Paper II — Q8
(a) Find a complete integral of the partial differential equation p = (z + qy)² by using Charpit's method. 15 marks (b) Derive…
Find a complete integral of the partial differential equation p = (z + qy)² by using Charpit's method. 15 marks
Derive Newton's backward difference interpolation formula and also do error analysis. 15 marks
Show that for the complex potential tan⁻¹z, the streamlines and equipotential curves are circles. Find the velocity at any point and check the singularities at z = ±i. 20 marks
हिंदी में प्रश्न पढ़ें
चारपिट विधि का उपयोग करके आंशिक अवकल समीकरण p = (z + qy)² का पूर्ण समाकल प्राप्त कीजिए। 15 अंक
न्यूटन के पश्चातर अंतर्वेशन सूत्र की व्युत्पत्ति कीजिए तथा त्रुटि-विश्लेषण भी कीजिए। 15 अंक
दर्शाइए कि सम्मिश्र विभव tan⁻¹z के लिए धारा-रेखाएँ तथा समविभव वक्र, वृत्त हैं। किसी भी बिंदु पर वेग निकालिए तथा z = ±i पर विचित्रता जाँचिए। 20 अंक
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) Complete integral by Charpit's method
Let p = ∂z/∂x, q = ∂z/∂y and F = p − (z + qy)² = 0. Put w = z + qy. Then F_p = 1, F_q = −2yw, F_z = −2w, F_x = 0, F_y = −2qw.
Charpit's equations dx/F_p = dy/F_q = dp/(−F_x − pF_z) = dq/(−F_y − qF_z) give dy/dx = −2yw and dp/dx = 2pw.
Hence d(py)/dx = y(2pw) + p(−2yw) = 0, so py = a, a constant, and p = a/y.
From F = 0, w² = p = a/y, so w = ε√a y^(−1/2), with ε = ±1. Since q = (w − z)/y,
dz = p dx + q dy = (a/y)dx + [ε√a y^(−3/2) − z/y]dy,
so y dz + z dy = d(yz) = a dx + ε√a y^(−1/2) dy. Integrating, yz = ax + 2ε√a √y + b.
Complete integral: z = (ax + b)/y + 2ε√(a/y), where a and b are arbitrary constants.
Check: p = a/y and z + qy = ε√(a/y), so p = (z + qy)².
(b) Newton's backward difference formula
Take equispaced nodes x₀, …, xₙ with step h. Define ∇fₖ = fₖ − fₖ₋₁, so E⁻¹ = 1 − ∇. Put x = xₙ + uh.
Then f(x) ≈ E^u fₙ = (1 − ∇)^(−u) fₙ. Expanding binomially,
Pₙ(x) = fₙ + u∇fₙ + [u(u+1)/2!]∇²fₙ + [u(u+1)(u+2)/3!]∇³fₙ + … + [u(u+1)…(u+n−1)/n!]∇ⁿfₙ.
This polynomial of degree n agrees with f at every node: at x = xₙ₋ₖ, u = −k, and the series reduces to (1 − ∇)^k fₙ = fₙ₋ₖ.
Error analysis: if f has a continuous (n+1)th derivative, then
f(x) − Pₙ(x) = [u(u+1)…(u+n) h^(n+1)/(n+1)!] f^(n+1)(ξ),
for some ξ in the interval spanned by the nodes and x. So the error is bounded by |u(u+1)…(u+n)| h^(n+1) M/(n+1)!, where M = max|f^(n+1)|. It is zero when f is a polynomial of degree ≤ n, and it is smallest for x near xₙ (small |u|).
(c) Complex potential W = tan⁻¹ z
Write W = φ + iψ, so z = tan(φ + iψ) = [sin 2φ + i sinh 2ψ]/[cos 2φ + cosh 2ψ].
Equipotentials (φ = C): eliminating ψ gives (x + cot 2C)² + y² = cosec² 2C. These are circles centred at (−cot 2C, 0), all passing through z = ±i.
Streamlines (ψ = C): eliminating φ gives x² + (y − coth 2C)² = cosech² 2C. These are circles centred at (0, coth 2C) on the y-axis, and they encircle the points z = ±i.
Velocity: u − iv = dW/dz = 1/(1 + z²). With z = x + iy and D = (1 + x² − y²)² + 4x²y²,
u = (1 + x² − y²)/D, v = 2xy/D, and the speed is 1/√D.
Singularities: at z = ±i, 1 + z² = 0, so dW/dz has simple poles and the speed becomes infinite there. Also tan⁻¹ z has logarithmic branch points at z = ±i, so these two points must be excluded from the flow region.
How this answer will be evaluated
What earns full marks: Complete, rigorous derivations with all steps shown and verified.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Find a complete integral of p = (z + qy)² by Charpit's method 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Charpit's auxiliary equations with the correct partial derivatives of F
- Integral py = a obtained from the auxiliary equations
- Complete integral z = (ax + b)/y + 2ε√(a/y)
Loses marks
- Wrong partial derivatives of F
- Dropping the ± branch
- Only one arbitrary constant in the answer
Earns more
- Derivation of w² = a/y from F = 0
- Reduction to d(yz) = a dx + ε√a y^(−1/2) dy
- Two arbitrary constants in the final integral
Extra mark
- Verifies the integral against the original equation
- (b) Derive Newton's backward difference interpolation formula and do the error analysis 15 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Backward difference operator and u = (x − xₙ)/h
- Formula Pₙ = fₙ + u∇fₙ + u(u+1)/2! ∇²fₙ + …
- Error term u(u+1)…(u+n)h^(n+1) f^(n+1)(ξ)/(n+1)!
Loses marks
- Mixing forward and backward differences
- Wrong sign convention for u
- Giving the formula without any derivation
Earns more
- Derivation through E^u = (1 − ∇)^(−u) or equivalent
- Check that the polynomial matches the nodal values
- Error bound with the maximum of the (n+1)th derivative
Extra mark
- Notes that the error is zero for polynomials of degree ≤ n and smallest near xₙ
- (c) Show that the streamlines and equipotentials of tan⁻¹ z are circles; find the velocity and test z = ±i 20 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Equipotentials as circles through z = ±i
- Streamlines as circles x² + (y − coth 2C)² = cosech² 2C
- u − iv = 1/(1 + z²) with singularities at z = ±i
Loses marks
- Taking the velocity as W instead of dW/dz
- Wrong centre or radius of the circles
- Not stating that the speed is infinite at z = ±i
Earns more
- Separation of x and y in terms of φ and ψ
- Elimination of the parameter to get circle equations
- Velocity components u and v written explicitly
Extra mark
- Identifies z = ±i as simple poles of the velocity and branch points of W
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
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