If y = y(x) is the solution of the differential equation dyd

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 Multiple Choice QuestionsMultiple Choice Questions

171.

Let y = y(x) be the solution of the differential equation, x2 + 1dydx + 2xx2 + 1y = 1 such that y(0) = 0. If ay1 = π32, then the value of 'a' is :

  • 14

  • 1

  • 116

  • 12


172.

The solution of the differential equation xdydx + 2y = x2, (x  0) with y(1)  = 1, is :

  • y = 45x3 + 15x2

  • y = 34x3 + 14x2

  • y = x24 + 34x2

  • y = x35 + 15x2


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173.

If y = y(x) is the solution of the differential equation dydx = tanx - ysec2x, x  - π2, π2, such that y(0) = 0, then y- π4 is equal to

  • 2 + 1e

  • e - 2

  • 12 - e

  • 1e - 2


B.

e - 2

Put tanx = t  dydx = t - y - ddtt - y = t - y - 1 dt - yt - y - 1 = - dt lnt - y - 1 = - t +CSo, the equation will betanx + ln- tanx + y + 1 = 0Put x = - π4 y + 2 = e  y = e - 2


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174.

Let y = y(x) be the solution of the differential equation dydx + ytanx = 2x +x2tanx, x  - π2, π2 such that if y(0) = 1, then

  • y'π4 - y'- π4 = π - 2

  • yπ4 - y- π4 = 2

  • yπ4 + y- π4 = π22 + 2

  • y'π4 + y'- π4 = - 2


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175.

The solution of the differential equation dxx + dyy = 0 is

  • xy = c

  • x + y = c

  • log(x)log(y) = c

  • x2 + y2 = c


176.

The differential equation obtained by eliminating arbitrary constants from y = a . ebx, is

  • yd2ydx2 + dydx = 0

  • yd2ydx2 - dydx = 0

  • yd2ydx2 - dydx2 = 0

  • yd2ydx2 + dydx2 = 0


177.

f(x) is a polynomial of degree 2, f(0) = 4, f'(0) = 3 and f''(0) = 4, then f(- 1) is equal to

  • 3

  • - 2

  • 2

  • - 3


178.

Solution of differential equation sec(x)dy - cosec(y)dx = 0 is

  • cos(x) + sin(y) = c

  • sin(x) + cos(y) = 0

  • sin(y) - cos(x) = c

  • cos(y) - sin(x) = c


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179.

The point P(9/2, 6) lies on the parabola y2 = 4ax, then parameter of the point P is

  • 3a2

  • 23a

  • 23

  • 32


180.

Solution of dydx = 3x +y is

  • 3x + y = c

  • 3x + 3y = c

  • 3- y

  • 3x + 3- y = c


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