The general solution of the differential equationdydx = 

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 Multiple Choice QuestionsMultiple Choice Questions

21.

If y = cos-1x, then it satisfies the differential equation

1 - x2d2ydx2 - xdydx = c, where c equal to 

  • 0

  • 3

  • 1

  • 2


22.

The solution of the differential equation ydydx = xy2x2 + ϕy2x2ϕ'y2x2 is (where, c is a constant)

  • ϕy2x2 = cx

  • y2x2 = c

  • ϕy2x2 = cx2

  • x2ϕy2x2 = c


23.

The curve y = cosx + y1/2 satisfies the differential equation

  • 2y - 1d2ydx2 + 2dydx2 + cosx = 0

  • d2ydx2 + 2dydx2 + cosx = 0

  • 2y - 1d2ydx2 -  2dydx2 + cosx = 0

  • 2y - 1d2ydx2 - dydx2 + cosx = 0


24.

The solution of the differential equation

dydx + yxlogex = 1x

under the condition y = 1 when x = e is

  • 2y = logex +1logex

  • y = logex +2logex

  • ylogex = logex +1

  • y = logex +e


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25.

If u(x) and u(x) are two independent solutions of the differential equation

d2ydx2 + b dydx + cy = 0,

then additional solution(s) of the given differential equation is(are)

  • y = 5u(x) + 8v(x)

  • y = c1{u(x) - v(x)} + c2v(x), c1 and c2 are arbitrary constants

  • y = c1u(x)v(x) + c2u(x)v(x), c1 and c2 are arbitrary constant

  • y = u(x)v(x)


26.

The solution of the differential equation y2 + 2xdydx = y satisfies x = 1, y = 1. Then, the solution is

  • x = y21 + logey

  • y = x21 + logex

  • x = y21 - logey

  • y = x21 + logex


27.

A family of curves is such that the length intercepted on the y-axis between the origin and the tangent at a point is three times the ordinate of the point of contact. The family of curves is

  • xy = C, C is a constant

  • xy2 = C, C is a constant

  • x2y = C, C is a constant

  • x2y2 = C, C is a constant


28.

The solution of the differential equation ysinxydx = xsinxy - ydy satisfying yπ4 = 1 is

  • cosxy = logey + 12

  • sinxy = logey + 12

  • sinxy = logex - 12

  • cosxy = - logex - 12


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29.

The general solution of the differential equation

dydx = x +y +12x +2y +1 is

  • loge3x + 3y + 2 + 3x + 6y = C

  • loge3x + 3y + 2 - 3x + 6y = C

  • loge3x + 3y + 2 - 3x - 6y = C

  • loge3x + 3y + 2 + 3x - 6y = C


D.

loge3x + 3y + 2 + 3x - 6y = C

Given differential equation

dydx = x +y +12x +2y +1       ...(i)

Put x + y = v

 1 + dydx = dvdx dydx = dvdx - 1 Eq. (i) becomes,  dvdx - 1 = v + 12v +1 dvdx = v + 12v +1 + 1 dvdx = v + 1 + 2v +12v +1 dvdx = 3v + 22v + 1 3v + 22v + 1dv = dx 233v +2 - 133v + 2dv = dx 23 - 133v +2dv = dx

Integrating both sides, we get

             23 - 13 × 13v +2dv = dx +C',Where c is a constant of integration.                  23v - 13log3v +23 = x + C' 23x +y - 19log3x +3y + 2 = x + C'     6x + 6y - log3x +3y + 2 = 9x +C     3x - 6y + log3x +3y + 2 = C


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30.

Let y be the solution of the differential equation

xdydx = y21 - ylogx satisfying y(1) = 1. Then, y satisfies

  • y = xy - 1

  • y = xy

  • y = xy + 1

  • y = xy + 2


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