Math Chapter Wise

EnglishUKPSC Lecturer (Mains) 2020

Q181.If f1 and f2 are integrating factors of dy x + 2y = 1 dx and dy 2 1 – y = dx x x respectively, then

A4 1 2 f x f = B3 1 2 f x f = C4 2 1 f x f = D3 2 1 f x f =

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EnglishUKPSC Lecturer (Mains) 2020

Q182.The solution of the differential equation 2 x 2 d y dy – 3 + 2y = e, dx dx if at x = 0, y = 3 and dy = 3, dx is:

A2 e (2) 2e + + x x x B2 e (1) 2e − + x x x C2 e (2) e − + x x x D2 2e (1) e − + x x x

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EnglishUP TGT 2013 UKPSC Lecturer (Mains) 2020

Q183.The degree of the differential equation 3 2 2 d y dy dy + +sin +1 = 0 dx dx dx is:

A3 B2 C1 Dnot defined

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EnglishUKPSC Lecturer (Mains) 2020

Q184.A particular solution of the differential equation: () 3 2 2 dy x + x + x +1 = 2x + x,y = 1 dx when x = 0 is:

A() 2 1 e e 3 1 1 y log 1 tan log 4 2 2 − = + − + x x () 1 1 + + x B() 2 1 e e 3 1 1 y log 1 tan log 4 2 2 − = + + + x x () 1 2 − + x C() 2 1 e 1 y log 1 tan 1 2 − = + − + x x D() 2 e 1 y log 1 1 2 = + + + x x

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EnglishUKPSC Lecturer (Mains) 2020

Q185.If 2 3 2 d y = 2y + 2y dx and it is given that dy = 1 dx and y = 0 at x = 0, then the value of y is:

Asin x B1 2 sin 2x Ccot x Dtan x

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EnglishUKPSC Lecturer (Mains) 2020

Q186.The general solution of the partial differential equation x(z2 – y2) p + y (x2 – z2) q = z(y2 – x2) is

Axyz = f (x2 + y2 + z2) B2 2 y z f (y) + + = + x x Cy z f = x D2 yz f y = x

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EnglishUKPSC Lecturer (Mains) 2020

Q187.The partial differential equation () 2 2 2 2 2 2 2 2 2 z z z z y y 3 y y y x ∂ ∂ ∂ ∂ + − − + = ∂ ∂ ∂ ∂ ∂ x x x x x is classified as:

AParabolic BHyperbolic CElliptic DNon Linear

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EnglishUKPSC Lecturer (Mains) 2020

Q188.The solution of the differential equation () () 2 3x dy x +1 – y = e x +1 dx subject to y = 1 at x = 0, is:

A3 y (1)e 1 = + + x B3 3y (1)(e 2) = + + x C3 3y (1)(e 1) = + + x DNone of these

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EnglishUKPSC Lecturer (Mains) 2020 UPPSC GIC 2015

Q189.If ya is an integrating factor of the differential equation 2xydx– (3x2 – y2) dy = 0 then the value of a is:

A–4 B4 C–1 D1

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EnglishUKPSC Lecturer (Mains) 2020

Q190.The solution of the partial different equation xzp + yzq = xy is:

A2 z y f z − = x B2 y z f y − = x C2 yz f z − = y x D2 2 f z − = x x y

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EnglishUKPSC Lecturer (Mains) 2020

Q191.The differential equation of the family of curves y = ex (A cos x + B sin x). where A and B are arbitrary constants, is:

A2 d y dy y 0 dx dx + + = B2 d y dy 2 2y 0 d d + + = x x C2 d y dy 2 2y 0 d d − + = x x D2 d y dy 7 2y 0 d d − + = x x

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EnglishUKPSC Lecturer (Mains) 2020

Q192.The functional 1 2 2 0 d dy 2 dt dt dt + + ∫ x x such that x(0)=0, y(0)=0, x(1)=1.5, y(1)=1 is stationary for

A2 t, y 1 t = + x B2 t 1, y 1 2 = − = x C2 t t,y t 2 = + = x D2 t, y 1 t = − x

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EnglishUKPSC Lecturer (Mains) 2020

Q193.The extremals of 1 2 2 0 I[y()] [y y' ]d, = + ∫ x x x y(0) 0,y(1) 1 =

Ay = x2 By = x Cy = x3 Dy = x2 + 1

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EnglishUKPSC Lecturer (Mains) 2020

Q194.The number of extremals s for the functional 1 2 2 0 I[y()] (y y 2y y')d,y(0) 1,y(1) 2 = + − = ∫ x x x is:

A0 B2 C3 D1

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EnglishUKPSC Lecturer (Mains) 2020

Q195.The solution of Euler's equation for the functional 1 0 x (y')y'd + ∫ x x is:

Ay = x2 + c1 x + c2 By= 2 4 − x + c1 x + c2 Cy 3 1 2 c c 3 = + + x Dy 2 1 2 c c 4 = + + x where c1 and c2 are arbitrary constants.

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EnglishUKPSC Lecturer (Mains) 2020

Q196.The extremals of the functional 1 0 x 2 3 x (y') d ∫ x x is:

Ay = c1x3 +c2 By = c1x 4 +c2 Cy = c1x2 +c2 Dy = c1 x + c2 where c1 and c2 are arbitrary constants.

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EnglishUKPSC Lecturer (Mains) 2020

Q197.Which of the following differential equations gives the extremals for the variation problem? [ ] ∫ 2 2 2 2 1 J y(x) = y + x (y') dx

A2xy'–y = 0 Bx2 y'' + 2xy' – y = 0 Cx2y"–y = 0 Dx2 y" – 2xy' + y = 0

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EnglishUKPSC Lecturer (Mains) 2020

Q198.The extremal of the functional I[y(x)] = ∫ π/8 2 2 0 π [(y') – 2yy' – 16y ]dx,y(0) = 0, y = 1, 8 is:

Ay = sin 2x By = sin 4x+ 2cos4 x Cy = sin 4x Dy = –2cos 4x

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EnglishUKPSC Lecturer (Mains) 2020

Q199.The curve along which functional b 2 a I[y] (y) d = − ∫ x x has minimum value, is:

Ay = x2 Bx = y2 Cy = x DNone of these

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EnglishHaryana PGT 2019

Q200.Solution of the differential equation 2 dy xy xy dx + = is:

A2 x 2 1 1 ce y − = + B2 x 2 1 1 ce y = + C2 x 2 1 1 ce y = − DNone of these

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