Q.1 (a) [03 Marks]

Exact First Order Differential Equations

Definition

A first-order differential equation of the form:

M(x, y) dx + N(x, y) dy = 0

is called an exact differential equation if there exists a scalar function u(x, y) such that:

du = M dx + N dy

Condition for Exactness

The necessary and sufficient condition for the equation to be exact is:

∂M/∂y = ∂N/∂x

where ∂ represents the partial derivative.

Standard Solution Procedure

Once an equation is verified to be exact, its general solution is given by:

∫ M dx (treating y as constant) + ∫ (terms of N not containing x) dy = C

where C is an arbitrary integration constant.

Worked Example (3 Marks)

Problem: Solve the differential equation:

(2xy + y³) dx + (x² + 3xy² - 1) dy = 0

1. Identify M and N:

• M(x, y) = 2xy + y³
• N(x, y) = x² + 3xy² - 1

1. Check for Exactness:

• ∂M/∂y = 2x + 3y²
• ∂N/∂x = 2x + 3y²
• Since ∂M/∂y = ∂N/∂x, the differential equation is exact.

1. Integrate M with respect to x (keeping y constant):

• ∫ (2xy + y³) dx = 2y (x²/2) + y³x = x²y + xy³

1. Integrate terms of N free from x with respect to y:

• Terms in N without x: -1
• ∫ (-1) dy = -y

1. Write the General Solution:

• x²y + xy³ - y = C

Final Answer: x²y + xy³ - y = C

Q.1 (b) [04 Marks]

First Order Linear Differential Equations

Standard Form and Integrating Factor

A first-order ordinary differential equation is linear when it can be expressed in the canonical form:

dy/dx + P(x) y = Q(x)

where P(x) and Q(x) are continuous functions of x alone (or constants).

Solution Steps

1. Identify P(x) and Q(x) from the standard form.
2. Compute the Integrating Factor (I.F.):

I.F. = e^(∫ P(x) dx)

1. Multiply both sides by the integrating factor. The left side becomes d/dx [y · (I.F.)].
2. Integrate both sides with respect to x:

y · (I.F.) = ∫ [Q(x) · (I.F.)] dx + C

Worked Example (4 Marks)

Problem: Solve the linear differential equation:

dy/dx + 2y = e^(-x)

1. Identify Functions:

• P(x) = 2
• Q(x) = e^(-x)

1. Calculate Integrating Factor:

• ∫ P(x) dx = ∫ 2 dx = 2x
• I.F. = e^(2x)

1. Set Up the General Solution Formula:

• y · e^(2x) = ∫ [e^(-x) · e^(2x)] dx + C
• y · e^(2x) = ∫ e^x dx + C

1. Evaluate the Integral:

• y · e^(2x) = e^x + C

1. Express y Explicitly:

• y = e^(-x) + C · e^(-2x)

Final Answer: y = e^(-x) + C e^(-2x)

Q.1 (c) kindly ignore the square box mentioned in upper limit and consider c as curve which is mentioned in lower limit [07 Marks]

Evaluation of Line Integrals along a Curve C

Context of the Question

The note clarifies that the integral symbol represents a contour/line integral:

∫_C F · dr

where C is a specified curve in the plane or space, rather than a standard definite integral with an upper numerical limit.

Definition of Line Integral

Let F(x, y) = M(x, y) i + N(x, y) j be a continuous vector field, and let C be a smooth curve parameterized by:

r(t) = x(t) i + y(t) j, where a ≤ t ≤ b

The line integral of F along C is defined as:

∫C F · dr = ∫C (M dx + N dy) = ∫_a^b [M(x(t), y(t)) (dx/dt) + N(x(t), y(t)) (dy/dt)] dt

---

Worked 7-Mark Example

Problem: Evaluate the line integral:

I = ∫_C [(x² - y²) dx + 2xy dy]

along the curve C given by the parabola y = x² from the point (0, 0) to (1, 1).

Step 1: Parameterize the Curve C

Let x = t. Since y = x², we have:

• x(t) = t
• y(t) = t²

Differential elements:

• dx = dt
• dy = 2t dt

Limits of parameter t:

• At (0, 0): t = 0
• At (1, 1): t = 1
• Therefore, t varies from 0 to 1.

Step 2: Substitute Parametric Expressions into the Integrand

• M(x, y) dx = (x² - y²) dx = (t² - (t²)²) dt = (t² - t⁴) dt
• N(x, y) dy = (2xy) dy = 2(t)(t²) (2t dt) = 4t⁴ dt

Summing the two terms:

• F · dr = (t² - t⁴) dt + 4t⁴ dt = (t² + 3t⁴) dt

Step 3: Integrate with Respect to t

I = ∫_0^1 (t² + 3t⁴) dt

Compute each antiderivative:

• ∫ t² dt = t³ / 3
• ∫ 3t⁴ dt = 3t⁵ / 5

Evaluating between t = 0 and t = 1: I = [t³ / 3 + 3t⁵ / 5]_0^1 I = (1/3 + 3/5) - (0 + 0) I = (5 + 9) / 15 = 14/15

Step 4: Verification using Path Independence Test

Let us check if F = (x² - y²) i + 2xy j is conservative:

• ∂M/∂y = ∂/∂y (x² - y²) = -2y
• ∂N/∂x = ∂/∂x (2xy) = 2y

Since ∂M/∂y ≠ ∂N/∂x (-2y ≠ 2y), the field is non-conservative, confirming that the value depends on the path y = x² and must be evaluated directly via parameterization.

Final Answer: The value of the line integral along curve C is 14/15.

Q.2 (a) [03 Marks]

Second-Order Linear Homogeneous ODEs with Constant Coefficients

General Form

A second-order homogeneous linear differential equation with constant coefficients has the form:

a (d²y/dx²) + b (dy/dx) + c y = 0

where a, b, and c are real constants and a ≠ 0.

Characteristic (Auxiliary) Equation

Substituting the trial solution y = e^(mx) yields the characteristic equation:

a m² + b m + c = 0

Let the roots of this quadratic equation be m₁ and m₂.

Three Cases of Complementary Functions

1. Real and Distinct Roots (m₁ ≠ m₂):
   y = C₁ e^(m₁x) + C₂ e^(m₂x)
2. Real and Equal Roots (m₁ = m₂ = m):
   y = (C₁ + C₂ x) e^(mx)
3. Complex Conjugate Roots (m = α ± iβ):
   y = e^(αx) [C₁ cos(βx) + C₂ sin(βx)]

Quick Example (3 Marks)

Solve: d²y/dx² - 5(dy/dx) + 6y = 0

• Auxiliary Equation: m² - 5m + 6 = 0
• Factoring: (m - 2)(m - 3) = 0 => m₁ = 2, m₂ = 3
• General Solution: y = C₁ e^(2x) + C₂ e^(3x)

Q.2 (b) [04 Marks]

Finding the Particular Integral using Operator D Method

Concept of Differential Operator D

Let D = d/dx, D² = d²/dx². A non-homogeneous equation is written as:

f(D) y = X(x)

The complete solution is:

y = yc + yp

where yc is the complementary function and yp = [1 / f(D)] X(x) is the particular integral.

Standard Rule for Exponential Functions

If X(x) = e^(ax), then:

y_p = [1 / f(D)] e^(ax) = [1 / f(a)] e^(ax), provided f(a) ≠ 0

If f(a) = 0 (case of failure), then:

y_p = x · [1 / f'(D)] e^(ax)

Worked Example (4 Marks)

Problem: Find the particular integral of:

(D² - 4D + 3) y = 2 e^(3x)

1. Identify f(D) and a:

• f(D) = D² - 4D + 3
• X(x) = 2 e^(3x), so a = 3

1. Test for Case of Failure:

• f(3) = 3² - 4(3) + 3 = 9 - 12 + 3 = 0 (Case of failure occurs)

1. Differentiate f(D) with Respect to D:

• f'(D) = d/dD (D² - 4D + 3) = 2D - 4

1. Apply Failure Rule:

• y_p = x · [1 / (2D - 4)] [2 e^(3x)]
• Substitute D = 3 into f'(D):
  f'(3) = 2(3) - 4 = 6 - 4 = 2
• y_p = x · [2 e^(3x) / 2] = x e^(3x)

Final Answer: y_p = x e^(3x)

Q.2 (c) [07 Marks]

Method of Variation of Parameters

Objective

The Method of Variation of Parameters is a general technique used to find the particular solution of second-order non-homogeneous linear differential equations:

y'' + P(x) y' + Q(x) y = R(x)

especially when R(x) is tan(x), sec(x), cosec(x), or 1/x, where the standard operator D method cannot be easily applied.

Method and Formulas

1. Find the complementary function:
   y_c = C₁ y₁ + C₂ y₂
   where y₁ and y₂ are two linearly independent solutions of the homogeneous equation.

1. Compute the Wronskian W of y₁ and y₂:
   W(y₁, y₂) = y₁ y₂' - y₁' y₂
   Since y₁ and y₂ are linearly independent, W ≠ 0.

1. The particular integral is assumed as:
   y_p = u(x) y₁ + v(x) y₂
   where u(x) and v(x) are determined by:

• u(x) = - ∫ [y₂ R(x) / W] dx
• v(x) = ∫ [y₁ R(x) / W] dx

1. The complete general solution is:
   y = yc + yp

---

Worked 7-Mark Problem

Solve: y'' + y = sec(x)

Step 1: Find the Complementary Function (y_c)

Homogeneous equation: y'' + y = 0

• Auxiliary equation: m² + 1 = 0 => m = ± i
• y_c = C₁ cos(x) + C₂ sin(x)
• Here, y₁ = cos(x) and y₂ = sin(x).

Step 2: Calculate the Wronskian (W)

W = y₁ y₂' - y₁' y₂

• y₁' = -sin(x)
• y₂' = cos(x)
• W = cos(x) [cos(x)] - [-sin(x)] [sin(x)]
• W = cos²(x) + sin²(x) = 1

Step 3: Compute u(x)

Here, R(x) = sec(x). u(x) = - ∫ [y₂ R(x) / W] dx u(x) = - ∫ [sin(x) · sec(x) / 1] dx Since sec(x) = 1/cos(x): u(x) = - ∫ [sin(x) / cos(x)] dx = - ∫ tan(x) dx u(x) = - ln|sec(x)| = ln|cos(x)|

Step 4: Compute v(x)

v(x) = ∫ [y₁ R(x) / W] dx v(x) = ∫ [cos(x) · sec(x) / 1] dx Since cos(x) · sec(x) = 1: v(x) = ∫ 1 dx = x

Step 5: Form the Particular Integral (y_p)

yp = u(x) y₁ + v(x) y₂ yp = [ln|cos(x)|] cos(x) + x sin(x)

Step 6: Write Complete General Solution

y = C₁ cos(x) + C₂ sin(x) + cos(x) ln|cos(x)| + x sin(x)

Final Answer: y = C₁ cos(x) + C₂ sin(x) + cos(x) ln|cos(x)| + x sin(x)

Q.2 (c) OR [07 Marks]

Cauchy-Euler Linear Differential Equations

Definition

A Cauchy-Euler equation of order 2 has the form:

x² (d²y/dx²) + a₁ x (dy/dx) + a₂ y = f(x)

where the power of x matches the order of each derivative.

Transformation to Constant Coefficients

Substitute:

• x = e^z or z = ln(x)
• Let θ = d/dz

Then derivatives transform as:

• x (dy/dx) = θ y
• x² (d²y/dx²) = θ(θ - 1) y

Substituting these converts the equation into a linear ODE with constant coefficients in independent variable z.

---

Worked 7-Mark Problem

Solve: x² (d²y/dx²) - 2x (dy/dx) - 4y = x²

Step 1: Substitute Transformations

Let x = e^z, so z = ln(x) and θ = d/dz.

• x (dy/dx) = θ y
• x² (d²y/dx²) = θ(θ - 1) y
• Right-hand side: x² = (e^z)² = e^(2z)

Substitute into the ODE: [θ(θ - 1) - 2θ - 4] y = e^(2z) (θ² - θ - 2θ - 4) y = e^(2z) (θ² - 3θ - 4) y = e^(2z)

Step 2: Find Complementary Function (y_c)

Auxiliary equation: m² - 3m - 4 = 0 (m - 4)(m + 1) = 0 => m₁ = 4, m₂ = -1

y_c = C₁ e^(4z) + C₂ e^(-z)

Step 3: Find Particular Integral (y_p)

y_p = [1 / (θ² - 3θ - 4)] e^(2z)

Substitute θ = 2: Denominator = 2² - 3(2) - 4 = 4 - 6 - 4 = -6 ≠ 0

y_p = e^(2z) / (-6) = - (1/6) e^(2z)

Step 4: Combine Solutions in Variable z

y(z) = C₁ e^(4z) + C₂ e^(-z) - (1/6) e^(2z)

Step 5: Convert Back to Variable x

Since e^z = x, e^(4z) = x⁴, e^(-z) = x^(-1), and e^(2z) = x²:

y(x) = C₁ x⁴ + C₂ x^(-1) - (1/6) x²

Final Answer: y = C₁ x⁴ + C₂ / x - (1/6) x²

Q.3 (a) [03 Marks]

Classification of Singular Points of Second Order ODEs

Definition

Consider the homogeneous linear ODE in standard form:

y'' + P(x) y' + Q(x) y = 0

1. Ordinary Point:
   A point x = x₀ is an ordinary point if both P(x) and Q(x) are analytic at x₀ (i.e., their Taylor series exist around x₀).

1. Singular Point:
   If either P(x) or Q(x) (or both) are not analytic at x = x₀, then x = x₀ is called a singular point.

1. Regular Singular Point:
   A singular point x = x₀ is a regular singular point if:

• lim_(x -> x₀) (x - x₀) P(x) is finite
• lim_(x -> x₀) (x - x₀)² Q(x) is finite

1. Irregular Singular Point:
   If either limit does not exist or is infinite, x = x₀ is an irregular singular point.

Example (3 Marks)

For the equation: x² y'' + x y' + (x² - 4) y = 0

• P(x) = 1/x, Q(x) = (x² - 4)/x²
• At x = 0, both P(x) and Q(x) are undefined; hence x = 0 is a singular point.
• Check regular condition:
• lim_(x -> 0) [x · (1/x)] = 1 (finite)
• lim_(x -> 0) [x² · ((x² - 4)/x²)] = -4 (finite)
• Therefore, x = 0 is a regular singular point.

Q.3 (b) [04 Marks]

Power Series Solution of ODEs around an Ordinary Point

Method Overview

When x = 0 is an ordinary point of y'' + P(x) y' + Q(x) y = 0, the solution can be expressed as a power series:

y = ∑(n=0)^∞ an x^n = a₀ + a₁ x + a₂ x² + a₃ x³ + ...

Worked Example (4 Marks)

Solve: y'' + y = 0 near x = 0

1. Differentiate Term by Term:

• y = ∑(n=0)^∞ an x^n
• y' = ∑(n=1)^∞ n an x^(n-1)
• y'' = ∑(n=2)^∞ n(n-1) an x^(n-2)

1. Substitute into ODE:
   ∑(n=2)^∞ n(n-1) an x^(n-2) + ∑(n=0)^∞ an x^n = 0

1. Shift Indices:
   Let k = n - 2 in the first sum (so n = k + 2):
   ∑(k=0)^∞ (k+2)(k+1) a(k+2) x^k + ∑(k=0)^∞ ak x^k = 0

1. Obtain Recurrence Relation:
   (k+2)(k+1) a(k+2) + ak = 0
   a(k+2) = - ak / [(k+1)(k+2)]

1. Compute Coefficients:

• For k = 0: a₂ = -a₀ / (1 · 2) = -a₀ / 2!
• For k = 1: a₃ = -a₁ / (2 · 3) = -a₁ / 3!
• For k = 2: a₄ = -a₂ / (3 · 4) = a₀ / 4!
• For k = 3: a₅ = -a₃ / (4 · 5) = a₁ / 5!

1. Group Terms:
   y = a₀ [1 - x²/2! + x⁴/4! - ...] + a₁ [x - x³/3! + x⁵/5! - ...]
   y = a₀ cos(x) + a₁ sin(x)

Final Answer: y = a₀ cos(x) + a₁ sin(x)

Q.3 (c) [07 Marks]

Legendre's Differential Equation and Legendre Polynomials

Definition of Legendre's Equation

Legendre's differential equation of order n is:

(1 - x²) (d²y/dx²) - 2x (dy/dx) + n(n + 1) y = 0

where n is a non-negative integer. Its solutions bounded on [-1, 1] are called Legendre Polynomials, denoted by P_n(x).

Rodrigues' Formula

Rodrigues' formula provides an explicit representation for P_n(x):

P_n(x) = [1 / (2^n · n!)] · (d^n/dx^n) [(x² - 1)^n]

Derivation of First Four Legendre Polynomials

1. For n = 0:
   P₀(x) = [1 / (2⁰ · 0!)] · (x² - 1)⁰ = 1

1. For n = 1:
   P₁(x) = [1 / (2¹ · 1!)] · (d/dx) [x² - 1] = (1/2)(2x) = x

1. For n = 2:
   P₂(x) = [1 / (2² · 2!)] · (d²/dx²) [(x² - 1)²]

• (x² - 1)² = x⁴ - 2x² + 1
• First derivative: 4x³ - 4x
• Second derivative: 12x² - 4
• P₂(x) = (1/8)(12x² - 4) = (3x² - 1) / 2

1. For n = 3:
   P₃(x) = [1 / (2³ · 3!)] · (d³/dx³) [(x² - 1)³]

• (x² - 1)³ = x⁶ - 3x⁴ + 3x² - 1
• 1st derivative: 6x⁵ - 12x³ + 6x
• 2nd derivative: 30x⁴ - 36x² + 6
• 3rd derivative: 120x³ - 72x
• P₃(x) = [1 / 48] (120x³ - 72x) = (5x³ - 3x) / 2

Generating Function for Legendre Polynomials

The generating function is defined as:

(1 - 2xt + t²)^(-1/2) = ∑(n=0)^∞ Pn(x) t^n

Key Properties

1. P_n(1) = 1 for all n
2. P_n(-1) = (-1)^n
3. Pn(-x) = (-1)^n Pn(x) (Even function for even n, odd for odd n)

Final Summary Table:

• P₀(x) = 1
• P₁(x) = x
• P₂(x) = (3x² - 1) / 2
• P₃(x) = (5x³ - 3x) / 2

Q.3 (a) OR [03 Marks]

Bessel's Differential Equation and Bessel Functions

Definition

Bessel's differential equation of order n is:

x² (d²y/dx²) + x (dy/dx) + (x² - n²) y = 0

where n is a non-negative constant.

Bessel Function of First Kind, J_n(x)

The solution bounded at x = 0 is called the Bessel function of the first kind of order n, defined by the power series:

Jn(x) = ∑(r=0)^∞ [(-1)^r / (r! · Γ(n + r + 1))] · (x / 2)^(n + 2r)

where Γ denotes the Gamma function.

For n = 0 (Bessel function of order zero):

J₀(x) = 1 - (x²/4) + (x⁴/64) - ... = ∑_(r=0)^∞ [(-1)^r / (r!)²] (x/2)^(2r)

Key Values

• J₀(0) = 1
• J_n(0) = 0 for n > 0

Q.3 (b) OR [04 Marks]

Recurrence Relations for Bessel Functions

Standard Recurrence Relations

Bessel functions satisfy several fundamental recurrence relations:

1. d/dx [x^n Jn(x)] = x^n J(n-1)(x)
2. d/dx [x^(-n) Jn(x)] = - x^(-n) J(n+1)(x)
3. J(n-1)(x) + J(n+1)(x) = (2n / x) J_n(x)
4. J(n-1)(x) - J(n+1)(x) = 2 J_n'(x)

Derivation of Recurrence Relation (3)

From relation (1): x^n Jn'(x) + n x^(n-1) Jn(x) = x^n J(n-1)(x) Dividing by x^n: Jn'(x) + (n/x) Jn(x) = J(n-1)(x) --- [Equation A]

From relation (2): x^(-n) Jn'(x) - n x^(-n-1) Jn(x) = - x^(-n) J(n+1)(x) Dividing by x^(-n): Jn'(x) - (n/x) Jn(x) = - J(n+1)(x) --- [Equation B]

Subtract Equation B from Equation A: [Jn'(x) + (n/x) Jn(x)] - [Jn'(x) - (n/x) Jn(x)] = J(n-1)(x) - [-J(n+1)(x)]

2 (n/x) Jn(x) = J(n-1)(x) + J_(n+1)(x)

Therefore:

J(n-1)(x) + J(n+1)(x) = (2n / x) J_n(x)

Conclusion: This fundamental recurrence relation allows computing higher-order Bessel functions from lower-order ones.

Q.3 (c) OR [07 Marks]

Orthogonality Property of Legendre Polynomials

Statement of Theorem

The Legendre polynomials Pm(x) and Pn(x) are orthogonal on the interval [-1, 1]:

∫(-1)^1 Pm(x) P_n(x) dx = 0, if m ≠ n

and for m = n:

∫(-1)^1 [Pn(x)]² dx = 2 / (2n + 1)

---

Proof for m ≠ n (7 Marks)

Step 1: Write Legendre Equations for m and n

Legendre's equation for Pn(x) is: (1 - x²) Pn''(x) - 2x Pn'(x) + n(n + 1) Pn(x) = 0 This can be rewritten in self-adjoint form: d/dx [(1 - x²) Pn'(x)] + n(n + 1) Pn(x) = 0 --- (Equation 1)

Similarly, for Pm(x): d/dx [(1 - x²) Pm'(x)] + m(m + 1) P_m(x) = 0 --- (Equation 2)

Step 2: Multiply and Subtract

Multiply Equation 1 by Pm(x) and Equation 2 by Pn(x): Pm d/dx [(1 - x²) Pn'] + n(n + 1) Pm Pn = 0 --- (3) Pn d/dx [(1 - x²) Pm'] + m(m + 1) Pm Pn = 0 --- (4)

Subtract Equation 4 from Equation 3: Pm d/dx [(1 - x²) Pn'] - Pn d/dx [(1 - x²) Pm'] + [n(n + 1) - m(m + 1)] Pm Pn = 0

Notice that: n(n + 1) - m(m + 1) = n² - m² + n - m = (n - m)(n + m) + (n - m) = (n - m)(n + m + 1)

Step 3: Integrate over [-1, 1]

∫(-1)^1 d/dx [(1 - x²) (Pm Pn' - Pn Pm')] dx + (n - m)(n + m + 1) ∫(-1)^1 Pm Pn dx = 0

Evaluate the first integral using the Fundamental Theorem of Calculus: [(1 - x²) (Pm Pn' - Pn Pm')] evaluated from -1 to 1

At x = 1: (1 - 1²) = 0 At x = -1: (1 - (-1)²) = 0

Thus, the boundary term vanishes identically: 0 + (n - m)(n + m + 1) ∫(-1)^1 Pm(x) P_n(x) dx = 0

Step 4: Conclusion

Since m ≠ n, (n - m) ≠ 0 and (n + m + 1) ≠ 0. Therefore:

∫(-1)^1 Pm(x) P_n(x) dx = 0

This completes the formal proof of orthogonality.

Q.4 (a) [03 Marks]

Fourier Series and Dirichlet's Conditions

Definition of Fourier Series

A periodic function f(x) with period 2π defined on [-π, π] can be represented as an infinite trigonometric series:

f(x) = a₀/2 + ∑(n=1)^∞ [an cos(nx) + b_n sin(nx)]

where the Euler-Fourier coefficients are:

• a₀ = (1/π) ∫_(-π)^π f(x) dx
• an = (1/π) ∫(-π)^π f(x) cos(nx) dx
• bn = (1/π) ∫(-π)^π f(x) sin(nx) dx

Dirichlet's Conditions (3 Marks)

For the Fourier series of f(x) to converge to f(x) (or to [f(x+) + f(x-)]/2 at points of discontinuity), f(x) must satisfy Dirichlet's conditions:

1. f(x) is single-valued and periodic.
2. f(x) is bounded and has at most a finite number of finite discontinuities in any single period.
3. f(x) has at most a finite number of local maxima and minima in any single period.

Q.4 (b) [04 Marks]

Half-Range Fourier Series

Concept

When a function f(x) is defined only over a half-interval [0, L], it can be expanded into either a series containing only cosine terms (even extension) or only sine terms (odd extension).

Half-Range Cosine Series

f(x) = a₀/2 + ∑(n=1)^∞ an cos(nπx / L)

where:

• a₀ = (2/L) ∫_0^L f(x) dx
• an = (2/L) ∫0^L f(x) cos(nπx / L) dx

Half-Range Sine Series

f(x) = ∑(n=1)^∞ bn sin(nπx / L)

where:

• bn = (2/L) ∫0^L f(x) sin(nπx / L) dx

Worked Example (4 Marks)

Problem: Find the half-range sine series for f(x) = 1 in [0, π].

1. Here L = π.
2. Calculate b_n:
   bn = (2/π) ∫0^π 1 · sin(nx) dx
   bn = (2/π) [-cos(nx) / n]0^π
   b_n = (2 / nπ) [1 - cos(nπ)] = (2 / nπ) [1 - (-1)^n]

1. Evaluate for even and odd n:

• When n is even: 1 - 1 = 0 => b_n = 0
• When n is odd: 1 - (-1) = 2 => b_n = 4 / (nπ)

1. Write the Series:
   f(x) = (4/π) [sin(x) + sin(3x)/3 + sin(5x)/5 + ...]

Final Answer: f(x) = (4/π) ∑_(k=1)^∞ [sin((2k-1)x) / (2k-1)]

Q.4 (c) [07 Marks]

Full Fourier Series Expansion and Deductions

Problem (7 Marks)

Find the Fourier series of f(x) = x² in the interval -π ≤ x ≤ π, and hence deduce that:

1/1² - 1/2² + 1/3² - 1/4² + ... = π² / 12

---

Step 1: Symmetry of the Function

f(-x) = (-x)² = x² = f(x) Since f(x) is an even function:

• b_n = 0 for all n ≥ 1.
• Only cosine terms survive.

Step 2: Compute Coefficient a₀

a₀ = (2/π) ∫0^π x² dx a₀ = (2/π) [x³ / 3]0^π = (2/π) (π³ / 3) = 2π² / 3

Therefore: a₀ / 2 = π² / 3

Step 3: Compute Coefficients a_n

an = (2/π) ∫0^π x² cos(nx) dx

Using integration by parts (generalized formula ∫ u v = u v₁ - u' v₂ + u'' v₃):

• u = x² => u' = 2x => u'' = 2 => u''' = 0
• v = cos(nx) => v₁ = sin(nx)/n => v₂ = -cos(nx)/n² => v₃ = -sin(nx)/n³

Evaluating from 0 to π: an = (2/π) [ x² sin(nx)/n - 2x (-cos(nx)/n²) + 2 (-sin(nx)/n³) ]0^π

Since sin(nπ) = sin(0) = 0: a_n = (2/π) [ 2π cos(nπ) / n² ] = (4/n²) cos(nπ) = 4(-1)^n / n²

Step 4: Write the Fourier Series

f(x) = π²/3 + ∑_(n=1)^∞ [4(-1)^n / n²] cos(nx) f(x) = π²/3 - 4 [cos(x)/1² - cos(2x)/2² + cos(3x)/3² - cos(4x)/4² + ...]

Step 5: Deduction of the Series

Substitute x = 0 into the Fourier series: At x = 0, f(0) = 0² = 0, and cos(0) = 1 for all n:

0 = π²/3 - 4 [1/1² - 1/2² + 1/3² - 1/4² + ...]

4 [1/1² - 1/2² + 1/3² - 1/4² + ...] = π²/3

Divide both sides by 4: 1/1² - 1/2² + 1/3² - 1/4² + ... = π² / 12

Final Answer: Fourier series: f(x) = π²/3 + 4 ∑_(n=1)^∞ [(-1)^n / n²] cos(nx) Deduction: 1/1² - 1/2² + 1/3² - 1/4² + ... = π²/12

Q.4 (a) OR [03 Marks]

Formation of Partial Differential Equations (PDE)

Methods of Formation

A partial differential equation can be formed either by:

1. Eliminating arbitrary constants.
2. Eliminating arbitrary functions.

Standard Notation

For a dependent variable z = z(x, y):

• p = ∂z/∂x
• q = ∂z/∂y

Worked Example (3 Marks)

Form a PDE by eliminating arbitrary constants a and b from:

z = a x + b y + a² + b²

1. Differentiate partially with respect to x:
   ∂z/∂x = a => p = a

1. Differentiate partially with respect to y:
   ∂z/∂y = b => q = b

1. Substitute a = p and b = q into the original equation:
   z = p x + q y + p² + q²

Final Answer: z = px + qy + p² + q² (Clairaut's form PDE)

Q.4 (b) OR [04 Marks]

Lagrange's Linear Partial Differential Equations

Standard Form

A quasi-linear first-order partial differential equation of the form:

P(x, y, z) p + Q(x, y, z) q = R(x, y, z)

is known as Lagrange's Linear PDE, where p = ∂z/∂x and q = ∂z/∂y.

Method of Solution: Lagrange's Auxiliary Equations

The corresponding subsidiary (auxiliary) equations are:

dx / P = dy / Q = dz / R

Two independent solutions u(x, y, z) = c₁ and v(x, y, z) = c₂ are obtained using the method of grouping or the method of multipliers.

The general solution is expressed as:

Φ(u, v) = 0 or u = f(v)

Worked Example (4 Marks)

Solve: y z p + z x q = x y

1. Write Auxiliary Equations:
   dx / (yz) = dy / (zx) = dz / (xy)

1. First Independent Solution (from 1st and 2nd ratios):
   dx / (yz) = dy / (zx) => x dx = y dy
   Integrating: x² - y² = c₁

1. Second Independent Solution (from 2nd and 3rd ratios):
   dy / (zx) = dz / (xy) => y dy = z dz
   Integrating: y² - z² = c₂

1. General Solution:
   Φ(x² - y², y² - z²) = 0

Final Answer: Φ(x² - y², y² - z²) = 0

Q.4 (c) OR [07 Marks]

Method of Separation of Variables for Partial Differential Equations

Principle of the Method

The Method of Separation of Variables assumes that the solution to a linear PDE can be written as a product of single-variable functions:

u(x, t) = X(x) · T(t)

Substituting this form converts the PDE into a system of independent ordinary differential equations.

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Worked 7-Mark Problem

Problem: Solve the one-dimensional heat conduction equation:

∂u/∂t = c² (∂²u/∂x²), for 0 < x < L, t > 0

subject to boundary conditions: u(0, t) = 0, u(L, t) = 0 for all t > 0.

Step 1: Assume Product Solution

Let u(x, t) = X(x) T(t).

• ∂u/∂t = X(x) T'(t)
• ∂²u/∂x² = X''(x) T(t)

Substitute into the PDE: X(x) T'(t) = c² X''(x) T(t)

Divide both sides by c² X(x) T(t): T'(t) / [c² T(t)] = X''(x) / X(x) = -λ² (separation constant)

(A negative separation constant -λ² is chosen so that solutions do not grow unbounded as t -> ∞).

Step 2: Set Up Ordinary Differential Equations

1. X''(x) + λ² X(x) = 0
2. T'(t) + λ² c² T(t) = 0

Step 3: Solve the Spatial Equation X(x)

The general solution of X'' + λ² X = 0 is: X(x) = C₁ cos(λx) + C₂ sin(λx)

Apply boundary conditions:

1. u(0, t) = X(0) T(t) = 0 => X(0) = 0
   X(0) = C₁ cos(0) + C₂ sin(0) = C₁ = 0
   Therefore, X(x) = C₂ sin(λx).

1. u(L, t) = X(L) T(t) = 0 => X(L) = 0
   C₂ sin(λL) = 0
   For non-trivial solutions (C₂ ≠ 0):
   sin(λL) = 0 => λL = nπ => λ_n = nπ / L, n = 1, 2, 3, ...

Eigenfunctions: X_n(x) = sin(nπx / L)

Step 4: Solve the Temporal Equation T(t)

T'(t) + (nπc / L)² T(t) = 0

Integrating: Tn(t) = Bn e^(- (nπc / L)² t)

Step 5: Superposition of Solutions

By the principle of linear superposition:

u(x, t) = ∑(n=1)^∞ An sin(nπx / L) e^(- (nπc / L)² t)

where coefficients A_n are determined by the initial condition u(x, 0) = f(x) using Fourier sine coefficients:

An = (2/L) ∫0^L f(x) sin(nπx / L) dx

Final Answer: u(x, t) = ∑(n=1)^∞ An sin(nπx / L) e^(- (nπc / L)² t)

Q.5 (a) [03 Marks]

Gradient of a Scalar Field and Directional Derivative

Gradient

Let φ(x, y, z) be a differentiable scalar field. The gradient of φ, denoted by grad φ or ∇φ, is the vector:

∇φ = (∂φ/∂x) i + (∂φ/∂y) j + (∂φ/∂z) k

Physical Meaning: ∇φ represents a vector pointing in the direction of maximum rate of increase of φ, and its magnitude |∇φ| gives the maximum rate of increase.

Directional Derivative

The directional derivative of φ in the direction of a unit vector u is:

D_u φ = ∇φ · u

Worked Example (3 Marks)

Find the directional derivative of φ = x²y + yz at point P(1, 2, 3) in the direction of vector a = 2i + j + 2k.

1. Compute Gradient ∇φ:

• ∂φ/∂x = 2xy
• ∂φ/∂y = x² + z
• ∂φ/∂z = y
• At (1, 2, 3): ∇φ = 2(1)(2) i + (1² + 3) j + 2 k = 4i + 4j + 2k

1. Find Unit Vector u:

• |a| = √(2² + 1² + 2²) = √(4 + 1 + 4) = √9 = 3
• u = (2i + j + 2k) / 3

1. Compute Dot Product:

• D_u φ = ∇φ · u = (4 · 2 + 4 · 1 + 2 · 2) / 3 = (8 + 4 + 4) / 3 = 16/3

Final Answer: Directional derivative = 16/3

Q.5 (b) [04 Marks]

Divergence and Curl of a Vector Field

Definitions

Let F = F₁ i + F₂ j + F₃ k be a continuously differentiable vector field.

1. Divergence of F:
   div F = ∇ · F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z

• If div F = 0, F is called solenoidal.

1. Curl of F:
   curl F = ∇ × F =
   | i j k |
   | ∂/∂x ∂/∂y ∂/∂z |
   | F₁ F₂ F₃ |
   = (∂F₃/∂y - ∂F₂/∂z) i - (∂F₃/∂x - ∂F₁/∂z) j + (∂F₂/∂x - ∂F₁/∂y) k

• If curl F = 0, F is called irrotational.

Worked Example (4 Marks)

Problem: Determine whether F = (y + z) i + (x + z) j + (x + y) k is solenoidal and irrotational.

1. Check Solenoidal Nature (div F):

• div F = ∂/∂x(y + z) + ∂/∂y(x + z) + ∂/∂z(x + y) = 0 + 0 + 0 = 0
• Since div F = 0, F is solenoidal.

1. Check Irrotational Nature (curl F):

• i-component: ∂/∂y(x + y) - ∂/∂z(x + z) = 1 - 1 = 0
• j-component: - [∂/∂x(x + y) - ∂/∂z(y + z)] = - [1 - 1] = 0
• k-component: ∂/∂x(x + z) - ∂/∂y(y + z) = 1 - 1 = 0
• curl F = 0
• Since curl F = 0, F is irrotational.

Final Answer: The vector field F is both solenoidal and irrotational.

Q.5 (c) [07 Marks]

Gauss Divergence Theorem

Statement

The Gauss Divergence Theorem states that the outward flux of a vector field F through a closed surface S bounding a volume V is equal to the volume integral of the divergence of F over V:

∬S (F · n) dS = ∭V (∇ · F) dV

where n is the outward-pointing unit normal vector to surface S.

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Worked 7-Mark Problem

Problem: Verify Gauss Divergence Theorem for:

F = 4x i - 2y² j + z² k

over the region V bounded by the cylinder x² + y² = 4 and the planes z = 0 and z = 3.

Step 1: Evaluate the Volume Integral (RHS)

Compute ∇ · F: ∇ · F = ∂/∂x(4x) + ∂/∂y(-2y²) + ∂/∂z(z²) ∇ · F = 4 - 4y + 2z

Volume integral: IV = ∭V (4 - 4y + 2z) dV

Use cylindrical coordinates (x = r cos θ, y = r sin θ, z = z, dV = r dr dθ dz):

• r varies from 0 to 2 (since r² = 4)
• θ varies from 0 to 2π
• z varies from 0 to 3

IV = ∫0^(2π) ∫0^2 ∫0^3 (4 - 4r sin θ + 2z) r dz dr dθ

1. Integrate with respect to z:
   ∫0^3 (4 - 4r sin θ + 2z) dz = [(4 - 4r sin θ) z + z²]0^3
   = 3(4 - 4r sin θ) + 9 = 12 - 12r sin θ + 9 = 21 - 12r sin θ

1. Multiply by r and integrate with respect to r:
   ∫0^2 (21r - 12r² sin θ) dr = [21(r²/2) - 12(r³/3) sin θ]0^2
   = 21(2) - 4(8) sin θ = 42 - 32 sin θ

1. Integrate with respect to θ from 0 to 2π:
   ∫0^(2π) (42 - 32 sin θ) dθ = [42θ + 32 cos θ]0^(2π)
   = 42(2π) + 32(cos 2π - cos 0) = 84π + 32(1 - 1) = 84π

Thus, Volume Integral = 84π.

Step 2: Evaluate Surface Integral over Closed Surface S (LHS)

The surface S consists of three parts:

1. Top face S₁ (z = 3): n = k, dS = dx dy

• F · n = z² = 3² = 9
• ∬_(S₁) 9 dx dy = 9 · (Area of circle of radius 2) = 9(π · 2²) = 36π

1. Bottom face S₂ (z = 0): n = -k, dS = dx dy

• F · n = -z² = 0
• ∬_(S₂) 0 dS = 0

1. Curved surface S₃ (x² + y² = 4):

• Normal vector: ∇(x² + y²) = 2x i + 2y j
• Unit normal: n = (x i + y j) / 2
• F · n = [4x(x/2) - 2y²(y/2)] = 2x² - y³
• In cylindrical coordinates on r = 2: x = 2 cos θ, y = 2 sin θ, dS = 2 dθ dz
• F · n = 2(4 cos² θ) - 8 sin³ θ = 8 cos² θ - 8 sin³ θ
• ∬(S₃) (F · n) dS = ∫0^3 dz ∫_0^(2π) (8 cos² θ - 8 sin³ θ) · 2 dθ
• ∫_0^3 dz = 3
• ∫_0^(2π) 8 sin³ θ dθ = 0
• ∫_0^(2π) 8 cos² θ dθ = 8 · π = 8π
• ∬_(S₃) = 3 · 2 · 8π = 48π

Total surface flux: ∬S (F · n) dS = ∬(S₁) + ∬(S₂) + ∬(S₃) = 36π + 0 + 48π = 84π

Conclusion

Since ∬S (F · n) dS = ∭V (∇ · F) dV = 84π, the Gauss Divergence Theorem is verified.

Q.5 (a) OR [03 Marks]

Conservative Vector Fields and Scalar Potential

Definition

A vector field F is called conservative if there exists a single-valued scalar potential function φ such that:

F = ∇φ

Properties

1. Irrotational: curl F = ∇ × F = 0.
2. Path Independence: The line integral ∫_C F · dr between any two points A and B depends only on the endpoints, not on the path chosen:
   ∫_A^B F · dr = φ(B) - φ(A)
3. Closed Loop Integral: For any closed contour C, ∮_C F · dr = 0.

Example (3 Marks)

Show that F = 2x y i + x² j is conservative and find φ.

• curl F = (∂/∂x(x²) - ∂/∂y(2xy)) k = (2x - 2x) k = 0 (Conservative)
• ∂φ/∂x = 2xy => φ = x²y + g(y)
• ∂φ/∂y = x² + g'(y) = x² => g'(y) = 0 => g(y) = C
• Scalar potential: φ(x, y) = x²y + C

Q.5 (b) OR [04 Marks]

Green's Theorem in a Plane

Statement

Let C be a positively oriented, piecewise-smooth, simple closed curve in a plane, and let R be the region bounded by C. If M(x, y) and N(x, y) have continuous partial derivatives on an open region containing R, then:

∮C (M dx + N dy) = ∬R (∂N/∂x - ∂M/∂y) dx dy

Application to Find Plane Area

Setting M = -y/2 and N = x/2 gives:

Area of region R = (1/2) ∮_C (x dy - y dx)

Worked Example (4 Marks)

Evaluate: ∮_C [(y - sin x) dx + cos x dy] along the triangle bounded by y = 0, x = π/2, and y = (2/π) x.

1. Identify M and N:

• M = y - sin x => ∂M/∂y = 1
• N = cos x => ∂N/∂x = -sin x

1. Apply Green's Theorem:
   ∮C (M dx + N dy) = ∬R (-sin x - 1) dx dy

1. Set Up Integrals with Limits:

• x goes from 0 to π/2
• y goes from 0 to (2/π)x

I = ∫0^(π/2) ∫0^((2/π)x) (-sin x - 1) dy dx I = ∫0^(π/2) (-sin x - 1) [(2/π)x] dx = -(2/π) ∫0^(π/2) (x sin x + x) dx

1. Evaluate Integrals:

• ∫0^(π/2) x sin x dx = [-x cos x + sin x]0^(π/2) = (0 + 1) - 0 = 1
• ∫0^(π/2) x dx = [x²/2]0^(π/2) = π² / 8

I = -(2/π) [1 + π²/8] = -2/π - π/4

Final Answer: ∮_C (M dx + N dy) = - (2/π + π/4)

Q.5 (c) OR [07 Marks]

Stokes' Theorem

Statement

Stokes' Theorem states that the line integral of a vector field F around a closed curve C bounding an open surface S is equal to the surface integral of the curl of F over S:

∮C F · dr = ∬S (∇ × F) · n dS

where C is traversed in the counterclockwise direction relative to the surface normal n.

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Worked 7-Mark Problem

Problem: Verify Stokes' Theorem for:

F = y i + z j + x k

where S is the upper half surface of the sphere x² + y² + z² = 1 (z ≥ 0) and C is its boundary in the xy-plane (x² + y² = 1, z = 0).

Step 1: Evaluate the Line Integral (LHS)

The boundary curve C is the unit circle in the xy-plane: x² + y² = 1, z = 0, dz = 0

Parameterize C:

• x = cos t
• y = sin t
• z = 0

where t varies from 0 to 2π.

Differentials:

• dx = -sin t dt
• dy = cos t dt
• dz = 0

Substitute into F · dr: F · dr = y dx + z dy + x dz Since z = 0 and dz = 0: F · dr = y dx = (sin t)(-sin t dt) = -sin² t dt

Evaluate line integral: IC = ∫0^(2π) (-sin² t) dt = - ∫0^(2π) [(1 - cos 2t) / 2] dt IC = - [t/2 - sin(2t)/4]_0^(2π) = - [2π/2 - 0] = -π

Thus, Line Integral = -π.

Step 2: Evaluate the Surface Integral (RHS)

1. Compute curl F:
   curl F = ∇ × F =
   | i j k |
   | ∂/∂x ∂/∂y ∂/∂z |
   | y z x |

• i-component: ∂(x)/∂y - ∂(z)/∂z = 0 - 1 = -1
• j-component: - [∂(x)/∂x - ∂(y)/∂z] = - [1 - 0] = -1
• k-component: ∂(z)/∂x - ∂(y)/∂y = 0 - 1 = -1

curl F = - i - j - k

1. Choice of Surface:
   By Stokes' Theorem, the surface integral depends only on the boundary contour C. We can choose the flat disk D in the xy-plane bounded by C: x² + y² ≤ 1, z = 0.
   For this planar disk in the xy-plane:

• Outward unit normal: n = k
• dS = dx dy

1. Compute (curl F) · n:
   (curl F) · n = (-i - j - k) · k = -1

1. Compute the Surface Integral:
   ∬S (curl F · n) dS = ∬D (-1) dx dy = - ∬_D dx dy
   Since ∬_D dx dy is the area of the unit circle (π · 1² = π):
   ∬_S (curl F · n) dS = - (π) = -π

Conclusion

• Line integral: ∮_C F · dr = -π
• Surface integral: ∬_S (∇ × F) · n dS = -π

Both sides are equal (-π = -π). Hence, Stokes' Theorem is verified.

Show that F = (y² - z² + 3yz - 2x)i + (3xz + 2xy)j + (3xy - 2xz + 2z)k is both solenoidal and irrotational. [03 Marks]

Definition

• A vector field F is solenoidal if its divergence is zero: div F = ∇ · F = 0.
• A vector field F is irrotational if its curl is zero: curl F = ∇ × F = 0.

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Step 1: Check Solenoidal (∇ · F)

Given:

• F₁ = y² - z² + 3yz - 2x
• F₂ = 3xz + 2xy
• F₃ = 3xy - 2xz + 2z

Compute the partial derivatives:

• ∂F₁/∂x = ∂/∂x (y² - z² + 3yz - 2x) = -2
• ∂F₂/∂y = ∂/∂y (3xz + 2xy) = 2x
• ∂F₃/∂z = ∂/∂z (3xy - 2xz + 2z) = -2x + 2

Add them together: ∇ · F = ∂F₁/∂x + ∂F₂/∂y + ∂F₃/∂z ∇ · F = -2 + 2x + (-2x + 2) = 0

Since ∇ · F = 0, the vector field F is solenoidal.

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Step 2: Check Irrotational (∇ × F)

Compute the curl using the determinant:

∇ × F = |   i          j          k     |
        |  ∂/∂x       ∂/∂y       ∂/∂z   |
        |   F₁         F₂         F₃    |

Calculate each component:

• i-component:
  ∂F₃/∂y - ∂F₂/∂z = ∂/∂y (3xy - 2xz + 2z) - ∂/∂z (3xz + 2xy)
  = 3x - 3x = 0

• j-component:
  ∂F₁/∂z - ∂F₃/∂x = ∂/∂z (y² - z² + 3yz - 2x) - ∂/∂x (3xy - 2xz + 2z)
  = (-2z + 3y) - (3y - 2z) = -2z + 3y - 3y + 2z = 0

• k-component:
  ∂F₂/∂x - ∂F₁/∂y = ∂/∂x (3xz + 2xy) - ∂/∂y (y² - z² + 3yz - 2x)
  = (3z + 2y) - (2y + 3z) = 3z + 2y - 2y - 3z = 0

Therefore: ∇ × F = 0i + 0j + 0k = 0

Since ∇ × F = 0, the vector field F is irrotational.

Conclusion: The vector field F is both solenoidal and irrotational.

Find Fourier integral representation of function f(x) = {1, |x| < 1; 0, |x| > 1}. Hence, Evaluate ∫ [sin(ω) cos(ωx) / ω] dω. [04 Marks]

Step 1: Identify Nature of Function

The given function is:

• f(x) = 1 for -1 < x < 1
• f(x) = 0 for |x| > 1

Notice that f(-x) = f(x), which means f(x) is an even function.

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Step 2: Fourier Cosine Integral Formulation

For an even function, the Fourier sine component B(ω) = 0, and the Fourier integral representation simplifies to the Fourier Cosine Integral: f(x) = (2/π) ∫₀^∞ A(ω) cos(ωx) dω where: A(ω) = ∫₀^∞ f(t) cos(ωt) dt

Substitute f(t) = 1 for 0 ≤ t < 1 and f(t) = 0 for t > 1: A(ω) = ∫₀¹ 1 · cos(ωt) dt = [sin(ωt) / ω]₀¹ = sin(ω) / ω

Therefore, the Fourier integral representation is: f(x) = (2/π) ∫₀^∞ [sin(ω) / ω] cos(ωx) dω

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Step 3: Evaluate the Integral

Rearranging the equation: ∫₀^∞ [sin(ω) cos(ωx) / ω] dω = (π/2) f(x)

Now consider the values based on x:

1. For |x| < 1:
   f(x) = 1 => ∫₀^∞ [sin(ω) cos(ωx) / ω] dω = π/2

1. For |x| > 1:
   f(x) = 0 => ∫₀^∞ [sin(ω) cos(ωx) / ω] dω = 0

1. At the points of discontinuity x = ±1:
   By Dirichlet's condition, the integral equals (π/2) · [f(1⁺) + f(1⁻)] / 2 = (π/2) · (0 + 1)/2 = π/4.

Final Answer: ∫₀^∞ [sin(ω) cos(ωx) / ω] dω =

• π/2, for |x| < 1
• π/4, for x = ±1
• 0, for |x| > 1

Verify Green’s theorem in the plane for ∮_C [(3x² - 8y²) dx + (4y - 6xy) dy], where C is the boundary of the region bounded by x = 0, y = 0, x + y = 1. [07 Marks]

Statement of Green's Theorem

Green's theorem relates a line integral around a simple closed curve C to a double integral over the plane region R bounded by C: ∮C (M dx + N dy) = ∬R (∂N/∂x - ∂M/∂y) dx dy

Here:

• M = 3x² - 8y²
• N = 4y - 6xy

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Part 1: Double Integral (Right-Hand Side)

Compute the partial derivatives:

• ∂M/∂y = ∂/∂y (3x² - 8y²) = -16y
• ∂N/∂x = ∂/∂x (4y - 6xy) = -6y

So: ∂N/∂x - ∂M/∂y = -6y - (-16y) = 10y

The region R is the triangle bounded by x = 0, y = 0, and y = 1 - x (from x = 0 to x = 1): ∬_R 10y dx dy = ∫₀¹ [ ∫₀^(1-x) 10y dy ] dx

Evaluate the inner integral: ∫₀^(1-x) 10y dy = [5y²]₀^(1-x) = 5(1 - x)² = 5(1 - 2x + x²)

Evaluate the outer integral: ∫₀¹ 5(1 - 2x + x²) dx = 5 [x - x² + x³/3]₀¹ = 5 [1 - 1 + 1/3] = 5/3

RHS = 5/3

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Part 2: Line Integral (Left-Hand Side)

The boundary curve C consists of three straight line segments oriented counter-clockwise:

1. C₁: Along y = 0 from (0, 0) to (1, 0)

• y = 0, dy = 0
• x goes from 0 to 1
• ∫_C₁ = ∫₀¹ (3x² - 0) dx + 0 = [x³]₀¹ = 1

1. C₂: Along x + y = 1 from (1, 0) to (0, 1)

• y = 1 - x => dy = -dx
• x goes from 1 to 0
• M = 3x² - 8(1 - x)² = 3x² - 8(1 - 2x + x²) = -5x² + 16x - 8
• N = 4(1 - x) - 6x(1 - x) = 4 - 4x - 6x + 6x² = 6x² - 10x + 4
• Integrand: M dx + N dy = (-5x² + 16x - 8) dx + (6x² - 10x + 4)(-dx)
  = (-5x² + 16x - 8 - 6x² + 10x - 4) dx
  = (-11x² + 26x - 12) dx
• ∫_C₂ = ∫₁⁰ (-11x² + 26x - 12) dx
  = [-11x³/3 + 13x² - 12x]₁⁰
  = 0 - [-11/3 + 13 - 12]
  = - [-11/3 + 1] = - [-8/3] = 8/3

1. C₃: Along x = 0 from (0, 1) to (0, 0)

• x = 0, dx = 0
• y goes from 1 to 0
• ∫_C₃ = ∫₁⁰ (4y - 0) dy = [2y²]₁⁰ = 0 - 2 = -2

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Sum of Line Integrals:

∮C (M dx + N dy) = ∫C₁ + ∫C₂ + ∫C₃ = 1 + 8/3 - 2 = -1 + 8/3 = 5/3

LHS = 5/3

Conclusion

Since LHS = RHS = 5/3, Green's theorem is verified.

Solve y e^x dx + (2y + e^x) dy = 0. [03 Marks]

Step 1: Test for Exactness

The equation is of the form M dx + N dy = 0, where:

• M = y e^x
• N = 2y + e^x

Find partial derivatives:

• ∂M/∂y = e^x
• ∂N/∂x = e^x

Since ∂M/∂y = ∂N/∂x = e^x, the differential equation is exact.

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Step 2: Integrate to Find the General Solution

The solution is given by: ∫ M dx (treating y as constant) + ∫ (terms of N independent of x) dy = C

• Term 1: ∫ y e^x dx = y e^x
• Term 2: Terms of N not containing x is 2y. So: ∫ 2y dy = y²

Therefore, the general solution is: y e^x + y² = C where C is an arbitrary constant.

Find the directional Derivative of ∅ = xy² + yz² at the point (2, -1, 1) in the direction of the vector i + 2j + 2k. [04 Marks]

Step 1: Find the Gradient ∇∅

The gradient is defined as: ∇∅ = (∂∅/∂x) i + (∂∅/∂y) j + (∂∅/∂z) k

Given ∅ = xy² + yz²:

• ∂∅/∂x = y²
• ∂∅/∂y = 2xy + z²
• ∂∅/∂z = 2yz

At the point P(2, -1, 1):

• ∂∅/∂x = (-1)² = 1
• ∂∅/∂y = 2(2)(-1) + (1)² = -4 + 1 = -3
• ∂∅/∂z = 2(-1)(1) = -2

Thus: ∇∅ at (2, -1, 1) = 1i - 3j - 2k

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Step 2: Find the Unit Vector along the Given Direction

Given vector: a = i + 2j + 2k Magnitude of a: |a| = √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3

Unit vector û: û = a / |a| = (i + 2j + 2k) / 3

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Step 3: Compute the Directional Derivative

The directional derivative is the dot product of ∇∅ and û: Directional Derivative = ∇∅ · û = (1i - 3j - 2k) · (1/3)(i + 2j + 2k) = (1/3) [ (1)(1) + (-3)(2) + (-2)(2) ] = (1/3) [ 1 - 6 - 4 ] = -9 / 3 = -3

Final Answer: The directional derivative is -3.

Find the power series solution of d²y/dx² + xy = 0. [07 Marks]

Step 1: Nature of the Ordinary Point

The equation is y'' + xy = 0. Here, x = 0 is an ordinary point because the coefficient of y'' is 1 (non-zero everywhere).

Let the power series solution about x = 0 be: y = ∑(m=0)^∞ am x^m = a₀ + a₁x + a₂x² + a₃x³ + ...

Differentiating term by term: y' = ∑(m=1)^∞ m am x^(m-1) y'' = ∑(m=2)^∞ m(m-1) am x^(m-2)

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Step 2: Substitute into Differential Equation

y'' + xy = 0 ∑(m=2)^∞ m(m-1) am x^(m-2) + x ∑(m=0)^∞ am x^m = 0 ∑(m=2)^∞ m(m-1) am x^(m-2) + ∑(m=0)^∞ am x^(m+1) = 0

Write out initial terms to equalize powers of x:

• For the first sum, at m = 2: 2(1) a₂ = 2a₂
• Shift index of first sum by letting m-2 = k => m = k+2:
  ∑(k=0)^∞ (k+2)(k+1) a(k+2) x^k
  = 2a₂ + ∑(k=1)^∞ (k+2)(k+1) a(k+2) x^k
• For the second sum, shift index by letting m+1 = k => m = k-1:
  ∑(k=1)^∞ a(k-1) x^k

Combining: 2a₂ + ∑(k=1)^∞ [ (k+2)(k+1) a(k+2) + a_(k-1) ] x^k = 0

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Step 3: Equating Coefficients to Zero

• Constant term (x⁰): 2a₂ = 0 => a₂ = 0
• Recurrence relation for k ≥ 1:
  a(k+2) = -a(k-1) / [(k+2)(k+1)]

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Step 4: Calculate Coefficients in Terms of a₀ and a₁

• For k = 1:
  a₃ = -a₀ / (3 · 2) = -a₀ / 6
• For k = 2:
  a₄ = -a₁ / (4 · 3) = -a₁ / 12
• For k = 3:
  a₅ = -a₂ / (5 · 4) = 0 (since a₂ = 0)
• For k = 4:
  a₆ = -a₃ / (6 · 5) = -(-a₀ / 6) / 30 = a₀ / 180
• For k = 5:
  a₇ = -a₄ / (7 · 6) = -(-a₁ / 12) / 42 = a₁ / 504

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Step 5: Write the General Solution

Grouping terms with a₀ and a₁: y = a₀ [ 1 - x³/6 + x⁶/180 - ... ] + a₁ [ x - x⁴/12 + x⁷/504 - ... ] where a₀ and a₁ are arbitrary constants.

Find the series solution of (1 - x²)y'' - 2xy' + 2y = 0. [07 Marks]

Step 1: Standard Legendre Form

Notice that this is Legendre's differential equation: (1 - x²)y'' - 2xy' + n(n+1)y = 0 with n = 1, since n(n+1) = 1(2) = 2. Here, x = 0 is an ordinary point.

Let the series solution be: y = ∑(m=0)^∞ am x^m y' = ∑(m=1)^∞ m am x^(m-1) y'' = ∑(m=2)^∞ m(m-1) am x^(m-2)

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Step 2: Substitute into the Equation

(1 - x²) ∑(m=2)^∞ m(m-1) am x^(m-2) - 2x ∑(m=1)^∞ m am x^(m-1) + 2 ∑(m=0)^∞ am x^m = 0

Expand the terms: ∑(m=2)^∞ m(m-1) am x^(m-2) - ∑(m=2)^∞ m(m-1) am x^m - ∑(m=1)^∞ 2m am x^m + ∑(m=0)^∞ 2 am x^m = 0

Shift index in the first summation by setting m - 2 = k => m = k + 2: ∑(k=0)^∞ (k+2)(k+1) a(k+2) x^k - ∑(k=2)^∞ k(k-1) ak x^k - ∑(k=1)^∞ 2k ak x^k + ∑(k=0)^∞ 2 ak x^k = 0

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Step 3: Equate Coefficients of Powers of x to Zero

• For x⁰ (k = 0):
  (2)(1) a₂ + 2 a₀ = 0 => 2a₂ + 2a₀ = 0 => a₂ = -a₀

• For x¹ (k = 1):
  (3)(2) a₃ - 2(1) a₁ + 2 a₁ = 0 => 6a₃ = 0 => a₃ = 0

• For x^k (k ≥ 2):
  (k+2)(k+1) a(k+2) - [k(k-1) + 2k - 2] ak = 0
  k(k-1) + 2k - 2 = k² - k + 2k - 2 = k² + k - 2 = (k+2)(k-1)

So: (k+2)(k+1) a(k+2) = (k+2)(k-1) ak a(k+2) = [(k - 1) / (k + 1)] ak

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Step 4: Calculate Subsequent Coefficients

• For k = 2:
  a₄ = (1/3) a₂ = (1/3)(-a₀) = -a₀/3
• For k = 3:
  a₅ = (2/4) a₃ = 0 (since a₃ = 0)
• For k = 4:
  a₆ = (3/5) a₄ = (3/5)(-a₀/3) = -a₀/5
• For all odd k ≥ 3:
  a_k = 0 because a₃ = 0.

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Step 5: Final Series Solution

y = a₀ [ 1 - x² - x⁴/3 - x⁶/5 - ... ] + a₁ x where a₀ and a₁ are arbitrary constants. Notice that y₁ = x is a polynomial solution (the Legendre polynomial P₁(x) = x).

Solve (4D² - 4D + 1)y = e^(x/2). [03 Marks]

Step 1: Complementary Function (y_c)

Auxiliary Equation (A.E.): 4m² - 4m + 1 = 0 (2m - 1)² = 0 => m = 1/2, 1/2 (real and repeated roots)

Therefore, the Complementary Function is: y_c = (c₁ + c₂ x) e^(x/2)

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Step 2: Particular Integral (y_p)

y_p = [1 / (4D² - 4D + 1)] e^(x/2) = [1 / (2D - 1)²] e^(x/2)

Since replacing D by 1/2 makes the denominator zero (case of failure with multiplicity 2): Using the standard formula: [1 / (D - a)^n] e^(ax) = (x^n / n!) e^(ax)

Rewrite: [1 / 4(D - 1/2)²] e^(x/2) = (1/4) · (x² / 2!) e^(x/2) = (x² / 8) e^(x/2)

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Step 3: General Solution

y = yc + yp y = (c₁ + c₂ x) e^(x/2) + (x² / 8) e^(x/2)

Solve (D² + 6D + 8)y = cos 2x. [04 Marks]

Step 1: Complementary Function (y_c)

Auxiliary Equation: m² + 6m + 8 = 0 (m + 2)(m + 4) = 0 => m = -2, -4

y_c = c₁ e^(-2x) + c₂ e^(-4x)

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Step 2: Particular Integral (y_p)

y_p = [1 / (D² + 6D + 8)] cos 2x

Substitute D² = -2² = -4: y_p = [1 / (-4 + 6D + 8)] cos 2x = [1 / (6D + 4)] cos 2x = [1 / (2(3D + 2))] cos 2x

Multiply numerator and denominator by conjugate (3D - 2): y_p = (3D - 2) / [2(9D² - 4)] cos 2x

Again substitute D² = -4: Denominator = 2 [ 9(-4) - 4 ] = 2 [ -36 - 4 ] = 2(-40) = -80

Numerator: (3D - 2) cos 2x = 3 D(cos 2x) - 2 cos 2x = 3(-2 sin 2x) - 2 cos 2x = -6 sin 2x - 2 cos 2x

So: y_p = (-6 sin 2x - 2 cos 2x) / (-80) = (3 sin 2x + cos 2x) / 40

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Step 3: General Solution

y = yc + yp y = c₁ e^(-2x) + c₂ e^(-4x) + (1/40) (3 sin 2x + cos 2x)

Solve (D² - 2D + 3)y = x³ + sin x by using the method of undetermined coefficients. [07 Marks]

Step 1: Complementary Function (y_c)

Auxiliary equation: m² - 2m + 3 = 0 m = [2 ± √(4 - 12)] / 2 = [2 ± √(-8)] / 2 = 1 ± i√2

y_c = e^x [ c₁ cos(√2 x) + c₂ sin(√2 x) ]

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Step 2: Form of the Particular Solution (y_p)

The right side is R(x) = x³ + sin x. Since no term in R(x) appears in yc, the trial solution is: yp = yp1 + yp2 where:

• y_p1 = A x³ + B x² + C x + D
• y_p2 = E cos x + F sin x

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Step 3: Determine y_p1 for R₁(x) = x³

yp1 = A x³ + B x² + C x + D yp1' = 3A x² + 2B x + C y_p1'' = 6A x + 2B

Substitute into (D² - 2D + 3)y = x³: (6A x + 2B) - 2(3A x² + 2B x + C) + 3(A x³ + B x² + C x + D) = x³ 3A x³ + (3B - 6A) x² + (3C - 4B + 6A) x + (3D - 2C + 2B) = x³

Equating coefficients:

1. x³: 3A = 1 => A = 1/3
2. x²: 3B - 6A = 0 => 3B = 6(1/3) = 2 => B = 2/3
3. x¹: 3C - 4B + 6A = 0 => 3C = 4(2/3) - 6(1/3) = 8/3 - 2 = 2/3 => C = 2/9
4. x⁰: 3D - 2C + 2B = 0 => 3D = 2(2/9) - 2(2/3) = 4/9 - 4/3 = -8/9 => D = -8/27

So: y_p1 = (1/3)x³ + (2/3)x² + (2/9)x - 8/27

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Step 4: Determine y_p2 for R₂(x) = sin x

yp2 = E cos x + F sin x yp2' = -E sin x + F cos x y_p2'' = -E cos x - F sin x

Substitute into (D² - 2D + 3)y = sin x: (-E cos x - F sin x) - 2(-E sin x + F cos x) + 3(E cos x + F sin x) = sin x (-E - 2F + 3E) cos x + (-F + 2E + 3F) sin x = sin x (2E - 2F) cos x + (2E + 2F) sin x = sin x

Equating coefficients:

• cos x: 2E - 2F = 0 => E = F
• sin x: 2E + 2F = 1 => 4E = 1 => E = 1/4, F = 1/4

So: y_p2 = (1/4) cos x + (1/4) sin x

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Step 5: General Solution

y = yc + yp1 + y_p2 y = e^x [ c₁ cos(√2 x) + c₂ sin(√2 x) ] + (1/3)x³ + (2/3)x² + (2/9)x - 8/27 + (1/4)(cos x + sin x)

Solve (D² + a²)y = cos ax. [03 Marks]

Step 1: Complementary Function (y_c)

Auxiliary Equation: m² + a² = 0 => m = ±ia y_c = c₁ cos(ax) + c₂ sin(ax)

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Step 2: Particular Integral (y_p)

y_p = [1 / (D² + a²)] cos ax

Since replacing D² with -a² gives zero in the denominator, this is a case of failure. We use the standard formula: [1 / (D² + a²)] cos ax = (x / 2a) sin ax

Derivation: [1 / (D² + a²)] cos ax = Real Part of [1 / (D² + a²)] e^(iax) = Real Part of e^(iax) [1 / ((D + ia)² + a²)] 1 = Real Part of e^(iax) [1 / (D² + 2iaD)] 1 = Real Part of e^(iax) (1 / 2ia) [1 / D] 1 = Real Part of e^(iax) [x / (2ia)] = Real Part of (-i x / 2a) (cos ax + i sin ax) = (x / 2a) sin ax

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Step 3: General Solution

y = yc + yp y = c₁ cos(ax) + c₂ sin(ax) + (x / 2a) sin ax

Solve d²y/dx² + 2 dy/dx + 2y = e^x sin x + 7. [04 Marks]

Step 1: Complementary Function (y_c)

Auxiliary Equation: m² + 2m + 2 = 0 m = [-2 ± √(4 - 8)] / 2 = -1 ± i

y_c = e^(-x) [ c₁ cos x + c₂ sin x ]

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Step 2: Particular Integral y_p1 (for e^x sin x)

y_p1 = [1 / (D² + 2D + 2)] (e^x sin x) Using the exponential shift property: [1 / f(D)] (e^(ax) V) = e^(ax) [1 / f(D + a)] V

Replace D with D + 1: y_p1 = e^x [1 / ((D + 1)² + 2(D + 1) + 2)] sin x = e^x [1 / (D² + 2D + 1 + 2D + 2 + 2)] sin x = e^x [1 / (D² + 4D + 5)] sin x

Substitute D² = -1² = -1: y_p1 = e^x [1 / (-1 + 4D + 5)] sin x = e^x [1 / (4D + 4)] sin x = (e^x / 4) [1 / (D + 1)] sin x

Multiply numerator and denominator by (D - 1): y_p1 = (e^x / 4) [(D - 1) / (D² - 1)] sin x = (e^x / 4) [(D - 1) / (-1 - 1)] sin x = (e^x / 4) [(D - 1) / (-2)] sin x = - (e^x / 8) [D(sin x) - sin x] = - (e^x / 8) [cos x - sin x] = (e^x / 8) [sin x - cos x]

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Step 3: Particular Integral y_p2 (for 7)

y_p2 = [1 / (D² + 2D + 2)] (7 e^(0x)) = 7 / (0 + 0 + 2) = 7/2

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Step 4: General Solution

y = yc + yp1 + y_p2 y = e^(-x) [ c₁ cos x + c₂ sin x ] + (e^x / 8) (sin x - cos x) + 7/2

Solve x² d²y/dx² + x dy/dx + y = log x · sin(log x) by Cauchy’s linear equation. [07 Marks]

Step 1: Transform Cauchy-Euler Equation into Constant Coefficients

Let x = e^z, so that z = log x. Define θ = d/dz. Then:

• x (dy/dx) = θ y
• x² (d²y/dx²) = θ(θ - 1) y

Substitute into the differential equation: [ θ(θ - 1) + θ + 1 ] y = z sin z [ θ² - θ + θ + 1 ] y = z sin z (θ² + 1) y = z sin z

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Step 2: Complementary Function (y_c)

Auxiliary equation: m² + 1 = 0 => m = ±i y_c = c₁ cos z + c₂ sin z

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Step 3: Particular Integral (y_p)

y_p = [1 / (θ² + 1)] (z sin z)

We know that z sin z = Imaginary Part of (z e^(iz)): y_p = Im { [1 / (θ² + 1)] (z e^(iz)) }

Using exponential shift property: = Im { e^(iz) [1 / ((θ + i)² + 1)] z } = Im { e^(iz) [1 / (θ² + 2iθ - 1 + 1)] z } = Im { e^(iz) [1 / (θ(θ + 2i))] z } = Im { e^(iz) [1 / (2iθ (1 + θ / 2i))] z } = Im { [e^(iz) / (2i)] (1/θ) [ 1 - θ/(2i) ] z } = Im { [-i e^(iz) / 2] (1/θ) [ z - 1/(2i) ] } = Im { [-i e^(iz) / 2] (1/θ) [ z + i/2 ] }

Integrating with respect to z: (1/θ) [ z + i/2 ] = z²/2 + iz/2

Now multiply: [-i e^(iz) / 2] [ z²/2 + iz/2 ] = (1/4) e^(iz) [ -i z² + z ] = (1/4) (cos z + i sin z) (z - i z²) = (1/4) [ z cos z + z² sin z + i (z sin z - z² cos z) ]

Taking the Imaginary Part: y_p = (1/4) [ z sin z - z² cos z ]

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Step 4: Express General Solution in Terms of x

Substitute z = log x into yc and yp: y = c₁ cos(log x) + c₂ sin(log x) + (1/4) [ log x · sin(log x) - (log x)² · cos(log x) ]

Find the Laplace Transform of f(t) = sinh at. [03 Marks]

Definition

The Laplace transform of a function f(t) is defined as: L{f(t)} = ∫₀^∞ e^(-st) f(t) dt

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Derivation

By definition, sinh at = (e^(at) - e^(-at)) / 2.

Applying linearity of the Laplace transform: L{sinh at} = L{ (e^(at) - e^(-at)) / 2 } = (1/2) [ L{e^(at)} - L{e^(-at)} ]

We know that: L{e^(at)} = 1 / (s - a), for s > a L{e^(-at)} = 1 / (s + a), for s > -a

Therefore: L{sinh at} = (1/2) [ 1/(s - a) - 1/(s + a) ] = (1/2) [ ( (s + a) - (s - a) ) / ((s - a)(s + a)) ] = (1/2) [ (2a) / (s² - a²) ] = a / (s² - a²), for s > |a|

Final Answer: L{sinh at} = a / (s² - a²).

Find the Laplace Transform of t e^(4t) cos 2t. [04 Marks]

Step 1: Laplace Transform of cos 2t

L{cos 2t} = s / (s² + 2²) = s / (s² + 4)

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Step 2: Multiplication by t Property

By property: L{t f(t)} = - d/ds [ L{f(t)} ]

L{t cos 2t} = - d/ds [ s / (s² + 4) ] = - [ ( (s² + 4)(1) - s(2s) ) / (s² + 4)² ] = - [ (s² + 4 - 2s²) / (s² + 4)² ] = - [ (4 - s²) / (s² + 4)² ] = (s² - 4) / (s² + 4)²

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Step 3: First Shifting Theorem

By the first shifting theorem: L{e^(at) g(t)} = G(s - a) Here, a = 4. Replace s with s - 4:

L{e^(4t) · (t cos 2t)} = [ (s - 4)² - 4 ] / [ (s - 4)² + 4 ]² = [ s² - 8s + 16 - 4 ] / [ s² - 8s + 16 + 4 ]² = (s² - 8s + 12) / (s² - 8s + 20)²

Final Answer: (s² - 8s + 12) / (s² - 8s + 20)²

Solve the differential equation using initial value problem: y'' + 6y' = 1, y(0) = 2, y'(0) = 0. [07 Marks]

Step 1: Apply Laplace Transform to Both Sides

L{y'' + 6y'} = L{1} L{y''} + 6 L{y'} = 1/s

Using initial value formulas:

• L{y'} = s Y(s) - y(0)
• L{y''} = s² Y(s) - s y(0) - y'(0)

Substitute y(0) = 2 and y'(0) = 0: [ s² Y(s) - 2s - 0 ] + 6 [ s Y(s) - 2 ] = 1/s (s² + 6s) Y(s) - 2s - 12 = 1/s s(s + 6) Y(s) = 2s + 12 + 1/s = (2s² + 12s + 1) / s

Y(s) = (2s² + 12s + 1) / [ s²(s + 6) ]

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Step 2: Partial Fraction Decomposition

Let: (2s² + 12s + 1) / [ s²(s + 6) ] = A/s + B/s² + C/(s + 6)

Multiply through by s²(s + 6): 2s² + 12s + 1 = A s (s + 6) + B (s + 6) + C s²

1. Put s = 0:
   1 = B(6) => B = 1/6

1. Put s = -6:
   2(-6)² + 12(-6) + 1 = C(-6)²
   2(36) - 72 + 1 = 36C
   72 - 72 + 1 = 36C => C = 1/36

1. Equate coefficients of s²:
   2 = A + C
   A = 2 - 1/36 = 71/36

Thus: Y(s) = (71/36)(1/s) + (1/6)(1/s²) + (1/36)(1/(s + 6))

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Step 3: Taking Inverse Laplace Transform

y(t) = L⁻¹{Y(s)} = (71/36) L⁻¹{1/s} + (1/6) L⁻¹{1/s²} + (1/36) L⁻¹{1/(s + 6)} = (71/36)(1) + (1/6) t + (1/36) e^(-6t)

Final Answer: y(t) = 71/36 + (1/6)t + (1/36)e^(-6t)

Find the inverse Laplace transform of log((s² + b²) / (s² + a²)). [03 Marks]

Step 1: Use Property of Derivative of Laplace Transform

Let F(s) = log((s² + b²) / (s² + a²)) = log(s² + b²) - log(s² + a²).

We know that: L⁻¹{ F'(s) } = -t f(t) => f(t) = - (1/t) L⁻¹{ F'(s) }

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Step 2: Differentiate F(s) with Respect to s

F'(s) = d/ds [ log(s² + b²) - log(s² + a²) ] = 2s / (s² + b²) - 2s / (s² + a²)

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Step 3: Take Inverse Laplace Transform

L⁻¹{ F'(s) } = 2 L⁻¹{ s / (s² + b²) } - 2 L⁻¹{ s / (s² + a²) } = 2 cos bt - 2 cos at

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Step 4: Compute f(t)

Since -t f(t) = 2(cos bt - cos at): f(t) = - [ 2(cos bt - cos at) ] / t = 2 (cos at - cos bt) / t

Final Answer: L⁻¹{ log((s² + b²) / (s² + a²)) } = 2(cos at - cos bt) / t

Find the Laplace Transform of [e^(-2t) sin 2t cosh t] / t. [04 Marks]

Step 1: Simplify the Numerator Function

Let g(t) = sin 2t cosh t. Using definition: cosh t = (e^t + e^(-t)) / 2 g(t) = (1/2) [ e^t sin 2t + e^(-t) sin 2t ]

We know that L{sin 2t} = 2 / (s² + 4). By shifting theorem: L{g(t)} = (1/2) [ 2 / ((s - 1)² + 4) + 2 / ((s + 1)² + 4) ] = 1 / ((s - 1)² + 4) + 1 / ((s + 1)² + 4)

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Step 2: Apply Division by t Property

L{ g(t) / t } = ∫s^∞ L{g(t)} du = ∫s^∞ [ 1 / ((u - 1)² + 4) + 1 / ((u + 1)² + 4) ] du

Using standard integral ∫ 1 / (x² + a²) dx = (1/a) arctan(x/a): = [ (1/2) arctan((u - 1)/2) + (1/2) arctan((u + 1)/2) ]_s^∞ = (1/2) [ (π/2 + π/2) - (arctan((s - 1)/2) + arctan((s + 1)/2)) ] = (1/2) [ π - arctan((s - 1)/2) - arctan((s + 1)/2) ]

Since π/2 - arctan(x) = arccot(x): = (1/2) [ arccot((s - 1)/2) + arccot((s + 1)/2) ]

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Step 3: Apply First Shifting for e^(-2t)

By shifting property, replace s with s + 2: L{ e^(-2t) [sin 2t cosh t / t] } = (1/2) [ arccot((s + 1)/2) + arccot((s + 3)/2) ]

Or equivalently in terms of arctan: = (1/2) [ π - arctan((s + 1)/2) - arctan((s + 3)/2) ]

Find the Inverse Laplace Transform of (5s + 3) / [ (s - 1)(s² + 2s + 5) ]. [07 Marks]

Step 1: Partial Fraction Decomposition

Let: (5s + 3) / [ (s - 1)(s² + 2s + 5) ] = A / (s - 1) + (Bs + C) / (s² + 2s + 5)

Multiplying by (s - 1)(s² + 2s + 5): 5s + 3 = A(s² + 2s + 5) + (Bs + C)(s - 1)

1. Set s = 1:
   5(1) + 3 = A(1 + 2 + 5) + 0
   8 = 8A => A = 1

1. Equate coefficients of s²:
   0 = A + B => B = -A => B = -1

1. Equate constant terms:
   3 = 5A - C
   3 = 5(1) - C => C = 5 - 3 => C = 2

So the expression becomes: 1 / (s - 1) + (-s + 2) / (s² + 2s + 5) = 1 / (s - 1) - (s - 2) / (s² + 2s + 5)

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Step 2: Complete the Square in Denominator

s² + 2s + 5 = (s + 1)² + 4 = (s + 1)² + 2²

Rewrite numerator in terms of (s + 1): s - 2 = (s + 1) - 3

Therefore: (s - 2) / [ (s + 1)² + 2² ] = (s + 1) / [ (s + 1)² + 2² ] - 3 / [ (s + 1)² + 2² ] = (s + 1) / [ (s + 1)² + 2² ] - (3/2) · 2 / [ (s + 1)² + 2² ]

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Step 3: Invert Term by Term

1. L⁻¹{ 1 / (s - 1) } = e^t
2. L⁻¹{ (s + 1) / [ (s + 1)² + 2² ] } = e^(-t) cos 2t
3. L⁻¹{ 2 / [ (s + 1)² + 2² ] } = e^(-t) sin 2t

Combining everything: f(t) = L⁻¹{ 1/(s - 1) } - [ L⁻¹{ (s + 1)/((s + 1)² + 2²) } - (3/2) L⁻¹{ 2/((s + 1)² + 2²) } ] = e^t - e^(-t) cos 2t + (3/2) e^(-t) sin 2t

Final Answer: y(t) = e^t + e^(-t) [ (3/2) sin 2t - cos 2t ]

Solve dy/dx + y cot x = 2 cos x. [03 Marks]

Step 1: Identify Equation Type

This is a standard first-order linear differential equation: dy/dx + P(x) y = Q(x) where:

• P(x) = cot x
• Q(x) = 2 cos x

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Step 2: Integrating Factor (I.F.)

I.F. = e^(∫ P(x) dx) = e^(∫ cot x dx) = e^(log(sin x)) = sin x

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Step 3: General Solution

The general solution is given by: y · (I.F.) = ∫ Q(x) · (I.F.) dx + C

y · sin x = ∫ (2 cos x) (sin x) dx + C y · sin x = ∫ sin 2x dx + C y · sin x = - (cos 2x) / 2 + C

Alternatively, since 2 sin x cos x = d(sin² x)/dx: y · sin x = sin² x + C y = sin x + C csc x

Solve 3x⁴p² - xp - y = 0. [04 Marks]

Step 1: Express in Solvable Form for y

The given differential equation is of first order and higher degree (where p = dy/dx): 3x⁴p² - xp - y = 0 => y = 3x⁴p² - xp

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Step 2: Substitution Method

Let X = 1/x and Y = y. (Alternatively, let u = 1/x, which transforms Clairaut-like equations). Let us substitute: u = x² or try transforming coordinates. Notice that 3x⁴p² = 3(x²p)². Let v = x² and w = y. Then: dy/dx = (dw/dv)(dv/dx) = P · 2x => p = 2x P (where P = dw/dv). Substitute into the equation: 3x⁴(2x P)² - x(2x P) - y = 0 => not Clairaut's form directly.

Let's rewrite the equation by putting y = 3(x²p)² - (xp): Let u = 1/x, then: p = dy/dx = (dy/du)(du/dx) = (-1/x²) (dy/du) = -u² Pu, where Pu = dy/du. Then x p = (1/u)(-u² Pu) = -u Pu. And x⁴ p² = (x² p)² = [ (1/u²) (-u² Pu) ]² = Pu².

Substitute into the original equation: 3(Pu)² - (-u Pu) - y = 0 => y = u Pu + 3 Pu²

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Step 3: Clairaut’s Form

This is in Clairaut's form: y = u Pu + f(Pu), where f(Pu) = 3 Pu².

The general solution of Clairaut's equation is obtained directly by replacing P_u with an arbitrary constant c: y = c u + 3 c²

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Step 4: Substitute Back u = 1/x

y = c (1/x) + 3c² xy = c + 3c² x or 3c²x - xy + c = 0 where c is an arbitrary constant.

Using the method variation of parameter solve the differential equation (D² + 1)y = x sin x. [07 Marks]

Step 1: Complementary Function (y_c)

(D² + 1)y = 0 => m² + 1 = 0 => m = ±i y_c = c₁ cos x + c₂ sin x

Here:

• y₁ = cos x
• y₂ = sin x
• R(x) = x sin x

---

Step 2: Wronskian (W)

W(y₁, y₂) = | y₁ y₂ | | y₁' y₂' | = | cos x sin x | | -sin x cos x | = cos² x - (-sin² x) = cos² x + sin² x = 1

---

Step 3: Method of Variation of Parameters

The particular solution is: y_p = u(x) y₁ + v(x) y₂ where: u(x) = - ∫ [y₂ R(x) / W] dx = - ∫ sin x (x sin x) / 1 dx = - ∫ x sin² x dx v(x) = ∫ [y₁ R(x) / W] dx = ∫ cos x (x sin x) / 1 dx = ∫ x sin x cos x dx = (1/2) ∫ x sin 2x dx

---

Step 4: Evaluate u(x)

Since sin² x = (1 - cos 2x) / 2: u(x) = - (1/2) ∫ x (1 - cos 2x) dx = - (1/2) [ ∫ x dx - ∫ x cos 2x dx ] = - (1/2) [ x²/2 - ( x (sin 2x / 2) - ∫ (sin 2x / 2) dx ) ] = - (1/2) [ x²/2 - (x sin 2x)/2 - (cos 2x)/4 ] = - x²/4 + (x sin 2x)/4 + (cos 2x)/8

---

Step 5: Evaluate v(x)

v(x) = (1/2) ∫ x sin 2x dx Using integration by parts: = (1/2) [ x (-cos 2x / 2) - ∫ 1 · (-cos 2x / 2) dx ] = (1/2) [ - (x cos 2x)/2 + (sin 2x)/4 ] = - (x cos 2x)/4 + (sin 2x)/8

---

Step 6: Form Particular Integral y_p

y_p = u(x) cos x + v(x) sin x = [ -x²/4 + (x sin 2x)/4 + (cos 2x)/8 ] cos x + [ - (x cos 2x)/4 + (sin 2x)/8 ] sin x = - (x²/4) cos x + (x/4) [ sin 2x cos x - cos 2x sin x ] + (1/8) [ cos 2x cos x + sin 2x sin x ]

Using trigonometric identities:

• sin 2x cos x - cos 2x sin x = sin(2x - x) = sin x
• cos 2x cos x + sin 2x sin x = cos(2x - x) = cos x

So: y_p = - (x²/4) cos x + (x/4) sin x + (1/8) cos x

Notice that (1/8) cos x can be absorbed into the complementary function c₁ cos x.

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Step 7: General Solution

y = c₁ cos x + c₂ sin x - (x²/4) cos x + (x/4) sin x

Solve (xy - 2y²) dx - (x² - 3xy) dy = 0. [03 Marks]

Step 1: Check Homogeneity

Rewrite in the form M dx + N dy = 0:

• M = xy - 2y² (degree 2)
• N = -(x² - 3xy) = -x² + 3xy (degree 2)

Since both M and N are homogeneous functions of degree 2, this is a homogeneous differential equation.

---

Step 2: Substitution Method

Put y = v x => dy = v dx + x dv

Substitute into the equation: [ x(vx) - 2(vx)² ] dx - [ x² - 3x(vx) ] (v dx + x dv) = 0 [ vx² - 2v²x² ] dx - [ x² - 3vx² ] (v dx + x dv) = 0

Divide by x² (for x ≠ 0): (v - 2v²) dx - (1 - 3v)(v dx + x dv) = 0 [ (v - 2v²) - (v - 3v²) ] dx - x (1 - 3v) dv = 0 [ v - 2v² - v + 3v² ] dx - x (1 - 3v) dv = 0 v² dx - x (1 - 3v) dv = 0

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Step 3: Separate Variables and Integrate

v² dx = x (1 - 3v) dv dx / x = [ (1 - 3v) / v² ] dv dx / x = [ 1/v² - 3/v ] dv

Integrate both sides: ∫ dx / x = ∫ v^(-2) dv - 3 ∫ (1/v) dv log|x| = - 1/v - 3 log|v| + C log|x| + 3 log|v| + 1/v = C log|x · v³| + 1/v = C

---

Step 4: Substitute Back v = y/x

log| x · (y/x)³ | + x/y = C log| y³ / x² | + x/y = C

Solve dy/dx + (1/x) y = y² / x². [04 Marks]

Step 1: Recognize Bernoulli’s Differential Equation

The equation is of the form: dy/dx + P(x) y = Q(x) yⁿ with n = 2, P(x) = 1/x, and Q(x) = 1/x².

---

Step 2: Divide by y²

(1/y²) dy/dx + (1/x)(1/y) = 1/x²

---

Step 3: Substitution

Let z = 1/y = y⁻¹ Then dz/dx = - y⁻² (dy/dx) = - (1/y²) (dy/dx) So: (1/y²) dy/dx = - dz/dx

Substitute into the equation:

• dz/dx + (1/x) z = 1/x²

dz/dx - (1/x) z = - 1/x²

This is a standard linear differential equation in z.

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Step 4: Integrating Factor (I.F.)

I.F. = e^(∫ - (1/x) dx) = e^(-log x) = 1/x

---

Step 5: Solve for z

z · (I.F.) = ∫ Q(x) · (I.F.) dx + C z · (1/x) = ∫ (-1/x²)(1/x) dx + C z / x = - ∫ x⁻³ dx + C z / x = - [ x⁻² / (-2) ] + C = 1 / (2x²) + C

Multiply through by x: z = 1 / (2x) + Cx

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Step 6: Substitute Back z = 1/y

1/y = 1 / (2x) + Cx = (1 + 2Cx²) / (2x)

y = 2x / (1 + 2Cx²) where C is an arbitrary constant.

Solve (D⁴ - 16)y = e^(2x) + x⁴. [07 Marks]

Step 1: Complementary Function (y_c)

Auxiliary Equation: m⁴ - 16 = 0 (m² - 4)(m² + 4) = 0 (m - 2)(m + 2)(m² + 4) = 0 Roots are: m = 2, -2, ±2i

y_c = c₁ e^(2x) + c₂ e^(-2x) + c₃ cos 2x + c₄ sin 2x

---

Step 2: Particular Integral y_p1 (for e^(2x))

y_p1 = [1 / (D⁴ - 16)] e^(2x)

Since m = 2 is a single root of the auxiliary equation, substituting D = 2 gives 0 (case of failure). Factor out the failing factor: 1 / (D⁴ - 16) = 1 / [ (D - 2)(D + 2)(D² + 4) ]

Substitute D = 2 into the non-zero terms: y_p1 = [1 / (D - 2)] [ 1 / ((2 + 2)(2² + 4)) ] e^(2x) = [1 / (D - 2)] [ 1 / (4 · 8) ] e^(2x) = (1/32) [1 / (D - 2)] e^(2x)

Using the standard formula [1 / (D - a)] e^(ax) = x e^(ax): y_p1 = (x / 32) e^(2x)

---

Step 3: Particular Integral y_p2 (for x⁴)

y_p2 = [1 / (D⁴ - 16)] x⁴ = [1 / -16(1 - D⁴/16)] x⁴ = - (1/16) [ 1 - D⁴/16 ]⁻¹ x⁴

Expanding binomially: (1 - u)⁻¹ = 1 + u + u² + ...: = - (1/16) [ 1 + D⁴/16 + ... ] x⁴

Apply the derivatives to x⁴:

• D(x⁴) = 4x³
• D²(x⁴) = 12x²
• D³(x⁴) = 24x
• D⁴(x⁴) = 24

So: y_p2 = - (1/16) [ x⁴ + 24/16 ] = - (1/16) [ x⁴ + 3/2 ] = - x⁴/16 - 3/32

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Step 4: General Solution

y = yc + yp1 + y_p2 y = c₁ e^(2x) + c₂ e^(-2x) + c₃ cos 2x + c₄ sin 2x + (x / 32) e^(2x) - x⁴/16 - 3/32

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