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Mathematics 2022 Paper I 50 marks Compulsory Prove

Paper I — Q5

(a) Show that the general solution of the differential equation dy/dx + Py = Q can be written in the form y = Q/P − e^(−∫P dx) {C…

(a)

Show that the general solution of the differential equation dy/dx + Py = Q can be written in the form y = Q/P − e^(−∫P dx) {C + ∫ e^(∫P dx) d(Q/P)}, where P, Q are non-zero functions of x and C, an arbitrary constant. 10 marks

(b)

Show that the orthogonal trajectories of the system of parabolas : x² = 4a(y + a) belong to the same system. 10 marks

(c)

A body of weight w rests on a rough inclined plane of inclination θ, the coefficient of friction, μ, being greater than tan θ. Find the work done in slowly dragging the body a distance 'b' up the plane and then dragging it back to the starting point, the applied force being in each case parallel to the plane. 10 marks

(d)

A projectile is fired from a point O with velocity √2gh and hits a tangent at the point P(x, y) in the plane, the axes OX and OY being horizontal and vertically downward lines through the point O, respectively. Show that if the two possible directions of projection be at right angles, then x² = 2hy and then one of the possible directions of projection bisects the angle POX. 10 marks

(e)

Show that A⃗ = (6xy + z³)î + (3x² - z)ĵ + (3xz² - y)k̂ is irrotational. Also find φ such that A⃗ = ∇φ. 10 marks

हिंदी में प्रश्न पढ़ें
(a)

दर्शाइए कि अवकल समीकरण dy/dx + Py = Q का व्यापक हल y = Q/P − e^(−∫P dx) {C + ∫ e^(∫P dx) d(Q/P)} के रूप में लिखा जा सकता है, जहाँ P, Q, x के शून्येतर फलन हैं तथा C एक स्वेच्छ अचर है। 10

(b)

दर्शाइए कि परवलयों के निकाय : x² = 4a(y + a) के लंबकोणीय संहेडी, उसी निकाय में स्थित होते हैं। 10

(c)

w भार का एक पिंड, θ कोण से झुके हुए एक रूक्ष समतल पर स्थित है, घर्षण गुणांक μ, tan θ से अधिक है। पिंड को समतल पर ऊपर की तरफ 'b' दूरी तक धीरे-धीरे खींचने तथा वापस आरंभिक बिंदु तक खींचने में किए गए कार्य को ज्ञात कीजिए, जहाँ लगाया गया बल प्रत्येक दशा में समतल के समांतर है। 10

(d)

एक प्रक्षेप्य √2gh वेग के साथ बिंदु O से प्रक्षेपित किया गया तथा समतल के बिंदु P(x, y) पर स्पर्शरेखा से टकराता है जहाँ अक्ष OX तथा OY क्रमशः बिंदु O से क्षैतिज तथा अधोमुखी उर्ध्वाधर रेखाएँ हैं। यदि प्रक्षेपण की दो संभव दिशाएँ समकोण पर हों, तो दर्शाइए कि x² = 2hy तथा प्रक्षेपण की संभव दिशाओं में से एक, कोण POX को द्विभाजित करती है। 10

(e)

दर्शाइए कि A⃗ = (6xy + z³)î + (3x² - z)ĵ + (3xz² - y)k̂ अघूर्णी है। φ को भी ज्ञात कीजिए जबकि A⃗ = ∇φ। 10

Q5 of the 2022 UPSC Mains Mathematics Paper I, as printed
The question as printed in the 2022 Mathematics paper
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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) General solution in the given form

Let I = ∫P dx. Multiplying dy/dx + Py = Q by the integrating factor e^I, and using d(e^I)/dx = P e^I, d/dx (y e^I) = Q e^I, so y e^I = C₁ + ∫Q e^I dx.

Since d(e^I) = P e^I dx, Q e^I dx = (Q/P) d(e^I). Integrating by parts (P ≠ 0, so Q/P is defined), ∫Q e^I dx = (Q/P) e^I − ∫e^I d(Q/P).

Hence y = Q/P + e^(−I)[C₁ − ∫e^I d(Q/P)]. Writing C₁ = −C (C arbitrary), y = Q/P − e^(−∫P dx) {C + ∫ e^(∫P dx) d(Q/P)}.

Final answer: the required form follows, with C an arbitrary constant.

(b) Self-orthogonal family of parabolas

x² = 4a(y + a). Differentiating, 2x = 4ap, where p = dy/dx, so a = x/(2p). Eliminating a: x² = (2x/p)(y + x/(2p)) ⟹ x²p² = 2xyp + x², so the differential equation of the family is p² − (2y/x)p − 1 = 0.

For orthogonal trajectories replace p by −1/p: 1/p² + (2y/x)(1/p) − 1 = 0 ⟹ p² − (2y/x)p − 1 = 0 (multiplying by −p²).

The equation is unchanged, so the orthogonal trajectories satisfy the same differential equation and are members of the same system x² = 4a(y + a) (with a different value of a). The family is self-orthogonal. Geometrically, all members are confocal parabolas with common focus at the origin and common axis, and two confocal coaxial parabolas opening in opposite directions cut orthogonally.

Final answer: the trajectories belong to x² = 4a(y + a) itself.

(c) Work done in dragging up and back

Diagram: Dragging a body up and down a rough incline

Normal reaction R = w cosθ, limiting friction μw cosθ. Since μ > tanθ, μ cosθ > sinθ, so the body does not slide down by itself.

Upward (slowly, so no acceleration): friction acts down the plane, P₁ = w sinθ + μw cosθ, W₁ = wb(sinθ + μ cosθ).

Downward: friction acts up the plane and exceeds w sinθ, so a force down the plane is needed, P₂ = μw cosθ − w sinθ, W₂ = wb(μ cosθ − sinθ).

Total work W = W₁ + W₂ = 2μwb cosθ. The gravity terms cancel over the closed path, while friction does negative work in both directions.

Final answer: W = 2μwb cosθ.

(d) Perpendicular directions of projection

Diagram: Two perpendicular directions of projection through P

u = √(2gh), so g/(2u²) = 1/(4h). Let the direction of projection make angle α with OX, measured towards OY, and m = tanα. Then x = u cosα·t, y = u sinα·t + ½gt², giving y = xm + x²(1 + m²)/(4h), i.e. x²m² + 4hxm + (x² − 4hy) = 0.

Its two roots m₁, m₂ give the two directions. Perpendicularity requires m₁m₂ = −1: (x² − 4hy)/x² = −1 ⟹ x² = 2hy.

With x² = 2hy (i.e. 4hy = 2x²), the quadratic becomes xm² + 4hm − x = 0. (1)

Let φ = ∠POX, so tanφ = y/x = x/(2h). If t = tan(φ/2), then 2t/(1 − t²) = x/(2h), i.e. xt² + 4ht − x = 0, which is exactly (1). Hence tan(φ/2) is one root m, so one direction of projection makes angle φ/2 with OX and bisects ∠POX.

Final answer: x² = 2hy and one direction bisects ∠POX.

(e) Irrotational field and its scalar potential

A = (6xy + z³)î + (3x² − z)ĵ + (3xz² − y)k̂.

curl A = (∂A₃/∂y − ∂A₂/∂z)î + (∂A₁/∂z − ∂A₃/∂x)ĵ + (∂A₂/∂x − ∂A₁/∂y)k̂ = (−1 + 1)î + (3z² − 3z²)ĵ + (6x − 6x)k̂ = 0. Hence A is irrotational.

For A = ∇φ: ∂φ/∂x = 6xy + z³ ⟹ φ = 3x²y + xz³ + f(y, z). ∂φ/∂y = 3x² + f_y = 3x² − z ⟹ f = −yz + g(z). ∂φ/∂z = 3xz² − y + g′(z) = 3xz² − y ⟹ g′(z) = 0, g = C.

Check: ∇(3x²y + xz³ − yz) = (6xy + z³, 3x² − z, 3xz² − y) = A.

Final answer: curl A = 0 and φ = 3x²y + xz³ − yz + C.

How this answer will be evaluated

What earns full marks: All parts fully derived with clear steps and verification.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Derive the general solution of dy/dx + Py = Q in the specified form. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Identify the equation as linear first-order ODE
    • Compute the integrating factor e^∫P dx
    • Integrate to find the general solution
    • Rearrange to match the required form

    Loses marks

    • Skipping the integrating factor step
    • Incorrect algebraic manipulation of the integral

    Earns more

    • Explicitly states the standard solution formula
    • Verifies the result by differentiation

    Extra mark

    • Notes the condition P, Q non-zero
  2. (b) Show that the orthogonal trajectories of x² = 4a(y + a) are of the same form. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Differentiate the given equation to find dy/dx
    • Replace dy/dx with -dx/dy for orthogonal trajectories
    • Solve the resulting differential equation
    • Show the new equation is identical, so trajectories are members of x² = 4a(y + a) with another value of a

    Loses marks

    • Incorrect differentiation of the implicit equation
    • Failing to solve the new differential equation

    Earns more

    • Uses a different parameter for the new family
    • Clearly states the condition for orthogonality

    Extra mark

    • Notes the family is confocal (focus at origin), hence self-orthogonal
  3. (c) Calculate the work done in dragging the body up and down the inclined plane. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Resolve forces parallel and perpendicular to the plane
    • Calculate the applied force for upward motion
    • Calculate the applied force for downward motion
    • Total work W = 2μwb cosθ

    Loses marks

    • Incorrect force resolution
    • Ignoring the direction of friction in one of the motions

    Earns more

    • Draws a free-body diagram
    • Explicitly states the condition μ > tan θ

    Extra mark

    • Notes that the net work is positive due to friction
  4. (d) Show that x² = 2hy and one direction bisects angle POX if the two directions are at right angles. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Write the trajectory equation for the projectile
    • Use the condition that the two angles are at right angles
    • Derive the relation x² = 2hy
    • Show that one angle is half of angle POX

    Loses marks

    • Incorrect use of the trajectory equation
    • Failing to apply the right-angle condition correctly

    Earns more

    • Uses the standard trajectory equation
    • Clearly defines the angles involved

    Extra mark

    • Provides a diagram showing the two directions
  5. (e) Show that the vector field is irrotational and find the potential function φ. 10 marks

    derive— given → assumptions → stepwise derivation → result → check

    Must cover

    • Calculate the curl of the vector field
    • Show that the curl is zero
    • Integrate the components to find φ
    • Obtain φ = 3x²y + xz³ − yz + C and verify ∇φ = A

    Loses marks

    • Incorrect calculation of the curl
    • Failing to verify the potential function

    Earns more

    • States the condition for a field to be irrotational
    • Clearly shows the integration steps

    Extra mark

    • Notes that the field is conservative
What "Prove" is asking you to do

Establish that the statement holds for every case it claims, not for one representative case. The argument must be closed: each line follows from a definition, a hypothesis, or a named theorem you are entitled to use.

Structure that answers it

Given and to prove, restated → theorem or construction to be used, named → the argument line by line → conclusion stated as proved

Where marks are lost

Testing one example, which illustrates but proves nothing. On an if and only if claim, proving one direction and stopping forfeits that half outright, and degenerate cases — zero, the empty set, the equality case — have to be disposed of rather than assumed away.

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