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GTU Mathematics 2 (3110015) Semester Semester 1 & 2 Summer 2025 Solved Paper

Complete step-by-step worked solutions for GTU Mathematics - 2 (3110015) Summer 2025 examination paper covering vector calculus, differential equations, series solutions, Fourier integrals, and Laplace transforms.

Introduction

This document contains fully worked, student-friendly solutions for the Gujarat Technological University (GTU) Mathematics - 2 (3110015) examination of Summer 2025. It provides detailed derivations, systematic steps for differential equations, clear integration limits for vector theorems, and rigorous methods for Laplace transforms and series solutions.

Paper Information

University: Gujarat Technological University
Department: First Year Engineering (All Branches)
Semester: Semester 1 & 2
Subject: Mathematics - 2
Subject Code: 3110015
Exam: Summer 2025
Year: 2025

Questions and Solutions

Q.1 (a)

Question

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

Solution

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.

---

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.

Q.1 (b)

Question

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

Solution

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.

---

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

---

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:

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

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

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

Q.1 (c)

Question

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]

Solution

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

---

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:

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

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

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

---

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.

Q.2 (a)

Question

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

Solution

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.

---

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.

Q.2 (b)

Question

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

Solution

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

---

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

---

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.

Q.2 (c)

Question

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

Solution

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)

---

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

---

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)]

---

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

---

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.

Q.2 (c) OR

Question

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

Solution

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)

---

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

---

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

---

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).

Q.3 (a)

Question

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

Solution

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)

---

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)

---

Step 3: General Solution

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

Q.3 (b)

Question

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

Solution

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)

---

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)

Q.3 (c)

Question

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

Solution

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) ]

---

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

---

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:

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

Q.3 (a) OR

Question

Solve (D² + a²)y = cos ax. [03 Marks]

Solution

Step 1: Complementary Function (y_c)

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

---

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

---

Step 3: General Solution

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

Q.3 (b) OR

Question

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

Solution

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 ]

---

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]

---

Step 3: Particular Integral y_p2 (for 7)

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

---

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

Q.3 (c) OR

Question

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

Solution

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

---

Step 2: Complementary Function (y_c)

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

---

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 ]

---

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) ]

Q.4 (a)

Question

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

Solution

Definition

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

---

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).

Q.4 (b)

Question

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

Solution

Step 1: Laplace Transform of cos 2t

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

---

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

---

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

Q.4 (c)

Question

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

Solution

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) ]

---

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

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

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

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))

---

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)

Q.4 (a) OR

Question

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

Solution

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) }

---

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)

---

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

---

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

Q.4 (b) OR

Question

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

Solution

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)

---

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) ]

---

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) ]

Q.4 (c) OR

Question

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

Solution

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)

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

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

---

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 ]

---

Step 3: Invert Term by Term

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

Q.5 (a)

Question

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

Solution

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

---

Step 2: Integrating Factor (I.F.)

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

---

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

Q.5 (b)

Question

Solve 3x⁴p² - xp - y = 0. [04 Marks]

Solution

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

---

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

---

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

---

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.

Q.5 (c)

Question

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

Solution

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

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

---

Step 7: General Solution

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

Q.5 (a) OR

Question

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

Solution

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

---

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

Q.5 (b) OR

Question

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

Solution

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

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

---

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

Frequently Asked Questions

What is the condition for a vector field to be both solenoidal and irrotational?

A vector field F is solenoidal if its divergence is zero (∇ · F = 0), meaning there are no sources or sinks. It is irrotational if its curl is zero (∇ × F = 0), meaning it has no rotational tendency and can be expressed as the gradient of a scalar potential.

How do you handle a case of failure when finding a particular integral with D² + a² and cos(ax)?

When substituting D² = -a² into 1/(D² + a²) cos(ax), the denominator becomes zero. To solve this, convert cos(ax) into the real part of e^(iax), apply the exponential shift property, and integrate to get the standard shortcut formula: [1 / (D² + a²)] cos(ax) = (x / 2a) sin(ax).

When does Green's Theorem apply?

Green's Theorem applies to a positively oriented, piecewise-smooth, simple closed curve C in the xy-plane that bounds a region R, provided the functions M(x, y) and N(x, y) have continuous partial derivatives in an open region containing R.

GTU Mathematics 1 (3110014) Semester 1 Summer 2026 Solved Paper

Worked solutions and paper analysis for GTU Mathematics - 1 (Subject Code: 3110014) Bachelor of Engineering Semester 1 Summer 2026 examination.

Introduction

This document provides the examination analysis and question paper layout for Gujarat Technological University (GTU) Bachelor of Engineering Semester 1, Subject: Mathematics - 1 (Subject Code: 3110014), scheduled for Summer 2026 (Exam Date: 13/07/2026). The paper structure carries a total of 70 marks distributed over five compulsory questions with internal choice options following GTU standard blueprint guidelines (3, 4, and 7 marks structure).

Paper Information

University: Gujarat Technological University
Department: Bachelor of Engineering
Semester: 1
Subject: Mathematics - 1
Subject Code: 3110014
Exam: Summer
Year: 2026

Questions and Solutions

Q.1 (a)

Question

(a) [Question text not present in source paper] (03 Marks)

Note on Question Content

The supplied paper text contains the question designation and mark allocation (7 Marks), but the question statement was omitted in the source paper template. Please consult the official GTU question bank or departmental portal for the full text of this problem.

Q.2 (a)

Question

(a) [Question text not present in source paper] (03 Marks)

Solution

Note on Question Content

Q.2 (b)

Question

Q.3 (a)

Question

(a) [Question text not present in source paper] (03 Marks)

Solution

Note on Question Content

Q.3 (b)

Question

(b) [Question text not present in source paper] (04 Marks)

Solution

The paper consists of 5 main questions carrying 14 marks each, totaling 70 marks. Each question typically contains sub-questions divided into (a) 3 marks, (b) 4 marks, and (c) 7 marks, with internal choices provided in Questions 2, 3, 4, and 5.

Are scientific calculators permitted in the GTU Mathematics - 1 exam?

Yes, simple and non-programmable scientific calculators (e.g., fx-82MS, fx-991ES Plus, or fx-991EX) are permitted as per GTU standard examination instructions.

What are the core topics covered in GTU Mathematics - 1 (3110014)?

The syllabus covers Indeterminate Forms and L'Hopital's Rule, Improper Integrals, Gamma and Beta Functions, Sequences and Infinite Series, Taylor and Maclaurin Series, Partial Derivatives, Euler's Theorem, Maxima and Minima of Functions of Several Variables, Multiple Integrals, and Matrices (including Rank, Eigenvalues, and Eigenvectors).

GTU Mathematics 1 (3110014) Semester 1 Summer 2025 Solved Paper

Complete worked solutions for GTU Mathematics - 1 (3110014) Summer 2025 examination. Step-by-step answers prepared in student-friendly plain text format.

Introduction

This document contains fully worked, step-by-step solutions for the Gujarat Technological University (GTU) Mathematics - 1 (Subject Code: 3110014) Semester 1 & 2 Examination held in Summer 2025. Every question and its sub-questions are answered comprehensively according to the allotted marks, following the official GTU syllabus guidelines.

Paper Information

University: Gujarat Technological University
Department: First Year Engineering
Semester: 1
Subject: Mathematics - 1
Subject Code: 3110014
Exam: Summer 2025
Year: 2025

Questions and Solutions

Q.1 (a)

Question

Evaluate: lim (x -> 0) [x^(-2) - cosec^2 x].

Solution

Problem Statement

Evaluate the limit: L = lim (x -> 0) [1/x^2 - cosec^2(x)]

Step 1: Identify the Indeterminate Form

Rewrite the expression using standard trigonometric functions: L = lim (x -> 0) [1/x^2 - 1/sin^2(x)]

As x -> 0, 1/x^2 -> infinity and 1/sin^2(x) -> infinity, which gives the indeterminate form (infinity - infinity).

Step 2: Combine into a Single Fraction

Combine over a common denominator: L = lim (x -> 0) [(sin^2(x) - x^2) / (x^2 * sin^2(x))]

Notice that lim (x -> 0) [sin(x)/x] = 1, so: x^2 sin^2(x) = x^4 (sin(x)/x)^2

Therefore, we can replace the denominator with x^4: L = lim (x -> 0) [(sin^2(x) - x^2) / x^4]

This is now of the form 0/0.

Step 3: Expand sin(x) using Taylor Series

The Taylor series expansion of sin(x) near x = 0 is: sin(x) = x - x^3/6 + x^5/120 - ...

Squaring both sides: sin^2(x) = [x - x^3/6 + O(x^5)]^2 sin^2(x) = x^2 - 2 x (x^3/6) + O(x^6) sin^2(x) = x^2 - x^4/3 + O(x^6)

Step 4: Substitute and Evaluate the Limit

Substitute this expansion back into the numerator: sin^2(x) - x^2 = (x^2 - x^4/3 + O(x^6)) - x^2 = -x^4/3 + O(x^6)

Now divide by x^4: L = lim (x -> 0) [(-x^4/3 + O(x^6)) / x^4] L = lim (x -> 0) [-1/3 + O(x^2)] L = -1/3

Final Answer

lim (x -> 0) [x^(-2) - cosec^2 x] = -1/3

Q.1 (b)

Question

Check the consistency of the system of linear equations. Solve it if consistent. 3x + y - 3z = 13, 2x - 3y + 7z = 5, 2x + 19y - 47z = 32.

Solution

1. Matrix Representation

The given system of linear equations is:

3x + y - 3z = 13
2x - 3y + 7z = 5
2x + 19y - 47z = 32

In matrix form, A X = B, where: A = [3, 1, -3] [2, -3, 7] [2, 19, -47]

X = [x] [y] [z]

B = [13] [ 5] [32]

The augmented matrix [A | B] is: [ 3 1 -3 | 13 ] [ 2 -3 7 | 5 ] [ 2 19 -47 | 32 ]

2. Row Reduction to Echelon Form

Make the leading entry of R1 equal to 1 using R1 -> R1 - R2: [ 1 4 -10 | 8 ] [ 2 -3 7 | 5 ] [ 2 19 -47 | 32 ]

Eliminate x from Row 2 and Row 3:

Perform R2 -> R2 - 2*R1:

R2 = [2 - 2(1), -3 - 2(4), 7 - 2(-10) | 5 - 2(8)] R2 = [0, -11, 27 | -11]

Perform R3 -> R3 - 2*R1:

R3 = [2 - 2(1), 19 - 2(4), -47 - 2(-10) | 32 - 2(8)] R3 = [0, 11, -27 | 16]

The matrix is now: [ 1 4 -10 | 8 ] [ 0 -11 27 | -11 ] [ 0 11 -27 | 16 ]

Eliminate y from Row 3:

Perform R3 -> R3 + R2:

R3 = [0 + 0, 11 + (-11), -27 + 27 | 16 + (-11)] R3 = [0, 0, 0 | 5]

The reduced matrix is: [ 1 4 -10 | 8 ] [ 0 -11 27 | -11 ] [ 0 0 0 | 5 ]

3. Consistency Check

Rank of coefficient matrix A: rank(A) = 2 (number of non-zero rows in A is 2).
Rank of augmented matrix [A | B]: rank([A | B]) = 3 (number of non-zero rows in [A | B] is 3, because row 3 has entry 5).

Since rank(A) != rank([A | B]) (2 != 3), the system is inconsistent.

Final Answer

The system of equations is inconsistent and has no solution.

Q.1 (c)

Question

Find the Fourier Series of the function f(x) = x^2 in the interval (-pi, pi).

Solution

1. Definition and Symmetry Analysis

The interval is (-pi, pi), so the half-period is L = pi. The given function is f(x) = x^2.

Check for symmetry: f(-x) = (-x)^2 = x^2 = f(x) Since f(-x) = f(x), f(x) is an even function.

For an even function on (-pi, pi):

bn = 0 for all n >= 1.
The Fourier series consists only of cosine terms:

f(x) = a0/2 + sum_{n=1}^infinity [an * cos(nx)]

2. Calculate Fourier Coefficient a0

a0 = (2/pi) integral from 0 to pi of x^2 dx a0 = (2/pi) [x^3 / 3] from 0 to pi a0 = (2/pi) [pi^3 / 3] a0 = 2pi^2 / 3

Therefore: a0 / 2 = pi^2 / 3

3. Calculate Fourier Coefficient an

an = (2/pi) integral from 0 to pi of [x^2 cos(nx)] dx

Apply integration by parts using the generalized formula (Bernoulli's rule): integral [u v] dx = u v1 - u' v2 + u'' v3 - ... Let: u = x^2 ==> u' = 2x ==> u'' = 2 ==> u''' = 0 v = cos(nx) ==> v1 = sin(nx)/n ==> v2 = -cos(nx)/n^2 ==> v3 = -sin(nx)/n^3

Evaluating the integral: an = (2/pi) [ x^2 (sin(nx)/n) - 2x (-cos(nx)/n^2) + 2 (-sin(nx)/n^3) ] from 0 to pi an = (2/pi) [ (x^2 sin(nx)/n) + (2x cos(nx)/n^2) - (2 sin(nx)/n^3) ] from 0 to pi

Substitute the limits:

At x = pi: sin(npi) = 0 and cos(npi) = (-1)^n
At x = 0: all terms vanish because sin(0) = 0 and x = 0

an = (2/pi) [ 0 + (2pi (-1)^n / n^2) - 0 ] an = (2/pi) [ 2pi (-1)^n / n^2 ] an = 4 * (-1)^n / n^2

4. Write the Complete Fourier Series

Substitute a0/2 and an into the Fourier series expansion: f(x) = pi^2 / 3 + sum_{n=1}^infinity [ 4 (-1)^n / n^2 cos(nx) ]

Expanding the first few terms: x^2 = pi^2/3 - 4 * [ cos(x)/1^2 - cos(2x)/2^2 + cos(3x)/3^2 - cos(4x)/4^2 + ... ]

Final Answer

f(x) = pi^2/3 + 4 sum_{n=1}^infinity [ ((-1)^n / n^2) cos(nx) ]

Q.2 (a)

Question

Define the improper integrals of the first kind and the second kind. State the relation between Beta and Gamma function.

Solution

1. Improper Integral of the First Kind

An improper integral of the first kind is a definite integral where at least one of the integration limits is infinite, while the integrand f(x) remains bounded on the entire interval.

Form:
integral from a to infinity of f(x) dx, or
integral from -infinity to b of f(x) dx, or
integral from -infinity to infinity of f(x) dx.
Example: integral from 1 to infinity of (1/x^2) dx.

2. Improper Integral of the Second Kind

An improper integral of the second kind is an integral over a finite interval [a, b], where the integrand f(x) becomes unbounded (has an infinite discontinuity) at one or more points in [a, b].

Form:
integral from a to b of f(x) dx, where lim (x -> c) |f(x)| = infinity for some c in [a, b].
Example: integral from 0 to 1 of (1/sqrt(x)) dx, where the integrand becomes infinite at the lower limit x = 0.

3. Relation between Beta and Gamma Functions

The relationship connecting the Beta function B(m, n) and the Gamma function Gamma(n) is:

B(m, n) = [ Gamma(m) * Gamma(n) ] / Gamma(m + n)

Where m > 0 and n > 0.

Q.2 (b)

Question

Find the area of the surface of revolution generated by revolving the curve x = y^3 from y = 0 to y = 2.

Solution

Given:

Curve: x = y^3
Limits: y ranges from y = 0 to y = 2
Standard revolution: revolved about the y-axis

1. Formula

The surface area S generated by revolving the curve x = g(y) about the y-axis between y = c and y = d is: S = 2 pi integral from c to d of [ x * sqrt(1 + (dx/dy)^2) ] dy

2. Compute Derivative and Arc Length Element

Given x = y^3: dx/dy = 3y^2

Compute 1 + (dx/dy)^2: 1 + (dx/dy)^2 = 1 + (3y^2)^2 = 1 + 9y^4

Thus, the arc length differential is: ds = sqrt(1 + 9y^4) dy

3. Set Up and Evaluate the Integral

S = 2 pi integral from 0 to 2 of [ y^3 * sqrt(1 + 9y^4) ] dy

Use substitution: Let u = 1 + 9y^4 du = 36y^3 dy ==> y^3 dy = du / 36

Change limits of integration:

When y = 0: u = 1 + 9(0)^4 = 1
When y = 2: u = 1 + 9(2)^4 = 1 + 9(16) = 1 + 144 = 145

Substitute u and du into the integral: S = 2 pi integral from 1 to 145 of [ sqrt(u) (du / 36) ] S = (2 pi / 36) integral from 1 to 145 of [ u^(1/2) ] du S = (pi / 18) [ (2/3) u^(3/2) ] from 1 to 145 S = (pi / 27) [ 145^(3/2) - 1^(3/2) ] S = (pi / 27) [ 145 sqrt(145) - 1 ]

(Note: If the revolution is considered about the x-axis, the formula would be S = 2piintegral ysqrt(1 + (dx/dy)^2) dy = 2piintegral_0^2 ysqrt(1 + 9y^4) dy. By standard convention in GTU calculus papers, when a function is given as x = g(y) with y-limits, revolving about the y-axis is the standard interpretation).

Final Answer

Surface Area S = (pi / 27) (145 sqrt(145) - 1) square units (approximately 203.04 square units).

Q.2 (c)

Question

Find eigen values and corresponding eigen vectors of the matrix ( 0 1 1 ) ( 1 0 1 ) ( 1 1 0 ).

Solution

1. Characteristic Equation

Let the given matrix be: A = [0 1 1] [1 0 1] [1 1 0]

Expanding the determinant: -lambda [(-lambda)(-lambda) - (1)(1)] - 1 [1(-lambda) - 11] + 1 [11 - 1(-lambda)] = 0 -lambda (lambda^2 - 1) - 1 (-lambda - 1) + 1 (1 + lambda) = 0 -lambda^3 + lambda + lambda + 1 + 1 + lambda = 0 -lambda^3 + 3lambda + 2 = 0 lambda^3 - 3*lambda - 2 = 0

2. Finding the Eigenvalues

Test for roots:

For lambda = -1: (-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0. So lambda = -1 is a root.

Factorize (lambda + 1): lambda^3 - 3lambda - 2 = (lambda + 1)(lambda^2 - lambda - 2) = 0 (lambda + 1)(lambda + 1)(lambda - 2) = 0 (lambda + 1)^2 (lambda - 2) = 0

Therefore, the eigenvalues are: lambda1 = 2, lambda2 = -1, lambda_3 = -1

3. Eigenvector for lambda = 2

Solve (A - 2*I) X = 0: [ -2 1 1 ] [ x1 ] [ 0 ] [ 1 -2 1 ] [ x2 ] = [ 0 ] [ 1 1 -2 ] [ x3 ] [ 0 ]

From Row 1 and Row 2: -2x1 + x2 + x3 = 0 x1 - 2x2 + x3 = 0

Subtracting the second equation from the first: -3x1 + 3x2 = 0 ==> x1 = x2

Substitute x1 = x2 into the first equation: -2*x1 + x1 + x3 = 0 ==> x3 = x1

Thus, x1 = x2 = x3. Choosing x1 = 1: X_1 = [ 1 ] [ 1 ] [ 1 ]

4. Eigenvectors for Repeated Eigenvalue lambda = -1

Solve (A - (-1)*I) X = (A + I) X = 0: [ 1 1 1 ] [ x1 ] [ 0 ] [ 1 1 1 ] [ x2 ] = [ 0 ] [ 1 1 1 ] [ x3 ] [ 0 ]

All three rows are identical, giving the single linear equation: x1 + x2 + x3 = 0 ==> x1 = -x2 - x3

This system has two linearly independent free variables (x2 and x3):

Setting x2 = 1, x3 = 0:

x1 = -1 X_2 = [ -1 ] [ 1 ] [ 0 ]

Setting x2 = 0, x3 = 1:

x1 = -1 X_3 = [ -1 ] [ 0 ] [ 1 ]

Final Answer

Eigenvalues: lambda = 2, -1, -1
Eigenvectors:
For lambda = 2: X_1 = [1, 1, 1]^T
For lambda = -1: X2 = [-1, 1, 0]^T and X3 = [-1, 0, 1]^T

Q.2 (c) OR

Question

Find the inverse of the matrix using Gauss-Jordan elimination method. A = ( 1 2 3 2 3 0 0 1 2 ).

Solution

1. Setup Augmented Matrix

Set up the matrix [A | I_3]: [ 1 2 3 | 1 0 0 ] [ 2 3 0 | 0 1 0 ] [ 0 1 2 | 0 0 1 ]

2. Apply Row Operations

Step 1: Eliminate the first entry of Row 2 using R2 -> R2 - 2*R1:

R2 = [2 - 2(1), 3 - 2(2), 0 - 2(3) | 0 - 2(1), 1 - 2(0), 0 - 2(0)]
R2 = [0, -1, -6 | -2, 1, 0]

The matrix becomes: [ 1 2 3 | 1 0 0 ] [ 0 -1 -6 | -2 1 0 ] [ 0 1 2 | 0 0 1 ]

Step 2: Multiply Row 2 by -1 (R2 -> -1 * R2): [ 1 2 3 | 1 0 0 ] [ 0 1 6 | 2 -1 0 ] [ 0 1 2 | 0 0 1 ]

Step 3: Eliminate entries above and below the pivot in column 2:

R1 -> R1 - 2*R2:
R1 = [1 - 0, 2 - 2(1), 3 - 2(6) | 1 - 2(2), 0 - 2(-1), 0 - 0]
R1 = [1, 0, -9 | -3, 2, 0]

R3 -> R3 - R2:
R3 = [0 - 0, 1 - 1, 2 - 6 | 0 - 2, 0 - (-1), 1 - 0]
R3 = [0, 0, -4 | -2, 1, 1]

The matrix becomes: [ 1 0 -9 | -3 2 0 ] [ 0 1 6 | 2 -1 0 ] [ 0 0 -4 | -2 1 1 ]

Step 4: Make the pivot in Row 3 equal to 1 (R3 -> R3 / (-4)): [ 1 0 -9 | -3 2 0 ] [ 0 1 6 | 2 -1 0 ] [ 0 0 1 | 1/2 -1/4 -1/4 ]

Step 5: Eliminate entries in column 3 above the pivot:

R1 -> R1 + 9*R3:
R1_1 = -3 + 9*(1/2) = -3 + 9/2 = 3/2
R1_2 = 2 + 9*(-1/4) = 2 - 9/4 = -1/4
R1_3 = 0 + 9*(-1/4) = -9/4
R1 = [1, 0, 0 | 3/2, -1/4, -9/4]

R2 -> R2 - 6*R3:
R2_1 = 2 - 6*(1/2) = 2 - 3 = -1
R2_2 = -1 - 6*(-1/4) = -1 + 6/4 = 1/2
R2_3 = 0 - 6*(-1/4) = 6/4 = 3/2
R2 = [0, 1, 0 | -1, 1/2, 3/2]

The augmented matrix is now [I | A^(-1)]: [ 1 0 0 | 3/2 -1/4 -9/4 ] [ 0 1 0 | -1 1/2 3/2 ] [ 0 0 1 | 1/2 -1/4 -1/4 ]

Taking 1/4 as a common factor: A^(-1) = (1/4) * [ 6 -1 -9 ] [ -4 2 6 ] [ 2 -1 -1 ]

3. Verification

Multiply A by A^(-1): [1 2 3] [ 6 -1 -9 ] [ 16 + 2(-4) + 32, 1(-1) + 22 + 3(-1), 1(-9) + 26 + 3(-1) ] [2 3 0] [ -4 2 6 ] = [ 26 + 3(-4) + 0, 2(-1) + 32 + 0, 2(-9) + 36 + 0 ] [0 1 2] [ 2 -1 -1 ] [ 0 + 1(-4) + 22, 0 + 12 + 2(-1), 0 + 16 + 2(-1) ]

= [ 6 - 8 + 6, -1 + 4 - 3, -9 + 12 - 3 ] [ 4 0 0 ] [ 12 - 12 + 0, -2 + 6 + 0, -18 + 18 + 0] = [ 0 4 0 ] [ -4 + 4, 2 - 2, 6 - 2 ] [ 0 0 4 ]

Multiplying by (1/4) yields I_3. The result is confirmed.

Final Answer

A^(-1) = [ 3/2 -1/4 -9/4 ] [ -1 1/2 3/2 ] [ 1/2 -1/4 -1/4 ]

Q.3 (a)

Question

Determine whether lim ((x, y) -> (0, 0)) (x^2 - y^2) / (x^2 + y^2) exists and find it if exists.

Solution

1. Approach

To determine if the multivariable limit exists, evaluate the limit as (x, y) approaches (0, 0) along different linear paths y = m*x.

2. Path Testing

Let y = m*x, where m is the slope of the straight line path passing through the origin.

Substitute y = mx into the given expression: L = lim (x -> 0) [ (x^2 - (mx)^2) / (x^2 + (mx)^2) ] L = lim (x -> 0) [ x^2 (1 - m^2) / (x^2 * (1 + m^2)) ]

Since x != 0 as x -> 0, cancel x^2: L = (1 - m^2) / (1 + m^2)

3. Conclusion

Along the path y = 0 (x-axis, m = 0):
L = (1 - 0) / (1 + 0) = 1

Along the path x = 0 (y-axis, m -> infinity):
L = -1

Along the path y = x (m = 1):
L = (1 - 1) / (1 + 1) = 0

Since the limit depends on the slope m of the path approaching the origin, the limit is not unique.

Final Answer

The limit does not exist.

Q.3 (b)

Question

Find the equation of tangent plane and normal line to the surface 2xz^2 - 3xy - 4x = 7 at the point (1, -1, 2).

Solution

1. Define the Surface Function

Let the implicit surface be: F(x, y, z) = 2xz^2 - 3xy - 4*x - 7 = 0 Given point P = (x0, y0, z0) = (1, -1, 2).

Check that P lies on the surface: F(1, -1, 2) = 2(1)(2^2) - 3(1)(-1) - 4(1) - 7 = 8 + 3 - 4 - 7 = 0 (Confirmed).

2. Compute the Gradient Vector (Normal Vector)

The normal vector to the surface is grad F = (dF/dx) i + (dF/dy) j + (dF/dz) k.

Calculate the partial derivatives:

dF/dx = 2z^2 - 3y - 4
dF/dy = -3*x
dF/dz = 4xz

Evaluate the partial derivatives at P(1, -1, 2):

dF/dx |_(1, -1, 2) = 2(2)^2 - 3(-1) - 4 = 8 + 3 - 4 = 7
dF/dy |_(1, -1, 2) = -3*(1) = -3
dF/dz |_(1, -1, 2) = 4(1)(2) = 8

Therefore, the normal vector is: N = grad F |_(1, -1, 2) = 7 i - 3 j + 8 k

3. Equation of the Tangent Plane

The equation of the tangent plane at (x0, y0, z0) is: (dF/dx)(x - x0) + (dF/dy)(y - y0) + (dF/dz)*(z - z0) = 0

Substitute the values: 7(x - 1) + (-3)(y - (-1)) + 8(z - 2) = 0 7(x - 1) - 3(y + 1) + 8(z - 2) = 0 7x - 7 - 3y - 3 + 8z - 16 = 0 7x - 3y + 8z - 26 = 0 7x - 3y + 8z = 26

4. Equation of the Normal Line

The symmetric equation of the normal line is: (x - x0) / (dF/dx) = (y - y0) / (dF/dy) = (z - z0) / (dF/dz)

Substitute the values: (x - 1) / 7 = (y + 1) / (-3) = (z - 2) / 8

Final Answer

Tangent Plane: 7x - 3y + 8z = 26
Normal Line: (x - 1)/7 = (y + 1)/(-3) = (z - 2)/8

Q.3 (c)

Question

Find the extreme values of the function x^3 + y^3 - 63(x + y) + 12xy.

Solution

1. Given Function

f(x, y) = x^3 + y^3 - 63x - 63y + 12xy

2. Find Critical Points

Compute first-order partial derivatives: fx = df/dx = 3x^2 - 63 + 12y fy = df/dy = 3y^2 - 63 + 12x

Set fx = 0 and fy = 0:

3x^2 + 12y - 63 = 0 ==> x^2 + 4y - 21 = 0 ==> 4y = 21 - x^2
3y^2 + 12x - 63 = 0 ==> y^2 + 4x - 21 = 0 ==> 4x = 21 - y^2

Subtract equation (2) from equation (1): (x^2 - y^2) + 4(y - x) = 0 (x - y)(x + y) - 4(x - y) = 0 (x - y) * (x + y - 4) = 0

This gives two cases:

Case 1: x = y
Case 2: x + y = 4 ==> y = 4 - x

Solving Case 1: x = y

Substitute y = x into x^2 + 4y - 21 = 0: x^2 + 4x - 21 = 0 (x + 7)(x - 3) = 0 So, x = 3 or x = -7. Since y = x, we get two critical points:

Point A: (3, 3)
Point B: (-7, -7)

Solving Case 2: y = 4 - x

Substitute y = 4 - x into x^2 + 4y - 21 = 0: x^2 + 4*(4 - x) - 21 = 0 x^2 + 16 - 4x - 21 = 0 x^2 - 4x - 5 = 0 (x - 5)(x + 1) = 0 So, x = 5 or x = -1.

If x = 5: y = 4 - 5 = -1 ==> Point C: (5, -1)
If x = -1: y = 4 - (-1) = 5 ==> Point D: (-1, 5)

Thus, the four critical points are: (3, 3), (-7, -7), (5, -1), and (-1, 5).

3. Second Derivative Test

Compute the second-order partial derivatives: r = fxx = d^2f/dx^2 = 6x s = fxy = d^2f/(dx dy) = 12 t = fyy = d^2f/dy^2 = 6y

Discriminant: D = rt - s^2 = (6x)(6y) - 12^2 = 36xy - 144

Now test each critical point:

At (3, 3):

r = 6(3) = 18 > 0
D = 36(3)(3) - 144 = 324 - 144 = 180 > 0
Since D > 0 and r > 0, f has a local minimum at (3, 3).
Value: f(3, 3) = 3^3 + 3^3 - 63(3 + 3) + 12(3)(3) = 27 + 27 - 378 + 108 = -216.

At (-7, -7):

r = 6(-7) = -42 < 0
D = 36(-7)(-7) - 144 = 36(49) - 144 = 1764 - 144 = 1620 > 0
Since D > 0 and r < 0, f has a local maximum at (-7, -7).
Value: f(-7, -7) = (-7)^3 + (-7)^3 - 63(-14) + 12(49) = -343 - 343 + 882 + 588 = 784.

At (5, -1):

D = 36(5)(-1) - 144 = -180 - 144 = -324 < 0
Since D < 0, (5, -1) is a saddle point.

At (-1, 5):

D = 36(-1)(5) - 144 = -180 - 144 = -324 < 0
Since D < 0, (-1, 5) is a saddle point.

Final Answer

Local Maximum Value: 784 at (-7, -7)
Local Minimum Value: -216 at (3, 3)
Saddle Points: (5, -1) and (-1, 5)

Q.3 (a) OR

Question

State chain rule for du/dx for u = f(v, w), where v = g(x, y), w = h(x, y). If u = f(x - 2y, 2y - 3z, 3z - x) then show that du/dx + (1/2)du/dy + (1/3)du/dz = 0.

Solution

1. Statement of Chain Rule

Let u = f(v, w) be a differentiable function of two variables v and w, where v = g(x, y) and w = h(x, y) are differentiable functions of x and y. Then the partial derivative of u with respect to x is given by the chain rule:

du/dx = (du/dv) (dv/dx) + (du/dw) (dw/dx)

2. Proof of Identity

(Note: The question paper has a minor typographical print '1/3 du/dy = 0'; the standard multivariable calculus identity is du/dx + (1/2)du/dy + (1/3)du/dz = 0).

Let the intermediate variables be defined as: r = x - 2y s = 2y - 3z t = 3z - x

Thus, u = f(r, s, t).

Step 2.1: Partial derivative with respect to x du/dx = (du/dr)(dr/dx) + (du/ds)(ds/dx) + (du/dt)(dt/dx) Since dr/dx = 1, ds/dx = 0, dt/dx = -1: du/dx = (du/dr)(1) + (du/ds)(0) + (du/dt)(-1) du/dx = du/dr - du/dt --- (Equation 1)

Step 2.2: Partial derivative with respect to y du/dy = (du/dr)(dr/dy) + (du/ds)(ds/dy) + (du/dt)(dt/dy) Since dr/dy = -2, ds/dy = 2, dt/dy = 0: du/dy = (du/dr)(-2) + (du/ds)(2) + (du/dt)(0) du/dy = -2(du/dr) + 2(du/ds)

Multiply by 1/2: (1/2) * (du/dy) = -du/dr + du/ds --- (Equation 2)

Step 2.3: Partial derivative with respect to z du/dz = (du/dr)(dr/dz) + (du/ds)(ds/dz) + (du/dt)(dt/dz) Since dr/dz = 0, ds/dz = -3, dt/dz = 3: du/dz = (du/dr)(0) + (du/ds)(-3) + (du/dt)(3) du/dz = -3(du/ds) + 3(du/dt)

Multiply by 1/3: (1/3) * (du/dz) = -du/ds + du/dt --- (Equation 3)

Step 2.4: Sum the three expressions Adding Equations (1), (2), and (3): du/dx + (1/2)du/dy + (1/3)du/dz = (du/dr - du/dt) + (-du/dr + du/ds) + (-du/ds + du/dt) = (du/dr - du/dr) + (du/ds - du/ds) + (du/dt - du/dt) = 0

Hence proved.

Q.3 (b) OR

Question

Find the directional derivative of f(x, y, z) = xy^2 + yz^2 at the point (2, -1, 1) in the direction of the vector i + 2j + 2k.

Solution

1. Formula

The directional derivative of a scalar function f in the direction of a vector a is given by: Du f = grad f . ucap where:

grad f = (df/dx) i + (df/dy) j + (df/dz) k
u_cap = a / |a| is the unit vector in the direction of a.

2. Compute the Gradient of f

Given f(x, y, z) = xy^2 + yz^2:

df/dx = y^2
df/dy = 2xy + z^2
df/dz = 2yz

Evaluate grad f at the point P(2, -1, 1):

df/dx = (-1)^2 = 1
df/dy = 2(2)(-1) + (1)^2 = -4 + 1 = -3
df/dz = 2(-1)(1) = -2

Thus: grad f |_(2, -1, 1) = 1 i - 3 j - 2 k

3. Compute the Unit Vector

Given vector a = i + 2 j + 2 k. Magnitude of a: |a| = sqrt(1^2 + 2^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3

Unit vector: u_cap = (1/3) * (i + 2 j + 2 k)

4. Calculate Directional Derivative

Du f = grad f . ucap = (1 i - 3 j - 2 k) . [(1/3) (i + 2 j + 2 k)] = (1/3) [ (1)(1) + (-3)(2) + (-2)(2) ] = (1/3) [ 1 - 6 - 4 ] = (1/3) (-9) = -3

Final Answer

The directional derivative is -3.

Q.3 (c) OR

Question

The temperature T(x, y, z) at any point in space is T = 400xy*z^2. Find the highest temperature on surface of the sphere x^2 + y^2 + z^2 = 1.

Solution

1. Method Setup (Lagrange Multipliers)

We want to maximize: T(x, y, z) = 400 x y * z^2 subject to the constraint: g(x, y, z) = x^2 + y^2 + z^2 - 1 = 0

Using the method of Lagrange multipliers: grad T = lambda * grad g

2. Form the System of Equations

Compute the gradients:

grad T = (400yz^2) i + (400xz^2) j + (800xy*z) k
grad g = (2x) i + (2y) j + (2z) k

Equating components:

400yz^2 = 2lambdax ==> 200yz^2 = lambda*x
400xz^2 = 2lambday ==> 200xz^2 = lambda*y
800xyz = 2lambdaz ==> 400xyz = lambda*z
x^2 + y^2 + z^2 = 1

3. Analyze the Equations

To maximize T, we need T > 0, so x != 0, y != 0, and z != 0.

Multiply (1) by x: lambdax^2 = 200xyz^2
Multiply (2) by y: lambday^2 = 200xyz^2
Multiply (3) by z: lambdaz^2 = 400xyz^2

Since the right-hand sides of the first two are identical and lambda != 0: lambdax^2 = lambday^2 ==> x^2 = y^2

Also, comparing lambdaz^2 with lambdax^2: lambdaz^2 = 2 (lambdax^2) ==> z^2 = 2x^2

4. Substitute into the Constraint Equation

Substitute y^2 = x^2 and z^2 = 2x^2 into x^2 + y^2 + z^2 = 1: x^2 + x^2 + 2x^2 = 1 4*x^2 = 1 x^2 = 1/4 ==> x = +- 1/2

Then: y^2 = 1/4 ==> y = +- 1/2 z^2 = 2*(1/4) = 1/2 ==> z = +- 1/sqrt(2)

5. Find the Maximum Value of T

For maximum temperature, we choose signs of x and y such that x*y > 0 (both positive or both negative):

x = 1/2, y = 1/2 (or x = -1/2, y = -1/2)
z^2 = 1/2

Substitute these into T: Tmax = 400 * (1/2) * (1/2) * (1/2) Tmax = 400 * (1/8) T_max = 50

Final Answer

The highest temperature on the surface of the sphere is 50 units.

Q.4 (a)

Question

Evaluate integral from 0 to 2 of integral from -1 to 1 of (1 - 6x^2y) dx dy.

Solution

1. Integral Setup

We are asked to evaluate the iterated double integral: I = integral (y = 0 to 2) [ integral (x = -1 to 1) (1 - 6x^2y) dx ] dy

2. Inner Integration (with respect to x)

Integrate with respect to x, treating y as a constant: inner_I = integral (x = -1 to 1) (1 - 6x^2y) dx = [ x - 6 (x^3 / 3) y ] from x = -1 to 1 = [ x - 2x^3y ] from x = -1 to 1

Substitute the limits: At x = 1: 1 - 2(1)^3y = 1 - 2y At x = -1: -1 - 2(-1)^3y = -1 + 2y

inner_I = (1 - 2y) - (-1 + 2y) = 1 - 2y + 1 - 2y = 2 - 4y

3. Outer Integration (with respect to y)

Now integrate the result with respect to y from 0 to 2: I = integral (y = 0 to 2) (2 - 4y) dy = [ 2y - 4 (y^2 / 2) ] from y = 0 to 2 = [ 2y - 2y^2 ] from y = 0 to 2

Substitute the limits: At y = 2: 2(2) - 2(2)^2 = 4 - 8 = -4 At y = 0: 2(0) - 2(0)^2 = 0

I = -4 - 0 = -4

Final Answer

The value of the integral is -4.

Q.4 (b)

Question

Find the value of double integral over R of (2x - y^2) dA over the triangular region R enclosed between the line y = -x + 1, y = x + 1 and y = 3.

Solution

1. Region Description

The bounding lines are:

y = -x + 1 ==> x = 1 - y
y = x + 1 ==> x = y - 1
y = 3

Find the vertices of the triangular region R:

Intersection of y = -x + 1 and y = x + 1:
-x + 1 = x + 1 ==> 2x = 0 ==> x = 0, y = 1. Vertex: (0, 1)
Intersection of y = -x + 1 and y = 3:
3 = -x + 1 ==> x = -2. Vertex: (-2, 3)
Intersection of y = x + 1 and y = 3:
3 = x + 1 ==> x = 2. Vertex: (2, 3)

2. Integration Limits

Since y goes from 1 to 3, for each fixed horizontal line y:

Left boundary: x = -(y - 1) = 1 - y
Right boundary: x = y - 1

Thus, the limits of integration are:

x runs from (1 - y) to (y - 1)
y runs from 1 to 3

3. Evaluate the Double Integral

I = integral (y = 1 to 3) [ integral (x = 1 - y to y - 1) (2x - y^2) dx ] dy

Inner integral with respect to x: integral (x = 1 - y to y - 1) (2x - y^2) dx = [ x^2 - y^2 * x ] from x = -(y - 1) to (y - 1)

Note that x^2 is an even function of x, so [x^2] from -(y-1) to (y-1) is 0. Specifically: = [ (y - 1)^2 - y^2(y - 1) ] - [ (-(y - 1))^2 - y^2(-(y - 1)) ] = [ (y - 1)^2 - y^2(y - 1) ] - [ (y - 1)^2 + y^2(y - 1) ] = -2 y^2 (y - 1) = -2y^3 + 2y^2

Outer integral with respect to y: I = integral (y = 1 to 3) (-2y^3 + 2y^2) dy = [ -2 (y^4 / 4) + 2 (y^3 / 3) ] from 1 to 3 = [ -y^4 / 2 + 2*y^3 / 3 ] from 1 to 3

Substitute upper limit y = 3: -3^4 / 2 + 2(3^3) / 3 = -81/2 + 2(27)/3 = -81/2 + 18 = -81/2 + 36/2 = -45/2

Substitute lower limit y = 1: -1^4 / 2 + 2*(1^3) / 3 = -1/2 + 2/3 = -3/6 + 4/6 = 1/6

Subtract: I = -45/2 - 1/6 = -135/6 - 1/6 = -136/6 = -68/3

Final Answer

The value of the integral is -68/3.

Q.4 (c)

Question

Change the order of integral and evaluate integral from 0 to 4a of integral from x^2/(4a) to 2*sqrt(ax) dy dx, (a > 0).

Solution

1. Identify the Original Limits and Curves

The given integral is: I = integral (x = 0 to 4a) [ integral (y = x^2 / (4a) to 2*sqrt(ax)) dy ] dx

Lower curve: y = x^2 / (4a) ==> x^2 = 4a*y (parabola opening upward)
Upper curve: y = 2sqrt(ax) ==> y^2 = 4ax (parabola opening to the right)

2. Find the Intersection Points

Substitute y = x^2 / (4a) into y^2 = 4ax: (x^2 / (4a))^2 = 4ax x^4 / (16a^2) = 4ax x^4 - 64a^3 x = 0 x (x^3 - 64a^3) = 0

This gives x = 0 and x = 4a.

At x = 0: y = 0 ==> (0, 0)
At x = 4a: y = (4a)^2 / (4a) = 4a ==> (4a, 4a)

3. Change the Order of Integration

In the original order, vertical strips are used (y varies from bottom parabola to top parabola, x from 0 to 4a).

To reverse the order, use horizontal strips:

y varies from 0 to 4a.
For a given y, x enters the region at the curve y^2 = 4ax (left curve) and exits at the curve x^2 = 4ay (right curve).
Left boundary: x = y^2 / (4a)
Right boundary: x = sqrt(4ay) = 2sqrt(a*y)

Therefore, with the order changed, the integral becomes: I = integral (y = 0 to 4a) [ integral (x = y^2 / (4a) to 2*sqrt(ay)) dx ] dy

4. Evaluate the Integral

Inner integral with respect to x: integral (x = y^2 / (4a) to 2sqrt(ay)) dx = [ x ] from y^2 / (4a) to 2sqrt(ay) = 2sqrt(a)y^(1/2) - y^2 / (4a)

Outer integral with respect to y: I = integral (y = 0 to 4a) [ 2sqrt(a)y^(1/2) - y^2 / (4a) ] dy = [ 2sqrt(a) (y^(3/2) / (3/2)) - (1 / (4a)) (y^3 / 3) ] from 0 to 4a = [ (4/3)sqrt(a)*y^(3/2) - y^3 / (12a) ] from 0 to 4a

Evaluate at y = 4a:

First term: (4/3)sqrt(a) (4a)^(3/2) = (4/3)sqrt(a) 8 a^(3/2) = (32/3) a^2
Second term: (4a)^3 / (12a) = 64a^3 / (12a) = (16/3) a^2

Subtract the terms: I = (32/3)a^2 - (16/3)a^2 I = (16/3) * a^2

Final Answer

After changing the order of integration, the value is (16/3) * a^2.

Q.4 (a) OR

Question

Calculate integral from 0 to 2 of integral from 0 to 1 of integral from 0 to yz of xyz dx dy dz.

Solution

1. Set Up the Triple Integral

I = integral (z = 0 to 2) [ integral (y = 0 to 1) [ integral (x = 0 to yz) (xyz) dx ] dy ] dz

2. First Integration (with respect to x)

Treat y and z as constants: inner_x = integral (x = 0 to yz) (xyz) dx = yz [ x^2 / 2 ] from x = 0 to yz = yz [ (yz)^2 / 2 - 0 ] = yz (y^2 z^2 / 2) = (1/2) y^3 * z^3

3. Second Integration (with respect to y)

Now integrate with respect to y from 0 to 1: inner_y = integral (y = 0 to 1) [ (1/2) y^3 z^3 ] dy = (1/2) z^3 [ y^4 / 4 ] from y = 0 to 1 = (1/2) z^3 (1/4 - 0) = (1/8) * z^3

4. Third Integration (with respect to z)

Now integrate with respect to z from 0 to 2: I = integral (z = 0 to 2) [ (1/8) z^3 ] dz = (1/8) [ z^4 / 4 ] from z = 0 to 2 = (1/32) [ 2^4 - 0 ] = (1/32) 16 = 1/2

Final Answer

The value of the triple integral is 1/2.

Q.4 (b) OR

Question

Evaluate integral from 0 to 1 of integral from x to 1 of sin(y^2) dy dx.

Solution

1. Observe the Given Integral

The given integral is: I = integral (x = 0 to 1) [ integral (y = x to 1) sin(y^2) dy ] dx

The integrand sin(y^2) cannot be integrated analytically with respect to y in terms of elementary functions. Therefore, we must change the order of integration.

2. Identify the Region of Integration

The limits in the given integral are:

x varies from 0 to 1
For a fixed x, y varies from the line y = x to y = 1

This describes a triangular region bounded by:

y = x
y = 1
x = 0 (the y-axis)

3. Reverse the Order of Integration

Using horizontal strips instead of vertical strips:

y varies from 0 to 1
For a given y, x varies from x = 0 to x = y

Thus, the integral becomes: I = integral (y = 0 to 1) [ integral (x = 0 to y) sin(y^2) dx ] dy

4. Evaluate the Integral

Inner integral with respect to x: integral (x = 0 to y) sin(y^2) dx = [ x sin(y^2) ] from x = 0 to y = y sin(y^2)

Outer integral with respect to y: I = integral (y = 0 to 1) y * sin(y^2) dy

Use substitution: Let u = y^2 ==> du = 2y dy ==> y dy = du / 2

When y = 0: u = 0
When y = 1: u = 1

Substitute: I = integral (u = 0 to 1) sin(u) (du / 2) = (1/2) [ -cos(u) ] from 0 to 1 = (1/2) [ -cos(1) - (-cos(0)) ] = (1/2) [ 1 - cos(1) ]

Final Answer

The value of the integral is (1 - cos(1)) / 2.

Q.4 (c) OR

Question

Compute integral from 0 to 2 of integral from 0 to sqrt(2x - x^2) of [ x / (x^2 + y^2) ] dy dx by transforming into polar coordinates.

Solution

1. Identify the Region in Cartesian Coordinates

The limits are:

y varies from 0 to sqrt(2x - x^2)
x varies from 0 to 2

The upper boundary is: y = sqrt(2x - x^2) y^2 = 2x - x^2 x^2 + y^2 - 2x = 0 (x - 1)^2 + y^2 = 1

This is the upper half of a circle of radius 1 centered at (1, 0), since y >= 0.

2. Transform into Polar Coordinates

The standard transformations are: x = r cos(theta) y = r sin(theta) x^2 + y^2 = r^2 dx dy = r dr dtheta

Transform the circle equation: x^2 + y^2 = 2x ==> r^2 = 2rcos(theta) ==> r = 2*cos(theta)

Determine the limits in polar coordinates:

Since the region lies entirely in the first quadrant (x >= 0 and y >= 0), theta ranges from 0 to pi/2.
For a given theta, r ranges from 0 to 2*cos(theta).

3. Transform the Integrand

The integrand is: x / (x^2 + y^2) = (r * cos(theta)) / r^2 = cos(theta) / r

Including the Jacobian r: [ x / (x^2 + y^2) ] dx dy = [ cos(theta) / r ] * r dr dtheta = cos(theta) dr dtheta

4. Evaluate the Integral

I = integral (theta = 0 to pi/2) [ integral (r = 0 to 2*cos(theta)) cos(theta) dr ] dtheta

Inner integral with respect to r: integral (r = 0 to 2cos(theta)) cos(theta) dr = cos(theta) [ r ] from 0 to 2cos(theta) = cos(theta) [ 2cos(theta) - 0 ] = 2 cos^2(theta)

Outer integral with respect to theta: I = integral (theta = 0 to pi/2) 2 * cos^2(theta) dtheta

Using the identity 2cos^2(theta) = 1 + cos(2theta): I = integral (theta = 0 to pi/2) [ 1 + cos(2theta) ] dtheta = [ theta + sin(2theta) / 2 ] from 0 to pi/2 = [ pi/2 + sin(pi) / 2 ] - [ 0 + sin(0) / 2 ] = pi/2 + 0 - 0 = pi/2

Final Answer

The value of the integral is pi / 2.

Q.5 (a)

Question

Define Monotonic sequence. Test the convergence of the sequence {2 - (-1)^n}.

Solution

1. Definition of Monotonic Sequence

A sequence {a_n} of real numbers is said to be monotonic if its terms either consistently increase or consistently decrease for all n in the natural numbers:

Monotonically Increasing: a(n+1) >= an for all n >= 1.
Monotonically Decreasing: a(n+1) <= an for all n >= 1.

If a sequence is either monotonically increasing or monotonically decreasing, it is called monotonic.

2. Test the Convergence of {a_n} = {2 - (-1)^n}

Write out the first few terms of the sequence for n = 1, 2, 3, 4, ...:

For n = 1: a_1 = 2 - (-1)^1 = 2 - (-1) = 3
For n = 2: a_2 = 2 - (-1)^2 = 2 - 1 = 1
For n = 3: a_3 = 2 - (-1)^3 = 2 - (-1) = 3
For n = 4: a_4 = 2 - (-1)^4 = 2 - 1 = 1

The sequence is: {3, 1, 3, 1, 3, 1, ...}

3. Conclusion on Convergence

As n -> infinity, the terms alternate infinitely between 3 and 1:

lim (n -> infinity, n is odd) a_n = 3
lim (n -> infinity, n is even) a_n = 1

Since the sequence oscillates between two distinct values, the limit lim (n -> infinity) a_n does not exist uniquely.

Final Answer

The sequence {2 - (-1)^n} is oscillatory (non-convergent) and does not converge.

Q.5 (b)

Question

Express the function f(x) = log(1 + x) in power series using the formula of Maclaurin’s series.

Solution

1. Maclaurin's Series Formula

The Maclaurin series expansion of a function f(x) about x = 0 is: f(x) = f(0) + f'(0)x + [f''(0)/2!]x^2 + [f'''(0)/3!]x^3 + [f^(4)(0)/4!]x^4 + ... + [f^(n)(0)/n!]*x^n + ...

2. Derivatives and Values at x = 0

Given f(x) = log(1 + x) (natural log):

f(0) = log(1 + 0) = log(1) = 0

First derivative:

f'(x) = 1 / (1 + x) = (1 + x)^(-1)
f'(0) = 1

Second derivative:

f''(x) = -1 * (1 + x)^(-2)
f''(0) = -1

Third derivative:

f'''(x) = 2 * (1 + x)^(-3)
f'''(0) = 2

Fourth derivative:

f^(4)(x) = -6 * (1 + x)^(-4)
f^(4)(0) = -6

Fifth derivative:

f^(5)(x) = 24 * (1 + x)^(-5)
f^(5)(0) = 24

In general, the n-th derivative at x = 0 is: f^(n)(0) = (-1)^(n-1) * (n - 1)!

3. Substitute into the Formula

Substitute the evaluated derivatives into the series: log(1 + x) = 0 + 1x + [(-1)/2!]x^2 + [2/3!]x^3 + [(-6)/4!]x^4 + [24/5!]*x^5 + ...

Simplify the coefficients:

x term: x
x^2 term: -1/2 * x^2
x^3 term: 2/6 x^3 = 1/3 x^3
x^4 term: -6/24 x^4 = -1/4 x^4
x^5 term: 24/120 x^5 = 1/5 x^5

Thus: log(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - ... for -1 < x <= 1

Final Answer

log(1 + x) = sum_{n=1}^infinity [ ((-1)^(n-1) / n) * x^n ] = x - x^2/2 + x^3/3 - x^4/4 + ...

Q.5 (c)

Question

(i) Test the convergence of sum (n = 1 to infinity) [ n! / (2n + 1)! ]. (ii) State Cauchy’s root test and discuss the convergence of sum (n = 1 to infinity) [ 1 / (log n)^n ].

Solution

Part (i): Convergence of sum_{n=1}^infinity [ n! / (2n + 1)! ]

1. Identify general term: un = n! / (2n + 1)! u(n+1) = (n + 1)! / (2(n + 1) + 1)! = (n + 1)! / (2n + 3)!

2. Apply D'Alembert's Ratio Test: Compute the ratio u(n+1) / un: u(n+1) / un = [ (n + 1)! / (2n + 3)! ] [ (2n + 1)! / n! ] = [ (n + 1) n! / ((2n + 3)(2n + 2)(2n + 1)!) ] * [ (2n + 1)! / n! ] = (n + 1) / [ (2n + 3)(2n + 2) ] = (n + 1) / [ 2(n + 1)(2n + 3) ] = 1 / [ 2(2n + 3) ]

3. Take the limit as n -> infinity: L = lim (n -> infinity) [ u(n+1) / un ] = lim (n -> infinity) [ 1 / (4n + 6) ] = 0

4. Conclusion: Since L = 0 < 1, by D'Alembert's Ratio Test, the series converges.

---

Part (ii): Cauchy's Root Test and sum_{n=2}^infinity [ 1 / (log n)^n ]

1. Statement of Cauchy's Root Test: Let sum un be a positive-term series and let: L = lim (n -> infinity) (un)^(1/n)

If L < 1, the series converges.
If L > 1, the series diverges.
If L = 1, the test is inconclusive.

2. Test the Series: Given u_n = 1 / (log n)^n for n >= 2 (for n=1, log 1 = 0, so the series naturally starts from n = 2):

Compute (un)^(1/n): (un)^(1/n) = [ 1 / (log n)^n ]^(1/n) = 1 / log(n)

3. Take the limit as n -> infinity: L = lim (n -> infinity) (u_n)^(1/n) = lim (n -> infinity) [ 1 / log(n) ] Since log(n) -> infinity as n -> infinity: L = 0

4. Conclusion: Since L = 0 < 1, by Cauchy's Root Test, the series converges.

Q.5 (a) OR

Question

State sandwich theorem for sequence. Show that the sequence u_n = (sin n) / n converges to zero.

Solution

1. Statement of Sandwich (Squeeze) Theorem for Sequences

Let {an}, {bn}, and {cn} be three sequences of real numbers such that: an <= bn <= cn for all n >= N (for some positive integer N).

If: lim (n -> infinity) an = lim (n -> infinity) cn = L

then the sequence {bn} also converges to the same limit: lim (n -> infinity) bn = L

2. Application to u_n = (sin n) / n

For any real number n, the sine function satisfies the inequality: -1 <= sin(n) <= 1

Since n is a positive integer (n >= 1), divide the entire inequality by n: -1/n <= (sin n) / n <= 1/n

Here:

Lower bounding sequence: a_n = -1/n
Middle sequence: u_n = (sin n) / n
Upper bounding sequence: c_n = 1/n

3. Take Limits

Evaluate the limits of the bounding sequences as n -> infinity:

lim (n -> infinity) a_n = lim (n -> infinity) (-1/n) = 0
lim (n -> infinity) c_n = lim (n -> infinity) (1/n) = 0

4. Conclusion

Since -1/n <= (sin n)/n <= 1/n for all n >= 1, and both outer limits equal 0, by the Sandwich Theorem: lim (n -> infinity) [ (sin n) / n ] = 0

Hence, the sequence converges to 0.

Q.5 (b) OR

Question

Using Taylor’s theorem find the approximate value of sqrt(10) up to three decimal places.

Solution

1. Setup Function and Expansion Point

We want to approximate sqrt(10). Let f(x) = sqrt(x) = x^(1/2).

Choose an expansion base point a close to 10 that is a perfect square: Let a = 9, so h = x - a = 10 - 9 = 1.

2. Taylor's Theorem

Taylor's expansion of f(a + h) is given by: f(a + h) = f(a) + h f'(a) + (h^2 / 2!) f''(a) + (h^3 / 3!) * f'''(a) + ...

3. Calculate Derivatives at a = 9

f(x) = x^(1/2)
f(9) = sqrt(9) = 3

f'(x) = (1/2) x^(-1/2) = 1 / (2 sqrt(x))
f'(9) = 1 / (2 * 3) = 1/6

f''(x) = (1/2) (-1/2) x^(-3/2) = -1 / (4 * x^(3/2))
f''(9) = -1 / (4 * 27) = -1 / 108

f'''(x) = (-1/4) (-3/2) x^(-5/2) = 3 / (8 * x^(5/2))
f'''(9) = 3 / (8 243) = 1 / (8 81) = 1 / 648

4. Substitute Values (with h = 1)

sqrt(10) = f(9 + 1) approx f(9) + 1 f'(9) + (1/2) f''(9) + (1/6) f'''(9) = 3 + (1/6) - (1/2) (1/108) + (1/6) * (1/648) = 3 + 1/6 - 1/216 + 1/3888

Convert each fraction to decimal form:

1/6 = 0.166667
1/216 = 0.004630
1/3888 = 0.000257

Calculate step by step: sqrt(10) approx 3 + 0.166667 - 0.004630 + 0.000257 = 3.166667 - 0.004630 + 0.000257 = 3.162037 + 0.000257 = 3.162294

Rounding to three decimal places: sqrt(10) approx 3.162

(Check: 3.162^2 = 9.998244, which is extremely accurate).

Final Answer

The approximate value of sqrt(10) is 3.162.

Q.5 (c) OR

Question

(i) Examine the convergence of the series 1/(12) - 1/(23) + 1/(34) - 1/(45) + ... (ii) State Cauchy’s Integral test for convergence of series and test the convergence of the series sum (n = 1 to infinity) [ n^2 * e^(-n^3) ].

Solution

Part (i): Convergence of 1/(12) - 1/(23) + 1/(3*4) - ...

1. Identify General Term: The given alternating series is: sum{n=1}^infinity (-1)^(n-1) * un where u_n = 1 / [n * (n + 1)].

2. Apply Leibniz's Test for Alternating Series: An alternating series sum (-1)^(n-1) * u_n converges if:

u_n > 0 for all n.
u(n+1) <= un for all n (terms are monotonically decreasing).
lim (n -> infinity) u_n = 0.

Verification of Conditions:

For all n >= 1, u_n = 1 / [n(n + 1)] > 0. (Condition satisfied)
Compare u(n+1) and un:
u_(n+1) = 1 / [(n + 1)(n + 2)]
Since (n + 1)(n + 2) > n(n + 1), it follows that:
1 / [(n + 1)(n + 2)] < 1 / [n(n + 1)] ==> u(n+1) < un.
Thus, {u_n} is strictly decreasing. (Condition satisfied)
Limit:
lim (n -> infinity) u_n = lim (n -> infinity) [ 1 / (n^2 + n) ] = 0. (Condition satisfied)

Conclusion: All three conditions of Leibniz's Test are satisfied. Therefore, the series is convergent (furthermore, it is absolutely convergent by comparison with 1/n^2).

---

Part (ii): Cauchy's Integral Test and sum_{n=1}^infinity [ n^2 * e^(-n^3) ]

1. Statement of Cauchy's Integral Test: Let f(x) be a continuous, positive, and monotonically decreasing function for x >= 1 such that f(n) = un. Then the infinite series sum{n=1}^infinity u_n and the improper integral integral from 1 to infinity of f(x) dx either both converge or both diverge.

2. Test the Series: Given the series: sum_{n=1}^infinity n^2 e^(-n^3) Let f(x) = x^2 e^(-x^3).

Verify Conditions for f(x) on [1, infinity):

Continuous: Product of polynomials and exponentials, hence continuous.
Positive: For x >= 1, x^2 > 0 and e^(-x^3) > 0, so f(x) > 0.
Monotonically decreasing:
f'(x) = 2x e^(-x^3) + x^2 (-3x^2 e^(-x^3)) = x e^(-x^3) * (2 - 3x^3).
For x >= 1, 2 - 3x^3 < 0, so f'(x) < 0. Thus, f(x) is strictly decreasing.

3. Evaluate the Improper Integral: I = integral (x = 1 to infinity) x^2 e^(-x^3) dx = lim (M -> infinity) [ integral (x = 1 to M) x^2 e^(-x^3) dx ]

Use substitution: Let u = -x^3 ==> du = -3x^2 dx ==> x^2 dx = -du / 3

When x = 1: u = -1
When x = M: u = -M^3

I = lim (M -> infinity) [ integral (u = -1 to -M^3) e^u (-du / 3) ] = (1/3) lim (M -> infinity) [ integral (u = -M^3 to -1) e^u du ] = (1/3) lim (M -> infinity) [ e^u ] from -M^3 to -1 = (1/3) lim (M -> infinity) [ e^(-1) - e^(-M^3) ]

As M -> infinity, e^(-M^3) -> 0: I = (1/3) * [ e^(-1) - 0 ] = 1 / (3e)

4. Conclusion: Since the improper integral converges to the finite value 1/(3e), by Cauchy's Integral Test, the series sum_{n=1}^infinity [ n^2 * e^(-n^3) ] is convergent.

Frequently Asked Questions

What is the passing mark for GTU Mathematics - 1 (3110014)?

The theory exam is conducted for 70 marks. The minimum passing threshold is 33% (equivalent to 23 marks out of 70).

How are eigenvalues and eigenvectors calculated for symmetric matrices?

Find the characteristic equation det(A - lambdaI) = 0 and solve for the roots lambda. For each distinct or repeated root, substitute lambda back into (A - lambdaI)X = 0 and use row operations or Gaussian elimination to determine linearly independent eigenvectors.

When is changing the order of integration necessary in double integrals?

Changing the order of integration is necessary when the inner integral cannot be integrated analytically in terms of elementary functions (e.g., sin(y^2)), or when reversing the order makes algebraic computation simpler.

C

Java

Mood

Introduction

Paper Information

Questions and Solutions

Q.1 (a)

Definition

Step 1: Check Solenoidal (∇ · F)

Step 2: Check Irrotational (∇ × F)

Q.1 (b)

Step 1: Identify Nature of Function

Step 2: Fourier Cosine Integral Formulation

Step 3: Evaluate the Integral

Q.1 (c)

Statement of Green's Theorem

Part 1: Double Integral (Right-Hand Side)

Part 2: Line Integral (Left-Hand Side)

Sum of Line Integrals:

Conclusion

Q.2 (a)

Step 1: Test for Exactness

Step 2: Integrate to Find the General Solution

Q.2 (b)

Step 1: Find the Gradient ∇∅

Step 2: Find the Unit Vector along the Given Direction

Step 3: Compute the Directional Derivative

Q.2 (c)

Step 1: Nature of the Ordinary Point

Step 2: Substitute into Differential Equation

Step 3: Equating Coefficients to Zero

Step 4: Calculate Coefficients in Terms of a₀ and a₁

Step 5: Write the General Solution

Q.2 (c) OR

Step 1: Standard Legendre Form

Step 2: Substitute into the Equation

Step 3: Equate Coefficients of Powers of x to Zero

Step 4: Calculate Subsequent Coefficients

Step 5: Final Series Solution

Q.3 (a)

Step 1: Complementary Function (y_c)

Step 2: Particular Integral (y_p)

Step 3: General Solution

Q.3 (b)

Step 1: Complementary Function (y_c)

Step 2: Particular Integral (y_p)

Step 3: General Solution

Q.3 (c)

Step 1: Complementary Function (y_c)

Step 2: Form of the Particular Solution (y_p)

Step 3: Determine y_p1 for R₁(x) = x³

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

Step 5: General Solution

Q.3 (a) OR

Step 1: Complementary Function (y_c)

Step 2: Particular Integral (y_p)

Step 3: General Solution

Q.3 (b) OR

Step 1: Complementary Function (y_c)

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

Step 3: Particular Integral y_p2 (for 7)

Step 4: General Solution

Q.3 (c) OR

Step 1: Transform Cauchy-Euler Equation into Constant Coefficients

Step 2: Complementary Function (y_c)

Step 3: Particular Integral (y_p)

Step 4: Express General Solution in Terms of x

Q.4 (a)

Definition

Derivation

Q.4 (b)

Step 1: Laplace Transform of cos 2t

Step 2: Multiplication by t Property

Step 3: First Shifting Theorem

Q.4 (c)

Step 1: Apply Laplace Transform to Both Sides

Step 2: Partial Fraction Decomposition

Step 3: Taking Inverse Laplace Transform

Q.4 (a) OR

Step 1: Use Property of Derivative of Laplace Transform