Math Chapter Wise

EnglishJSSC PGT 2018

Q61.The steady state solution µ(x, t) = µs (x) of heat equation µt = a2 µxx satisfying the conditions µ (a, t) = –1, µ (2, t) = 1 is:

A() x 1– x µ = B() s x –1 x µ = + C() s x –1 3x µ = + D() s x 1 2x µ = +

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EnglishJSSC PGT 2018

Q62.The order and degree of the differential equation x2 (dx)2 + 2xydxdy + y2 (dy)2 – z2 (dz)2 = 0 are respectively:

A1 and 1 B1 and 2 C2 and 2 D2 and 1

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EnglishJSSC PGT 2018

Q63.Which of the following can be reduced to a Clairaut's equation?

Ay' = ex + ey + y2 By' = sin x + sin y Cy = x4 p – p2 x, where p = dy dx Dy = 2xp + y2 p3, where p = dy dx

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EnglishJSSC PGT 2018

Q64.The equation of the characteristic curve of the one-parameter family (x –a)2 + (y – a)2 + z2 = 1 is:

A(x – y)2 = 2 (1 – z2) B(x – y)2 = 1 – z2 C(x – z)2 = 2(1 – y2) D(x – z)2 = 1 – y2

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EnglishJSSC PGT 2018

Q65.The partial differential equation by eliminating the arbitrary function f from z = emx f (x + y) is:

Ap – q = mz Bq – r = z Cp – t = z Dp + q = mz

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EnglishJSSC PGT 2018

Q66.If x and e–x are two linearly independent solutions of the equation () 2 2 2 d y dy 1+ x + x – y dx dx () 2 = 1+ x, then by the method of variation of parameter, the particular integral is:

Ax + 1 – x2 Bx2/– 1 – ex Cx2/+ 1 – x Dx + 1 – ex

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EnglishJSSC PGT 2018

Q67.The integral surface of of yp + xq – z = 0 passing through the curve z = x3, y = 0 is:

A() ()() 2 2 2 2 z 1+ x – y = x – y x + y + z B()() 2 2 2 2 1+ x – y x – y x + y + z = C() ()() 2 2 2 2 z 1– x – y = x – y x + y + z D() ()() 2 2 2 2 z 1+ x – y = x – y x + y + z

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EnglishJSSC PGT 2018

Q68.The particular integral of the equation (2D2 – D')z = 10ex–3y is:

A2ex – 3y Bex – 3y C2ex + 3y Dex + 3y

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EnglishJSSC PGT 2018

Q69.The integral curves of the equation () 2 3 3 dx = x y – z () () 2 3 3 2 3 3 dy dz = y z – x z x – y are given by: Where c1 and c2 are arbitrary constants.

Ax2 + y2 – z2 = c1 and x–1 + y–1 – z–1 = c2 Bx2 – y2 + z2 = c1 and x–1 + y–1 + z–1 = c2 Cx2 – y2 – z2 = c1 and x–1 + y–1 –z–1 = c2 Dx2 + y2 + z2 = c1 and x–1 + y–1 +z–1 = c2

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EnglishJSSC PGT 2018

Q70.The differential equation for the family of surfaces x3 z + x2 y = c, where c is a parameter, is:

A(3xz + 2y)dx – xdy – x2 dz = 0 B(2xz + y)dx + xdy + x2 dz = 0 C(3xz + 2y)dx + xdy + x2 dz = 0 D(xz + 2y)dx + xdy + xdz = 0

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EnglishJSSC PGT 2018

Q71.The general solution of the differential equation x 1 y" – y'+ y = 0 x –1 x –1 is:

Ay = c2x – c1 cos x By = c1 (x2 – 1) + c2x Cy = c1ex – c2x2 ex Dy = c1x – c2ex

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EnglishJSSC PGT 2018

Q72.If the roots of the auxiliary equation of the differential equation 4 2 2 4 3 2 d y d y d y dy – 5 + 6 + 4 dx dx dx dx –8y = 0 are 2, 2, 2 and –1, then the general solution is:

Ay = c1e2x + c2e–x By = c1e2x + c2e–x + c3ex + c4e–2x Cy = (c1 + c2x + c3x2)e–x + c4e2x Dy = (c1 + c2x + c3x2)e2x + c4e–x

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EnglishJSSC PGT 2018

Q73.Which of the following is a Cauchy-Euler equation?

A3 2 d y d y dy x – 4x +8x – 8y = 4 dx dx dx lnx B3 2 3 3 3 2 d y d y x – 2x +8y = 2x dx dx C4 3 2 3 3 x 4 3 2 d y d y d y x – x + x e dx dx dx = D2 3 2 d y + 2xy x dx =

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EnglishJSSC PGT 2018

Q74.If y = y1 (x) is not a known solution, then by the removal of the first derivative, the general solution of y" + P (x) y' + Q(x) y = R(x) is obtained, where y1 is given by:

Ae∫Pdx B1 Pdx 2 e ∫ C1 – Pdx 2 e ∫ D– Pdx/e ∫

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EnglishKVS TGT 2017

Q75.The solution of the differential equation dy 2x y 3 dx represents–

Aa straight line Bcircles Cparabola Dellipse

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EnglishKVS TGT 2017

Q76.Differential equation for y = A cos αx + B sin αx, where A and B are arbitrary constants, is–

A2 d y y 0 dx B2 d y y 0 dx C2 d y y 0 dx D2 d y y 0 dx

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EnglishKVS TGT 2017

Q77.The integrating factor of 3 dy y x 3 dx x is–

A–x Blog x Cx Dex

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EnglishKVS TGT 2017

Q78.The solution of 1 3 dy y dx x is–

A2 2 3 3 x y C B1 1 3 3 x y C C2 2 3 3 y x C D1 1 3 3 y x C

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EnglishOPSC PGT 2017

Q79.Shooting method is used to solve:

AAny differential equation BOnly initial value problems COnly boundary value problems DSystem of differential equations

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EnglishOPSC PGT 2017

Q80.In recurrence relation r 2 r 1 r a 2a a, 0 a 2, 1 a 1, the ar is given by:

Ar/1 2r 2 Br/1 2r 2 Cr/1 2r 2 Dr/1 2r 2

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