R] f(x) dx + ∫[C_R] f(z) dz According to Jordan’s Lemma, if |g(z)| → 0 as |z| → ∞ in the upper half-plane, then ∫[C_R] g(z)e^(ikz) dz → 0 as R → ∞ (for k>0). In our case, g(z) = 1/(z² + a²) and |g(z)| ≈ 1/R² → 0 as R→∞. Therefore, the integral over the semicircular arc vanishes: lim[R→∞] ∫[C_R] e^(iz) / (z² + a²) dz = 0 Thus, as R → ∞: ∮_C f(z) dz = ∫[-∞ to ∞] e^(ix) / (x² + a²) dx 5. Final Result: Combining the results: ∫[-∞ to ∞] e^(ix) / (x² + a²) dx = πe^(-a) / a Now, we take the real part: ∫[-∞ to ∞] cos(x) / (x² + a²) dx = Re [ πe^(-a) / a ] = πe^(-a) / a The original integral I is: I = ∫[0 to ∞] cos(x) / (x² + a²) dx = (1/2) ∫[-∞ to ∞] cos(x) / (x² + a²) dx I = (1/2) * (πe^(-a) / a) Therefore, ∫[0 to ∞] cos(x) / (x² + a²) dx = πe^(-a) / (2a) .

Q8. (a) Solve the initial value problem: 7 y” + 5y’ + 4y = f(t) y(0)=0, y'(0)=0 with f(t) = {1, 0<t<1; 0, otherwise} using the methods of Laplace transforms. (b) Define equivalence classes of a group. 3

Ans. (a) Solution of the Initial Value Problem The given differential equation is: y” + 5y’ + 4y = f(t), with y(0)=0, y'(0)=0. The function f(t) can be written in terms of the Heaviside step function u(t): f(t) = 1 * [u(t) – u(t-1)] = u(t) – u(t-1). 1. Taking the Laplace Transform: Take the Laplace transform of both sides of the equation. Let L{y(t)} = Y(s). L{y”} = s²Y(s) – sy(0) – y'(0) = s²Y(s) L{y’} = sY(s) – y(0) = sY(s) L{f(t)} = L{u(t) – u(t-1)} = L{u(t)} – L{u(t-1)} = 1/s – e^(-s)/s = (1 – e^(-s))/s The transformed equation is: s²Y(s) + 5sY(s) + 4Y(s) = (1 – e^(-s))/s (s² + 5s + 4)Y(s) = (1 – e^(-s))/s (s+1)(s+4)Y(s) = (1 – e^(-s))/s Y(s) = (1 – e^(-s)) / (s(s+1)(s+4)) 2. Partial Fraction Decomposition: Let G(s) = 1 / (s(s+1)(s+4)). We decompose this into partial fractions: G(s) = A/s + B/(s+1) + C/(s+4) A = [1/((s+1)(s+4))] at s=0 = 1/4 B = [1/(s(s+4))] at s=-1 = 1/(-1*3) = -1/3 C = [1/(s(s+1))] at s=-4 = 1/(-4*-3) = 1/12 So, G(s) = (1/4)/s – (1/3)/(s+1) + (1/12)/(s+4) 3. Taking the Inverse Laplace Transform: First, find g(t) = L⁻¹{G(s)}: g(t) = (1/4)L⁻¹{1/s} – (1/3)L⁻¹{1/(s+1)} + (1/12)L⁻¹{1/(s+4)} g(t) = 1/4 – (1/3)e^(-t) + (1/12)e^(-4t) Now, Y(s) = G(s) – e^(-s)G(s). The inverse Laplace transform y(t) = L⁻¹{Y(s)} is: y(t) = L⁻¹{G(s)} – L⁻¹{e^(-s)G(s)} Using the second time-shifting theorem, L⁻¹{e^(-as)F(s)} = f(t-a)u(t-a). y(t) = g(t) – g(t-1)u(t-1) 4. Final Solution: The solution can be written piecewise:

• For 0 < t < 1: u(t-1) = 0, so

y(t) = g(t) =

1/4 – (1/3)e^(-t) + (1/12)e^(-4t)

• For t ≥ 1: u(t-1) = 1, so

y(t) = g(t) – g(t-1)

y(t) = (1/4 – (1/3)e^(-t) + (1/12)e^(-4t)) – (1/4 – (1/3)e^(-(t-1)) + (1/12)e^(-4(t-1)))

y(t) =

-(1/3)(e^(-t) – e

e^(-t)) + (1/12)(e^(-4t) – e⁴

e^(-4t))

y(t) =

(e-1)/3

e^(-t) – (e⁴-1)/12

e^(-4t)

(b) Equivalence Classes In mathematics, an equivalence relation is a binary relation on a set that is reflexive, symmetric, and transitive. Any such equivalence relation partitions the set into disjoint subsets called equivalence classes . A relation ~ on a set S is an equivalence relation if for all a, b, and c in S: 1. Reflexivity: a ~ a 2. Symmetry: if a ~ b, then b ~ a 3. Transitivity: if a ~ b and b ~ c, then a ~ c The equivalence class of an element a in S, denoted [a], is the set of all elements x that are equivalent to a: [a] = {x ∈ S | x ~ a}. In the context of Group Theory: There are two important examples of equivalence classes in group theory: 1. Conjugacy Classes: In a group G, ‘being conjugate’ is an equivalence relation. Two elements a, b ∈ G are called conjugate if there exists an element g in G such that b = g⁻¹ag . The equivalence classes resulting from this relation are called the conjugacy classes. The conjugacy class of a is Cl(a) = {g⁻¹ag | g ∈ G}. The group G is a disjoint union of its conjugacy classes. 2. Cosets: Let H be a subgroup of a group G. We can define a relation ~ on G by a ~ b if a⁻¹b ∈ H. This is an equivalence relation. The equivalence classes of this relation are the left cosets of H in G. The equivalence class [a] containing element a is the set {x ∈ G | x ~ a} = {x ∈ G | a⁻¹x ∈ H} = {ah | h ∈ H} = aH. Similarly, right cosets arise from another equivalence relation (ab⁻¹ ∈ H).