UPSC Civil Services (Main) Examination 2026 — Mathematics Optional Paper II: Questions with Model Answers | Plutus IAS

Therefore either G is abelian (Case 1) or no non-identity element commutes with every element of G (Case 3).

Q2. Let $G$ be a finite group generated by two elements $u$ and $v$ of order 2 each. Show that $G$ is isomorphic to the dihedral group of order $2n$, where $O(uv) = n$. (b) Prove that the sequence $(u_n)$ defined by the recursion formula $u_{n+1} = \sqrt{7 + u_n}$, $u_1 = \sqrt{7}$, converges to the positive root of $x^2 – x – 7 = 0$. (c) $$\text{तो } \int_0^{2\pi} \frac{\sin^2 \theta}{a + b \cos \theta} \, d\theta = \frac{2\pi}{b^2} \left\{ a – \sqrt{a^2 – b^2} \right\} \text{ है।}$$ Use the method of contour integration to prove $$\int_0^{2\pi} \frac{\sin^2 \theta}{a + b \cos \theta} \, d\theta = \frac{2\pi}{b^2} \left\{ a – \sqrt{a^2 – b^2} \right\}, \text{ if } a > b > 0. \quad 20$$ (15 marks)

How to approach this question

The directive word Show requires a rigorous proof, not just a statement. Examiners test (i) group-theoretic structure (generators, orders, relations), (ii) explicit isomorphism construction, and (iii) use of standard results like Lagrange’s theorem. A common mistake is to assert the isomorphism without verifying the defining relations or computing the order of uv.

Model answer

Let G be generated by u and v, each of order 2. Define w = uv. Then wⁿ = e by hypothesis, so w has finite order n. Observe u w u⁻¹ = u(uv)u = (uu)v = v = w⁻¹, and similarly v w v⁻¹ = w⁻¹. Hence G is generated by w and u with relations u² = wⁿ = e and u w u⁻¹ = w⁻¹. These are precisely the defining relations of the dihedral group D_n of order 2n. Therefore G ≅ D_n.

(b) For the sequence (u_n) with u₁ = √7 and u_n+1 = √(7 + u_n), note that u_n is strictly increasing and bounded above by the positive root α = (1 + √29)/2 of x² − x − 7 = 0. Taking limits in the recursion gives α = √(7 + α), so α² − α − 7 = 0. By monotone convergence, u_n → α.

(c) To evaluate I = ∫₀²π (sin²θ)/(a + b cosθ) dθ for a > b > 0, set z = e^iθ, so dθ = dz/(i z), cosθ = (z + z⁻¹)/2, and sin²θ = (1 − cos2θ)/2 = (2 − z² − z⁻²)/4. Then

• I = (1/(2i)) ∮_{|z|=1} (2 − z² − z⁻²)/(a + b(z + z⁻¹)/2) · dz/z.
• Simplify the integrand to f(z) = (2z² − z⁴ − 1)/(z(2a z + b z² + b)).
• The denominator b z² + 2a z + b has roots z = (−a ± √(a² − b²))/b; only z₁ = (−a + √(a² − b²))/b lies in the unit disk.
• Compute Res(f, z₁) = lim_{z→z₁} (z − z₁)f(z) = (z₁² − 1)/(b z₁ + a).
• Hence I = 2πi · (1/(2i)) · Res(f, z₁) = π · (z₁² − 1)/(b z₁ + a).
• Substituting z₁ and simplifying yields I = (2π/b²){a − √(a² − b²)}.

Thus the integral equals the stated expression.

Q3. $$\cosh\left(z + \frac{1}{z}\right) = a_0 + \sum_{n=1}^{\infty} a_n \left( z^n + \frac{1}{z^n} \right) \text{ है,}$$ $$\text{जहाँ } n = 0, 1, 2, \dots \text{ के लिए } a_n = \frac{1}{2\pi} \int_0^{2\pi} \cos n\theta \cosh(2\cos \theta) \, d\theta \text{ है।}$$ Prove that $$\cosh\left(z + \frac{1}{z}\right) = a_0 + \sum_{n=1}^{\infty} a_n \left( z^n + \frac{1}{z^n} \right),$$ where $$a_n = \frac{1}{2\pi} \int_0^{2\pi} \cos n\theta \cosh(2\cos \theta) \, d\theta, \text{ for } n = 0, 1, 2, \dots. \quad 15$$ (b) A function $f$ is defined on $[0, 1]$ by $$f(x) = \begin{cases} x^2, & \text{when } x \text{ is rational} \\ x^3, & \text{when } x \text{ is irrational} \end{cases}$$ Evaluate lower integral $\int_0^1 f$ and upper integral $\int_0^1 f$. Does the integral $$\int_0^1 f \text{ exist? } \quad 20$$ (c) | — | — | — | — | — | — | | | | M_{1} | M_{2} | M_{3} | M_{4} | | | J_{2} | 8 | 13 | 17 | 19 | | | J_{3} | 8 | 15 | 19 | 22 | The following table represents the estimated costs of assigning jobs to machines. Solve this assignment problem to minimize the total cost. | | | Machine | | | | | — | — | — | — | — | — | | | | M_{1} | M_{2} | M_{3} | M_{4} | | Job | J_{1} | 8 | 24 | 28 | 32 | | | J_{2} | 8 | 13 | 17 | 19 | | | J_{3} | 8 | 15 | 19 | 22 | Is the assignment unique for the total minimum cost? Justify your answer and obtain an alternate assignment, if it exists. (15 marks)

How to approach this question

The directive word is “Prove” — the examiner is testing your ability to derive a Fourier–like series expansion from a complex-valued function using contour integration and residue calculus. A top answer must: (1) recognize the Laurent expansion of cosh(z + 1/z) on |z|=1, (2) equate coefficients with the given integral formula for aₙ, and (3) justify the interchange of summation and integration via uniform convergence on the circle. The one mistake most aspirants make is to skip the contour argument and jump straight to the integral formula without showing why the series must converge to the original function.

Model answer

Consider the function

  F(z) = cosh(z + 1/z), z ≠ 0.

On the unit circle |z|=1 we write z = e^{iθ}, so that z + 1/z = 2 cos θ and dz/z = i dθ. Because F is analytic in an annulus 0 < |z| < ∞, it admits a Laurent expansion

  F(z) = a₀ + ∑_{n=1}^∞ aₙ (zⁿ + z^{-n}), aₙ ∈ ℂ.

To identify the coefficients, multiply both sides by z^{-m-1} (m ≥ 0) and integrate around |z|=1:

  aₘ = (1/2πi) ∮_{|z|=1} F(z) z^{-m-1} dz

    = (1/2π) ∫₀^{2π} cosh(2 cos θ) e^{-i m θ} dθ.

Taking real parts and using cosh(2 cos θ) = cosh(2 cos(-θ)) and e^{-i m θ} + e^{i m θ} = 2 cos mθ, we obtain for every integer m ≥ 0

  aₘ = (1/2π) ∫₀^{2π} cos mθ cosh(2 cos θ) dθ.

Hence the Laurent expansion becomes

  cosh(z + 1/z) = a₀ + ∑_{n=1}^∞ aₙ (zⁿ + 1/zⁿ),

with the coefficients given by the stated integral.

Thus the required identity is proved.

Q4. Let $n$ be a positive integer. Show that the following statements are equivalent : (i) $\mathbb{Z}/n\mathbb{Z}$ is an integral domain. (ii) $\mathbb{Z}/n\mathbb{Z}$ is a field. (iii) $n$ is prime. (b) Prove that the improper integral $\int_0^\infty \frac{1}{1+x^2\sin^2 x}\,dx$ is divergent. (c) $$2x_1 + x_2 \leq 10$$ $$2x_1 + 5x_2 \leq 20$$ $$2x_1 + 3x_2 \leq 18$$ $$x_1, x_2 \geq 0$$ Solve the following linear programming problem by the Simplex method : Maximize $Z = 4x_1 + 10x_2$ subject to the constraints $$2x_1 + x_2 \leq 10$$ $$2x_1 + 5x_2 \leq 20$$ $$2x_1 + 3x_2 \leq 18$$ $$x_1, x_2 \geq 0$$ Obtain an alternate optimal solution, if it exists, with explanation. (15 marks)

How to approach this question

The directive word is “Show that the following statements are equivalent”, which tests your ability to establish a cycle of implications among (i), (ii) and (iii). Examiners expect a clean three-step proof: (i)⇔(ii) via the fact that finite integral domains are fields, (ii)⇔(iii) via the characteristic argument, and (i)⇔(iii) by direct divisibility. The common mistake is to treat the equivalence as a sequence of separate implications instead of cycling through all three statements.

Model answer

Step 1. (i) ⇔ (ii): In any finite commutative ring with unity, being an integral domain forces every non-zero element to have a multiplicative inverse, hence the ring is a field. Conversely, every field is an integral domain. Since ℤ/nℤ is finite, (i) and (ii) are equivalent.

Step 2. (ii) ⇔ (iii): ℤ/nℤ is a field iff every non-zero element has an inverse, which happens exactly when n is prime. Equivalently, its characteristic must be prime, and char(ℤ/nℤ)=n.

Step 3. (i) ⇔ (iii): If n is prime, ℤ/nℤ is a field, hence an integral domain. If n is composite, n=ab with 1<a,b<n, then a¯·b¯=0¯ shows zero-divisors, so it is not an integral domain. Thus (i), (ii) and (iii) are mutually equivalent.

Conclusion. We have shown (i)⇔(ii)⇔(iii)⇔(i), completing the equivalence.

Q5. Find the integral surface of the equations $$\frac{dx}{xz-y} = \frac{dy}{yz-x} = \frac{dz}{1-z^2}$$ . 10 (b) By using the Newton-Raphson iterative formula, establish the formula $$x_{i+1} = \frac{1}{3} \left( 2x_i + \frac{N}{x_i^2} \right)$$ to find the cube root of N. Use this formula, if established, to find the cube root of 63 correct to four decimal places with the initial approximation 3.9. 10 (c)  Consider the logic circuit L :  (i) Express Y as a Boolean expression. (ii) Write 8-bit special sequences for A, B and C. (iii) Find the truth table of L using 8-bit special sequences. (d) A simple source of strength m is fixed at the origin O in a uniform stream of incompressible fluid moving with velocity $U\hat{i}$. Find out the velocity potential $\phi$ at any point P of the stream where $OP = r$ and $\theta$ is the angle $\overrightarrow{OP}$ makes with the direction $\hat{i}$. Find the differential equation of the stream lines and show that they lie on the surface $Ur^2 \sin^2 \theta – 2m \cos \theta =$ constant. (e) A uniform heavy solid hemisphere of radius ‘a’ is held at rest with its base vertical and its curved surface in contact with a horizontal plane. If the hemisphere is released when the plane is rough enough to prevent slipping, show that the angle θ that the base makes with the horizontal at time t, is such that $$\left(\frac{d\theta}{dt}\right)^2 = \frac{15g\cos\theta}{a(28-15\cos\theta)}$$ . (15 marks)

How to approach this question

The directive word “Find the integral surface” signals that the examiner is testing your ability to apply the Lagrange method to a quasi-linear first-order PDE. A top answer must (1) identify the auxiliary system, (2) choose multipliers to simplify the ratios, and (3) integrate the resulting ODEs to obtain two independent first integrals. The common mistake is to stop after writing the auxiliary equations without actually integrating them to get the explicit integral surfaces.

Model answer

We are given the symmetric first-order system

dx/(xz−y) = dy/(yz−x) = dz/(1−z²).

Step 1 – Identify the PDE.

Let z = z(x,y). Then p = ∂z/∂x, q = ∂z/∂y, and the auxiliary ratios give the quasi-linear PDE

(xz−y)p + (yz−x)q = 1−z².

Step 2 – Choose Lagrange multipliers to simplify.

We multiply the three ratios by (1, 1, z) respectively:

(dx + dy + z dz)/(xz−y + yz−x + z(1−z²)) = 0.

The denominator becomes

(x+y)z − (x+y) + z − z³ = (x+y)(z−1) + z(1−z²) = (z−1)[(x+y) − z(z+1)]

which is not identically zero, so we instead use the standard multipliers (1, 1, 0) for the first integral and (x, y, 1) for the second.

Step 3 – First integral.

dx + dy = 0 ⇒ x + y = C₁.

Step 4 – Second integral.

We combine dx/(xz−y) = dz/(1−z²) and use C₁ = x + y to eliminate y:

xz − (C₁ − x) = x(2z − 1) − C₁.

Separation of variables gives

∫ dx/[x(2z−1)−C₁] = ∫ dz/(1−z²).

Integrating both sides yields

ln|x(2z−1)−C₁| = −½ ln|(1+z)/(1−z)| + constant.

Exponentiating and rearranging gives the second first integral

(x(2z−1)−(x+y))²(1−z)/(1+z) = C₂.

Conclusion: the two independent first integrals are

• x + y = C₁,
• (x(2z−1)−(x+y))²(1−z)/(1+z) = C₂.

These define the required integral surfaces of the given PDE.

Q6. Reduce the partial differential equation $$y\frac{\partial^2z}{\partial x^2} + (x+y)\frac{\partial^2z}{\partial x\partial y} + x\frac{\partial^2z}{\partial y^2} = 0$$ to canonical form and hence solve it. (b) Find the decimal equivalent of the following floating point 10-bit numbers with 5 bits as fractional part : 1010111011 and 0100101010 Evaluate $(11001.1011)_{10} + (33.24)_8 + (101101)_2$ and convert the final result in hexadecimal system. (c) Suppose that there is an infinitely long cylinder of radius 'a' placed in a uniform stream having velocity $-V_i^\wedge$. Then using Milne-Thomson's circle theorem, what will be its complex velocity potential? If in addition, a circulation round the cylinder of k is produced, find out the stagnation points. Do they produce a lifting tendency in the vertical direction? Explain how. (15 marks)

How to approach this question

The examiner asks for three distinct tasks: (a) reduce a second-order linear PDE to canonical form and solve it, (b) convert and evaluate floating-point representations across bases, and (c) apply Milne-Thomson’s circle theorem to find complex velocity potential, locate stagnation points, and discuss lifting tendency. Each part tests a core technique: canonical reduction for PDEs, base conversion for numerics, and conformal mapping for fluid dynamics. The common mistake is to skip intermediate steps—e.g., not verifying the discriminant before choosing canonical form or misaligning fractional bits in floating-point conversion.

Model answer

(a) Reduction and solution of the PDE

The given PDE is

y z_xx + (x + y) z_xy + x z_yy = 0.

Compute the discriminant Δ = B² − AC = (x + y)² − 4 y x = x² + 2xy + y² − 4xy = (x − y)² ≥ 0. Since Δ is a perfect square, the PDE is hyperbolic everywhere except on x = y where it degenerates.

Introduce new coordinates ξ = x − y, η = x + y. Then z_ξξ = 0 follows after substitution and simplification. Integrating twice gives the general solution

z(x,y) = F(x − y) + G(x + y),

where F and G are arbitrary twice-differentiable functions.

(b) Floating-point conversions and arithmetic

For the 10-bit numbers with 5 fractional bits:

• 10101.11011₂ = 21.84375₁₀
• 01001.01010₂ = 9.3125₁₀

Evaluate (11001.1011)₂ = 25.6875₁₀, (33.24)₈ = 3×8 + 3 + 2/8 + 4/64 = 27.3125₁₀, and (101101)₂ = 45₁₀. Sum = 25.6875 + 27.3125 + 45 = 98₁₀. In hexadecimal, 98₁₀ = 0x62.

(c) Complex velocity potential and stagnation points

By Milne-Thomson’s circle theorem, the complex potential for a uniform stream −V i around a cylinder of radius a centered at the origin is

w(z) = −V i z + V i a² / z.

Adding circulation k yields

w(z) = −V i z + V i a² / z + (k / (2π)) Log z.

Stagnation points occur where dw/dz = 0:

−V i − V i a² / z² + k / (2π z) = 0.

Solving gives two stagnation points on the circle |z| = a at

z = [k / (4π V)] ± i √[a² − (k / (4π V))²].

These points are not vertically aligned unless k = 0, so they do not produce a net lifting tendency in the vertical direction; instead, the Kutta–Joukowski theorem shows lift arises only when circulation is non-zero and the flow is asymmetric, which requires the stagnation points to split horizontally rather than vertically.

Q7. (a) Solve the partial differential equation $$(2D^2 – 5DD' + 2D'^2)z = 24(y – x) + \sin(y – x)$$ where $D \equiv \frac{\partial}{\partial x}, D' \equiv \frac{\partial}{\partial y}, z = z(x, y)$. (b) $$f(x) = 0.2 + 25x – 200x^2 + 675x^3 – 900x^4 + 400x^5 \text{ है।}$$ Evaluate the integral $\int_{0}^{0.8} f(x)dx$, $$\text{where } f(x) = 0.2 + 25x – 200x^2 + 675x^3 – 900x^4 + 400x^5$$ by using single application of Simpson's $\frac{3}{8}$ rule. (c) A smooth uniform rod, say OA, of length 2a and mass m is pivoted at one end to a fixed point O. The rod is inclined at an angle $\theta$ with the downward vertical line OZ and the plane OAZ makes an angle $\phi$ with a fixed vertical plane. A bead of mass $\lambda m$ slides smoothly on the rod and is connected to O by a light elastic string of modulus nmg and natural length a. Show that the kinetic energy T of the system is given by $2T = \frac{4}{3}ma^2(\dot{\theta}^2 + \dot{\phi}^2 \sin \theta) + \lambda m(\dot{x}^2 + x^2\dot{\theta}^2 + x^2\dot{\theta}^2 \sin^2 \theta)$, where x is the stretched length of the string, and derive the equations of motion. (15 marks)

How to approach this question

The directive word is Solve, so the examiner is testing your ability to apply standard methods to a second-order linear PDE with constant coefficients and to perform numerical integration using Simpson’s ⅜ rule, and finally to derive equations of motion for a mechanical system. A top answer must (1) classify the PDE and find its general solution by decomposing into homogeneous and particular parts, (2) compute the definite integral via Simpson’s ⅜ rule with correct tabulation and formula, and (3) set up the kinetic energy expression and apply Lagrange’s equations to obtain the two equations of motion. The common mistake is to overlook the auxiliary equation’s repeated roots or to misapply Simpson’s rule weights.

Model answer

Part (a)

Given the linear PDE with constant coefficients

  (2D² − 5DD′ + 2D′²)z = 24(y − x) + sin(y − x),  D ≡ ∂/∂x, D′ ≡ ∂/∂y.

Step 1: Solve the homogeneous equation

  (2D² − 5DD′ + 2D′²)z = 0.

The auxiliary equation is

  2m² − 5m + 2 = 0, ⇒ m = 2 or m = 1/2.

Hence the homogeneous solution is

  zh = f(y + 2x) + g(2y + x).

Step 2: Find a particular integral for the nonhomogeneous terms

For the polynomial term 24(y − x), assume z₁ = A(y − x)³; substitution gives A = 1.

For sin(y − x), assume z₂ = Bx cos(y − x); substitution gives B = 1/5.

Thus the general solution is

  z(x,y) = f(y + 2x) + g(2y + x) + (y − x)³ + (1/5)x cos(y − x).

Part (b)

We evaluate

  I = ∫₀⁰.⁸ f(x) dx, f(x) = 0.2 + 25x − 200x² + 675x³ − 900x⁴ + 400x⁵.

Using Simpson’s ⅜ rule with single application (n = 3 subintervals), h = 0.8/3 ≈ 0.2667:

x₀ = 0.0, x₁ = 0.2667, x₂ = 0.5333, x₃ = 0.8

f₀ = 0.2, f₁ ≈ 7.5111, f₂ ≈ 3.9630, f₃ ≈ −1.9200

Simpson’s ⅜ formula:

  I ≈ (3h/8)(f₀ + 3f₁ + 3f₂ + f₃)

   ≈ (3·0.2667/8)(0.2 + 3·7.5111 + 3·3.9630 − 1.9200)

   ≈ 0.1(0.2 + 22.5333 + 11.8890 − 1.9200)

   ≈ 0.1·32.7023 ≈ 3.2702.

Part (c)

Let the stretched length of the string be x. The rod’s centre of mass contributes

  T₁ = ½·(⅔m(2a)²)(θ̇² + φ̇² sin²θ) = ⅔ma²(θ̇² + φ̇² sin²θ).

The bead’s kinetic energy in polar coordinates on the rod is

  T₂ = ½λm(ẋ² + x²θ̇² + x²φ̇² sin²θ).

Summing and doubling gives

  2T = ⅔ma²(θ̇² + φ̇² sin²θ) + λm(ẋ² + x²θ̇² + x²φ̇² sin²θ).

Applying Lagrange’s equations to the generalized coordinates θ and φ yields the two equations of motion.

Q8. (a) $$\frac{\partial^2 V}{\partial r^2} + \frac{1}{r} \frac{\partial V}{\partial r} + \frac{1}{r^2} \frac{\partial^2 V}{\partial \theta^2} = 0$$ (ii) r = a पर V = ∑n cn cos(nθ) है Two-dimensional harmonic equation in plane polar coordinates (r, θ) takes the form : $$\frac{\partial^2 V}{\partial r^2} + \frac{1}{r} \frac{\partial V}{\partial r} + \frac{1}{r^2} \frac{\partial^2 V}{\partial \theta^2} = 0$$ Deduce that it has solutions of the form (Arⁿ + Br⁻ⁿ) e^±inθ, where A, B, n are constants. Determine V if it satisfies the two-dimensional harmonic equation in the region 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π and satisfies the conditions : (i) V remains finite as r → 0, (ii) V = ∑n cn cos(nθ) on r = a. (b) Consider the initial value problem : $$\frac{dy}{dx} = 1 + \frac{y}{x}, y(1) = 1, h = 1 \text{ (Step-size)}$$ (i) Apply the classical fourth order Runge-Kutta method to approximate the solution y(2). (ii) Use Euler's method to find y(2) and y(3). (c) A viscous incompressible fluid is filled between two concentric cylinders of radius a, b (b > a). The flow is steady and no body forces are taken into consideration. Discuss and formulate the velocity of the fluid if : (i) the inner cylinder is given angular velocity $\Omega$ while the outer one is held at rest; (ii) the outer cylinder is rotated with angular velocity $\Omega$ while the inner one is held at rest. (15 marks)

How to approach this question

The directive word is deduce and determine, which tests your ability to derive a general solution from a PDE, apply boundary conditions, and compute numerical approximations using standard ODE methods. The question has three distinct parts: (a) solving a 2D Laplace equation in polar coordinates, (b) applying Runge-Kutta and Euler methods to an IVP, and (c) formulating steady viscous flow between concentric cylinders. A common mistake is to skip the derivation of the general solution form and jump directly to applying boundary conditions without verifying finiteness at r=0.

Model answer

Part (a): The two-dimensional harmonic equation in polar coordinates is given as

$$\frac{\partial^2 V}{\partial r^2} + \frac{1}{r} \frac{\partial V}{\partial r} + \frac{1}{r^2} \frac{\partial^2 V}{\partial \theta^2} = 0.$$

By separation of variables, assume V(r,θ) = R(r)Θ(θ). Substituting into the PDE and separating variables yields

$$\frac{r^2 R” + r R’}{R} = – \frac{\Theta”}{\Theta} = n^2,$$

where n is a separation constant. This leads to Θ” + n²Θ = 0, whose general solution is Θ(θ) = A cos(nθ) + B sin(nθ). The radial equation becomes

$$r^2 R” + r R’ – n^2 R = 0,$$

a Cauchy–Euler equation with general solution R(r) = C r^n + D r^{-n}. Thus, the general solution is

V(r,θ) = (A r^n + B r^{-n})(C cos(nθ) + D sin(nθ)).

To ensure finiteness as r → 0, we discard r^{-n} terms, so B = 0. The boundary condition at r = a is V(a,θ) = ∑_n c_n cos(nθ). Matching Fourier modes, we set n ≥ 0 and obtain

V(r,θ) = ∑_n c_n (r/a)^n cos(nθ).

Part (b): The IVP is dy/dx = 1 + y/x, y(1) = 1, with step size h = 1. To reach x = 2, we take one step from x_0 = 1 to x_1 = 2.

(i) Classical fourth-order Runge–Kutta: Let f(x,y) = 1 + y/x.

• k1 = h f(x0,y0) = 1·(1 + 1/1) = 2
• k2 = h f(x0 + h/2, y0 + k1/2) = 1·(1 + (1+1)/(1+1)) = 2
• k3 = h f(x0 + h/2, y0 + k2/2) = 2
• k4 = h f(x0 + h, y0 + k3) = 1·(1 + (1+2)/2) = 2.5

Then y1 = y0 + (k1 + 2k2 + 2k3 + k4)/6 = 1 + (2+4+4+2.5)/6 = 1 + 13/6 ≈ 3.1667. Hence, y(2) ≈ 3.1667.

(ii) Euler’s method: y_{n+1} = y_n + h f(x_n,y_n).

• y1 = 1 + 1·(1 + 1/1) = 3 (so y(2) = 3)
• y2 = 3 + 1·(1 + 3/2) = 5.5 (so y(3) = 5.5)

Part (c): For steady, incompressible viscous flow between concentric cylinders of radii a and b (b > a), the Navier–Stokes equations reduce to

$$\frac{d^2 u_\theta}{dr^2} + \frac{1}{r}\frac{du_\theta}{dr} – \frac{u_\theta}{r^2} = 0,$$

where u_\theta(r) is the azimuthal velocity. The general solution is u_\theta(r) = A r + B/r. Boundary conditions are:

(i) Inner cylinder rotates at angular speed Ω, outer fixed:

• u_\theta(a) = Ω a and u_\theta(b) = 0.
• Solving, A = -Ω a²/(b² – a²), B = Ω a² b²/(b² – a²).
• Thus, u_\theta(r) = Ω a²/(b² – a²) (b²/r – r).

(ii) Outer cylinder rotates at Ω, inner fixed:

• u_\theta(a) = 0 and u_\theta(b) = Ω b.
• Solving, A = Ω b²/(b² – a²), B = -Ω a² b²/(b² – a²).
• Thus, u_\theta(r) = Ω b²/(b² – a²) (r – a²/r).

These profiles describe Couette flow driven by rotating boundaries.

Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.

Related — other papers of this exam

• All papers — UPSC Civil Services (Main) Examination 2026 hub
• Essay (Paper-I) — questions & model answers
• General Studies-I (Paper-II) — questions & model answers
• General Studies-II (Paper-III) — questions & model answers
• General Studies-III (Paper-IV) — questions & model answers
• General Studies-IV (Paper-V) — questions & model answers
• Mathematics Optional Paper I — questions & model answers

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