The solution of the differential equation d2ydx2 = 

Subject

Mathematics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

41.

010πsinxdx is equal to

  • 20

  • 8

  • 10

  • 18


42.

The general solution of the differential equation (x + y)dx + xdy = 0 is

  • x2 + y2 = c

  • 2x2 - y2

  • x2 + 2xy = c

  • y2 + 2xy = c


43.

The order and degree of the differential 1 + 3dydx23 = 4d3ydx3 are

  • 1, 2/3

  • 3, 1

  • 3, 3

  • 1, 2


44.

The differential equation of all straight lines passing through the point (1, - 1)is

  • y = x + 1dydx + 1

  • y = x + 1dydx - 1

  • y = x - 1dydx + 1

  • y = x - 1dydx - 1


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45.

The solution of the differential equation d2ydx2 = e- 2x is

  • y = e- 2x4

  • y = e- 2x4 + cx + d

  • y = e- 2x4 + cx2 + d

  • y = e- 2x4 + c + d


B.

y = e- 2x4 + cx + d

Given differential equation isd2ydx2 = e- 2xOn integrating, we get  dydx = e- 2x- 2 + cAgain, on integrating both sides, we get     y = e- 2x4 + cx + d


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46.

The solution of the differential equation dydx + sin2y = 0 is

  • x = coty + c

  • y = cotx + c

  • x = 2cscycoty + c

  • y = 2sinycosy + c


47.

There are four letters and four addressed envelopes. The chance that all letters are not despatched in the right envelope is

  • 19/24

  • 21/23

  • 23/24

  • None of these


48.

If I = x0x0 + nhydx, then by Trapezoidal rule I is equal to

  • hy0 + yn + 2y1 + y2 + ... + yn - 1

  • h12y0 + yn + 2y1 + y2 + ... + yn - 1

  • h2y0 + yn + 2y1 + y2 + ... + yn - 1

  • hy0 + yn + 2y1 + y2 + ... + yn - 1


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49.

By graphical method, the solution of linear programming problem maxirmze z = 3x1 + 5x2 subject to 3x1 + 2x 18, x1  4, x2  6, x1  0, x2  0

  • x1 = 2, x2 = 0, z = 6

  • x1 = 2, x2 = 6, z = 36

  • x1 = 4, x2 = 3, z = 36

  • x1 = 4, x2 = 6, z = 42


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