Q5 50M Compulsory prove Differential equations, mechanics and vector calculus
(a) Show that the general solution of the differential equation $\frac{dy}{dx} + Py = Q$ can be written in the form $y = \frac{Q}{P} - e^{-\int P \, dx}\left\{C + \int e^{\int P \, dx} \, d\left(\frac{Q}{P}\right)\right\}$, where P, Q are non-zero functions of x and C, an arbitrary constant. 10
(b) Show that the orthogonal trajectories of the system of parabolas : $x^2 = 4a(y + a)$ belong to the same system. 10
(c) A body of weight w rests on a rough inclined plane of inclination $\theta$, the coefficient of friction, $\mu$, being greater than tan $\theta$. 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
(d) A projectile is fired from a point O with velocity $\sqrt{2gh}$ 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 $x^2 = 2hy$ and then one of the possible directions of projection bisects the angle POX. 10
(e) Show that $\vec{A} = (6xy + z^3)\hat{i} + (3x^2 - z)\hat{j} + (3xz^2 - y)\hat{k}$ is irrotational. Also find $\phi$ such that $\vec{A} = \nabla\phi$. 10
हिंदी में पढ़ें
(a) दर्शाइए कि अवकल समीकरण $\frac{dy}{dx} + Py = Q$ का व्यापक हल $$y = \frac{Q}{P} - e^{-\int P \, dx}\left\{C + \int e^{\int P \, dx} \, d\left(\frac{Q}{P}\right)\right\}$$ 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 : $x^2 = 4a(y + a)$ के लंबकोणीय संहेडी, उसी निकाय में स्थित होते हैं। 10
(c) w भार का एक पिंड, $\theta$ कोण से झुके हुए एक रूक्ष समतल पर स्थित है, घर्षण गुणांक $\mu$, tan $\theta$ 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 $\sqrt{2gh}$ 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 $x^2 = 2hy$ तथा प्रक्षेपण की संभव दिशाओं में से एक, कोण POX को द्विभाजित करती है। 10
(e) दर्शाइए कि $\vec{A} = (6xy + z^3)\hat{i} + (3x^2 - z)\hat{j} + (3xz^2 - y)\hat{k}$ अघूर्णी है। $\phi$ को भी ज्ञात कीजिए जबकि $\vec{A} = \nabla\phi$। 10
Answer approach & key points
Prove the required results across all five sub-parts with rigorous mathematical derivations. For (a), derive the alternative form of the general solution using the standard integrating factor method; for (b), find the differential equation of the family and its orthogonal trajectories; for (c), analyze forces and calculate work done in both directions; for (d), use projectile equations and the condition of perpendicular directions; for (e), compute curl and integrate to find the scalar potential. Allocate approximately 20% time to each part given equal marks distribution.
- Part (a): Derivation of the alternative form using integrating factor e^(∫Pdx), recognition that d(Q/P)/dx expansion leads to the required structure, and proper handling of the constant of integration
- Part (b): Formation of differential equation by eliminating parameter 'a', obtaining dy/dx = x/(2a), substitution to get orthogonal trajectories differential equation, and verification that the resulting equation represents the same family of parabolas
- Part (c): Correct force analysis showing applied force = w(sinθ + μcosθ) upward and w(μcosθ - sinθ) downward, integration to find work done as wb(sinθ + μcosθ) + wb(μcosθ - sinθ) = 2wbμcosθ
- Part (d): Use of trajectory equation y = xtanα - (gx²sec²α)/(4gh), condition that product of slopes m₁m₂ = -1 for perpendicular directions leading to x² = 2hy, and verification that one bisector condition holds
- Part (e): Calculation of curl A showing all components vanish (∇×A = 0), and integration to find φ = 3x²y + xz³ - yz + C with correct verification that ∇φ = A
Q6 50M prove Mechanics, differential equations and vector calculus
(a) A cable of weight $w$ per unit length and length $2l$ hangs from two points P and Q in the same horizontal line. Show that the span of the cable is $2l\left(1 - \dfrac{2h^2}{3l^2}\right)$, where $h$ is the sag in the middle of the tightly stretched position. 20
(b) Solve the following differential equation by using the method of variation of parameters : $(x^2 - 1)\dfrac{d^2y}{dx^2} - 2x\dfrac{dy}{dx} + 2y = (x^2 - 1)^2$, given that $y = x$ is one solution of the reduced equation. 15
(c) Verify Green's theorem in the plane for $\displaystyle\oint_C (3x^2 - 8y^2)\,dx + (4y - 6xy)\,dy$, where C is the boundary curve of the region defined by $x = 0$, $y = 0$, $x + y = 1$. 15
हिंदी में पढ़ें
(a) $2l$ लम्बाई का एक तार (केबल) जिसका भार $w$ प्रति इकाई (यूनिट) लम्बाई है, एक क्षैतिज रेखा के दो बिन्दुओं P तथा Q से लटकी हुई है। दर्शाइए कि तार की विस्तृति (स्पैन) $2l\left(1 - \dfrac{2h^2}{3l^2}\right)$ है, जहाँ $h$ तार के कसकर खींची हुई स्थिति में मध्य का झोल है। 20
(b) प्राचल-विचरण विधि का उपयोग करके, निम्नलिखित अवकल समीकरण : $$(x^2 - 1)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = (x^2 - 1)^2$$ को हल कीजिए, जहाँ समानीत समीकरण का एक हल $y = x$ दिया गया है। 15
(c) समतल में ग्रीन के प्रमेय को $\displaystyle\oint_C (3x^2 - 8y^2)\,dx + (4y - 6xy)\,dy$ के लिए सत्यापित कीजिए, जहाँ C, $x = 0$, $y = 0$, $x + y = 1$ द्वारा परिभाषित क्षेत्र का सीमा वक्र है। 15
Answer approach & key points
This question demands rigorous mathematical proof and solution across three distinct areas: cable mechanics, differential equations, and vector calculus. Allocate approximately 40% of time to part (a) as it carries 20 marks, requiring careful derivation of the catenary approximation; spend 30% each on parts (b) and (c). Begin each part with clear statement of given data, proceed through systematic derivation/solution, and conclude with explicit verification of results.
- Part (a): Correct setup of catenary equation y = c cosh(x/c), Taylor expansion for small sag-to-span ratio, and derivation of span formula 2l(1 - 2h²/3l²) using arc length constraint
- Part (b): Identification of second solution y₂ using reduction of order or Wronskian, correct construction of particular integral via variation of parameters formula, and complete integration with arbitrary constants
- Part (c): Proper parameterization of triangular boundary C with three segments, correct evaluation of line integral ∮(Mdx + Ndy), computation of double integral ∬(∂N/∂x - ∂M/∂y)dA over region, and explicit equality verification
- Clear labeling of all three parts with logical flow between steps, proper use of mathematical notation and conventions
- Explicit statement of assumptions: tightly stretched cable (small sag approximation), continuous differentiability for Green's theorem, and non-homogeneous term handling in (b)
Q7 50M verify Vector calculus, Laplace transforms and mechanics
(a) Verify Stokes' theorem for $\vec{F} = x\hat{i} + z^2\hat{j} + y^2\hat{k}$ over the plane surface : $x + y + z = 1$ lying in the first octant. 20
(b) Solve the following initial value problem by using Laplace's transformation $\frac{d^2y}{dt^2} - 3\frac{dy}{dt} + 2y = h(t)$, where $$h(t) = \begin{cases} 2, & 0 < t < 4, \\ 0, & t > 4, \end{cases} \quad y(0) = 0, \quad y'(0) = 0$$ 15
(c) Suppose a cylinder of any cross-section is balanced on another fixed cylinder, the contact of curved surfaces being rough and the common tangent line horizontal. Let $\rho$ and $\rho'$ be the radii of curvature of the two cylinders at the point of contact and $h$ be the height of centre of gravity of the upper cylinder above the point of contact. Show that the upper cylinder is balanced in stable equilibrium if $h < \frac{\rho\rho'}{\rho+\rho'}$. 15
हिंदी में पढ़ें
(a) स्टोक्स प्रमेय को $\vec{F} = x\hat{i} + z^2\hat{j} + y^2\hat{k}$ के लिए प्रथम अष्टांशक में स्थित समतल पृष्ठ : $x + y + z = 1$ पर सत्यापित कीजिए। 20
(b) लाप्लास रूपांतरण का उपयोग करके निम्नलिखित प्रारंभिक मान समस्या : $$\frac{d^2y}{dt^2} - 3\frac{dy}{dt} + 2y = h(t), \text{ जहाँ } h(t) = \begin{cases} 2, & 0 < t < 4, \\ 0, & t > 4, \end{cases} \quad y(0) = 0, \quad y'(0) = 0$$ को हल कीजिए । 15
(c) माना किसी भी अनुप्रस्थ-काट का एक बेलन दूसरे स्थिर बेलन पर संतुलित है, जहाँ वक्रीय पृष्ठों का संपर्श रूक्ष है तथा उभयनिष्ठ स्पर्श-रेखा क्षैतिज है । माना दोनों बेलनों के स्पर्श बिंदु पर उनकी वक्रता त्रिज्याएं $\rho$ तथा $\rho'$ हैं और संपर्श बिंदु से ऊपरी बेलन के गुरुत्व केंद्र की ऊँचाई $h$ है । दर्शाइए कि स्थायी साम्य में ऊपरी बेलन संतुलित है यदि $h < \frac{\rho\rho'}{\rho+\rho'}$ । 15
Answer approach & key points
Verify Stokes' theorem in part (a) by computing both surface and line integrals; solve the IVP in part (b) using Laplace transforms with proper handling of the piecewise forcing function; prove the stability condition in part (c) using virtual work and energy methods. Allocate approximately 40% time to part (a) as it carries 20 marks, 30% each to parts (b) and (c). Structure each part with clear statement of method, step-by-step execution, and final verification or conclusion.
- For (a): Correct parameterization of the triangular boundary C in the first octant with vertices (1,0,0), (0,1,0), (0,0,1) and computation of curl F = (2y-2z)i - 0j + 0k
- For (a): Evaluation of surface integral ∫∫(curl F)·n̂ dS over the plane x+y+z=1 and line integral ∮F·dr around the triangular boundary, showing both equal 1/2
- For (b): Correct application of Laplace transform to the piecewise function h(t) using Heaviside step function: L{h(t)} = 2(1-e^{-4s})/s
- For (b): Proper partial fraction decomposition of Y(s) = 2(1-e^{-4s})/[s(s-1)(s-2)] and inversion to obtain y(t) for 0<t<4 and t>4 with continuity at t=4
- For (c): Setting up virtual displacement analysis with upper cylinder's center of gravity at height h above contact, using radii of curvature ρ and ρ'
- For (c): Deriving the stability condition by requiring the potential energy to be minimum, leading to h < ρρ'/(ρ+ρ') = 1/(1/ρ + 1/ρ')
- For (c): Physical interpretation that the equivalent radius of curvature for the combined surfaces must exceed the height of the center of gravity
Q8 50M solve Differential equations, mechanics and vector calculus
(a) (i) Find the general and singular solutions of the differential equation : $(x^2 - a^2)p^2 - 2xyp + y^2 + a^2 = 0$, where $p = \frac{dy}{dx}$. Also give the geometric relation between the general and singular solutions. 10
(ii) Solve the following differential equation : $$(3x + 2)^2\frac{d^2y}{dx^2} + 5(3x + 2)\frac{dy}{dx} - 3y = x^2 + x + 1$$ 10
(b) A chain of n equal uniform rods is smoothly jointed together and suspended from its one end A₁. A horizontal force $\vec{P}$ is applied to the other end Aₙ₊₁ of the chain. Find the inclinations of the rods to the downward vertical line in the equilibrium configuration. 15
(c) Using Gauss' divergence theorem, evaluate $\iint\limits_{S} \vec{F}.\vec{n}$ dS, where $\vec{F} = x\hat{i} - y\hat{j} + (z^2-1)\hat{k}$ and S is the cylinder formed by the surfaces z = 0, z = 1, x² + y² = 4. 15
हिंदी में पढ़ें
(a) (i) अवकल समीकरण : $(x^2 - a^2)p^2 - 2xyp + y^2 + a^2 = 0$, जहाँ $p = \frac{dy}{dx}$, के व्यापक व विचित्र हलों को ज्ञात कीजिए । व्यापक व विचित्र हलों के बीच ज्यामितीय संबंध को भी दीजिए । 10
(ii) निम्नलिखित अवकल समीकरण को हल कीजिए : $$(3x + 2)^2\frac{d^2y}{dx^2} + 5(3x + 2)\frac{dy}{dx} - 3y = x^2 + x + 1$$ 10
(b) n बराबर एकसमान छड़ों की एक श्रृंखला एक-दूसरे के साथ चिकने रूप से जुड़ी हुई है तथा इसके एक सिरे A₁ से लटकी हुई है । एक क्षैतिज बल $\vec{P}$ श्रृंखला के दूसरे सिरे Aₙ₊₁ पर लगाया गया है । साम्य विन्यास में अधोमुखी उद्वाधर रेखा से छड़ों के झुकाव ज्ञात कीजिए । 15
(c) गॉस के अपसरण प्रमेय का उपयोग करके $\iint\limits_{S} \vec{F}.\vec{n}$ dS का मान निकालिए, जहाँ $\vec{F} = x\hat{i} - y\hat{j} + (z^2-1)\hat{k}$ तथा S, पृष्ठों z = 0, z = 1, x² + y² = 4 द्वारा बना हुआ बेलन है । 15
Answer approach & key points
Solve each sub-part systematically: spend ~40% time on (a) covering both differential equations (Clairaut's form recognition and Euler-Cauchy transformation), ~30% on (b) for the static equilibrium of linked rods using virtual work or force balance, and ~30% on (c) for applying divergence theorem with careful handling of cylindrical surface and end caps. Present solutions with clear identification of method, step-by-step working, and boxed final answers.
- (a)(i) Recognize the equation as Clairaut's form, extract general solution y = px + f(p), find singular solution by eliminating p, and state that singular solution is the envelope of general solution family
- (a)(ii) Apply substitution 3x+2 = e^t to convert to constant coefficient linear ODE, solve homogeneous part, find particular integral for quadratic RHS, and back-substitute
- (b) Set up equilibrium conditions for n-rod chain using tension propagation, horizontal/vertical force balance at each joint, derive recurrence for inclinations θ_k, and obtain closed form tan θ_k = P/((2k-1)W/2) where W is rod weight
- (c) Compute div F = 1 - 1 + 2z = 2z, evaluate volume integral ∫∫∫ 2z dV over cylinder 0≤z≤1, x²+y²≤4, and verify by direct surface integration on three surfaces (curved wall, top disk, bottom disk)
- Geometric interpretation for (a)(i): singular solution touches each member of general solution family (parabola envelope)
- For (b), recognize the pattern forms arithmetically progressing horizontal reactions and express final inclination angles in terms of P, W, and n