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 Multiple Choice QuestionsMultiple Choice Questions

61.

The differential equation of the family of curves y = e2x(acos(x) + bsin(x)), where a and b are arbitrary constants, is given by

  • y2 - 4y1 + 5y = 0

  • 2y2 - y1 + 5y = 0

  • y2 + 4y1 - 5y = 0

  • y2 - 2y1 + 5y = 0


62.

The solution of the differential equation xdy - ydx = x2 + y2dx is

  • x + x2 + y2 = cx2

  • y - x2 + y2 = cx2

  • x - x2 + y2 = cx

  • y + x2 + y2 = cx2


63.

The differential equation of all non-vertical lines in a plane is

  • d2ydx2 = 0

  • d2xdy2 = 0

  • dydx = 0

  • dxdy = 0


64.

The differential equation of all parabolas whose axes are parallel to y-axis is

  • d3ydx3 = 0

  • d2xdy2 = 0

  • d3ydx3+ d2xdy2 = 0

  • d2ydx2 + 2dydx = 0


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65.

Solution of Given equation is    dydx = xlogx2 + xsiny + ycosy siny + ycosydy = xlogx2 + xdxOn integrating both sides, we get  siny + ycosydy = xlogx2 + xdx - cosy + ysiny + cosy = x22logx2 is

  • ysiny = x2logx + C

  • ysiny = x2 + C

  • ysiny = x2 + logx + C

  • ysiny = xogx + C


66.

If xpyq = (x + y)p + q, then dydx is equal to

  • yx

  • pyqx

  • xy

  • qypx


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67.

The solution of dydx + ytanx = secx is

  • ysecx = tanx + C

  • ytanx = secx + C

  • tanx = ytanx + C

  • xsecx = ytany + C


A.

ysecx = tanx + C

Given equation is dydx + ytanx = secx

Here, P = tan(x) and Q = sec(x)

 IF = etanxdx         = elogsecx = secx Solution is ysecx = sec2x + C                  ysecx = tanx + C


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68.

If y = tan-1sinx + cosxcosx - sinx, then dydx is

  • 12

  • π4

  • 0

  • 1


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69.

The differential equation dydx + y2x2 = y/x has the solution

  • x = y(log(x) + C)

  • y = x(log(y) + C)

  • x = (y + C)log(x)

  • y = (x + C)log(y)


70.

The degree of the differential equation satisfying 1 - x2 + 1 - y2 =ax - y

  • 1

  • 2

  • 3

  • None


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