Math Chapter Wise

EnglishUP Higher Asst. Prof. 2021

Q161.The general solution of the Partial Differential Equation () () 2 2 y zx p x yz q x y + − + = − is

Aφ(x2 + y2 + z2, xy + z) = 0 Bφ(x2 + y2 – z2, xy + z) = 0 Cφ(x2 + y2 + z2, xyz) = 0 Dφ(xy + yz + zx, xyz) = 0

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EnglishUP Higher Asst. Prof. 2021

Q162.The complete integral of q = (z + px)2 is

Axy = 2 a x + ax + b Bxy = 2 a x + ay + b Cyz = 2 a y + ay + b Dxz = 2 a x + ay + b

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EnglishUP Higher Asst. Prof. 2021

Q163.The singular solution of the differential equation p3 – 4xyp + 8y2 = 0

A4y3 = 27x B27y3 = 4x C27y = 4x3 D4y = 27x3

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EnglishUP Higher Asst. Prof. 2021

Q164.Singular solution of the differential equation () 3 2 8p 27 x 12p y − = is

A4y3 + x3 = 0 B4y3 + 27x3 = 0 C4y3 – 27x3 = 0 D4y3 – x3 = 0

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EnglishUP Higher Asst. Prof. 2021

Q165.The solution of the Volterra integral equation () () () ∫ x 0 y x = 1+ x - t y t dt, when y0(x) = 0, is

Ay (x) = sinh x By(x) = cos x Cy(x) = sin x Dy(x) = cosh x

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EnglishUP Higher Asst. Prof. 2021

Q166.Laplace transform of Bessel function J1(x) is -

A2 s 1 s 1 + B() 3/ 2 2 1 s 1 + C() 3/ 2 2 s 1 s 1 + D2 s 1 s 1 − +

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EnglishUP Higher Asst. Prof. 2021

Q167.Inverse Laplace transform of () 4s 3 e s 2 − + is ––

A() () () 2 2 x 4 1 x 4 e H x 4 2 − − B() () () 2 2 x 4 1 x 4 e H x 4 2 − − − + C() () 2 2 x 4 x 4 e− − − D() () 2 x 4 x 4 e− − −

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EnglishUP Higher Asst. Prof. 2021

Q168.The solution of the Fredholm linear integral equation () () () 1 xt x 0 u x x e 1 u t dt e x + − = − ∫ is

A() u x 1 = B() x u x e x = − C() u x 3 = D() x u x e =

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EnglishUP Higher Asst. Prof. 2021

Q169.The Homogeneous integral equation () () () 1 0 x 3x 2 t t dt 0 φ − λ − φ = ∫ has

AOne characteristic number. BTwo characteristic number. CThree characteristic number DNo characteristic number

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EnglishUP Higher Asst. Prof. 2021

Q170.The Resolvent Kernel R (x, t; λ) of the Volterra integral equation () () () x 3 x t 0 y x 1 e y t dt − = + λ∫ shall be,

Ae(3–λ)(x–t) Be(3–λ)(x+t) Ce(3+λ)(x+t) De(3+λ)(x–t)

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EnglishUP Higher Asst. Prof. 2021

Q171.Which is not the Euler-Lagrange equation for variation problems

Ad 0 dx y' y ∂ρ ∂ρ − = ∂ ∂ B2 d 0 dx y y' ∂ρ ∂ρ − = ∂ ∂ Cd y' 0 x dx y' ∂ρ ∂ρ − ρ − = ∂ ∂ D2 2 2 2 y' y" 0 y x y' y y' y' ∂ρ ∂ ρ ∂ ρ ∂ ρ − − − = ∂ ∂ ∂ ∂ ∂ ∂

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EnglishUP Higher Asst. Prof. 2021

Q172.Solution of partial differential equation u u 2 u x t ∂ ∂ = + ∂ ∂,where u(x,0) = 6e–3x is

A6e–(2x+3t) B6e–(3x+4t) C6e (3x+4t) D6e –(3x+2t)

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EnglishUP Higher Asst. Prof. 2021

Q173.Transformation of Laplace differential equation 2 2 d d 0 dx dy ν ν + = in to polar from r, θ in R2. where ν ν ν ≡ ≡ ≡ ν ν ν (x, y), is

A2 2 2 2 2 1 1 0 r r r r ∂ ν ∂ν ∂ ν + + = ∂ ∂ ∂θ B2 2 2 2 2 1 1 0 r r r r ∂ ν ∂ν ∂ ν − + = ∂ ∂ ∂θ C2 2 2 2 2 1 1 0 r r r r ∂ ν ∂ν ∂ ν − − = ∂ ∂ ∂θ D2 2 2 2 2 1 1 0 r r r r ∂ ν ∂ν ∂ ν + − = ∂ ∂ ∂θ

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EnglishUPPSC Ashram Paddhati 2021

Q174.If c is an arbitrary constant, the general solution of the differential equation () 3 dy x + 2y = y dx is

Ax = y3 – cy2 Bx = y3 + cy Cy = x3 + cx Dy = x3 – cx2

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EnglishUPPSC Ashram Paddhati 2021

Q175.Putting x = et and denoting d dt by θ, the differential equation 3 2 d y d y dy x + 3x + x + y = logx dx dx dx is transformed into the equation

Aθ3 y = t B(θ3 + 1)y = t C(θ3 – 1)y = t D(θ3 + 3θ + 1)y = t

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EnglishUKPSC Lecturer (Mains) 2020

Q176.Which of the following partial differential equations is/are quasi-linear? I. 3 2 z z y y z y y ∂ ∂ + = + ∂ ∂ x x x x II. 2 2 2 2 z z (z) yz z y z y ∂ ∂ + = − ∂ ∂ x + x x x

AI BII CBoth I and II DNeither I nor II

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EnglishUKPSC Lecturer (Mains) 2020

Q177.Partial differential equation 2 2 z z 1 z y 0 y 2 y ∂ ∂ ∂ − − ∂ ∂ ∂ = x is: I. hyperbolic if y < 0 II. parabolic if y = 0 III. elliptic if y < 0 Which of the following options is correct?

AI and II both are correct BII and III both are correct CI and III both are correct DNone of these is correct

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EnglishUKPSC Lecturer (Mains) 2020

Q178.The solution of the differential equation 2 e 2 2 12log x d y 1 dy + = dx x dx x is: Where C1 and C2 are arbitrary constants.

A2 1 y c c = + + x x B2 2 3 1 y c c = + + x x C2 2 1 c y c = + + x x D2 3 1 y c c log 2(log) = + + x x

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EnglishUKPSC Lecturer (Mains) 2020

Q179.If y satisfies x+y dy = e dx and it if given that for x = 1, y = 1, then the value of y when x = –1 is:

A1 B0 C–1 D2

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EnglishUKPSC Lecturer (Mains) 2020

Q180.The solution curve of the different equation xdx– dy = 0 is:

AHyperbola BEllipse CCircle DNone of these

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