The general solution of the differential equation dydx 

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 Multiple Choice QuestionsMultiple Choice Questions

661.

The equation of the curve through the point (1, 0), whose slope is y - 1x2 + x, is

  • 2x(y - 1) + x + 1 = 0

  • (x + 1)(y - 1) + 2x = 0

  • x(y - 1)(x + 1) + 2 = 0

  • None of these


662.

The solution of the equation dydx = 1x + y + 1 is

  • x + y = Cey - 2

  • x + y = Clog(y) - 4

  • log(x+ y + 2) = Cy

  • log(x + y + 2) = C + y


663.

The solution of differential equation (ylog(x) - 1)ydx = xdy is

  • ylogex + Cx = 1

  • logxe + Cxx = y

  • logCx2 + ex2y = x

  • None of these


664.

The solution of the differential equation a + xdydx + xy = 0 is

  • y = Ce232a  - xx + a

  • y = Ce23a  - xx + a

  • y = Ce232a  + xx + a

  • y = Ce- 232a  - xx + a


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665.

The general solution of the differential equation dydx = ytanx - y2secx is

  • tanx = C + secxy

  • secy = C + tanyx

  • secx = C + tanxy

  • tany = C + secxx


C.

secx = C + tanxy

Given differential equation is,                           dydx = ytanx - y2secx      dydx - ytanx = - y2secx 1y2dydx - tanxy = - secx      ...iPut - 1y = uFrom Eq. (i),      dudx + tanx . u = - secx        ...iiThis is linear differental equaton of the formdudx + P . u = Q, where P = tanx, Q = - secx IF = etanxdx = elogsecx = secxHence, general solution isu . IF = IF . Qdx + C1 u . secx = secx . - secxdx + C1 u . secx = - sec2xdx +C1 u . secx = - tanx + C1 - secxy = - tanx + C1     u = - 1y      secx = ytanx - C1      secx = C + tanxy     where, C = - C1


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666.

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

  • 1

  • 2

  • 3

  • None of the above


667.

The solution of the differential equation y - xdydx = ay2 + dydx is

  • y = C(x + a)(1 - ay)

  • y = C(x + a)(a + ay)

  • y = C(x - a)(1 - ay)

  • None of the above


668.

The solution of differential equation (2y - 1)dx - (2x + 3)dy = 0 will be

  • 2x - 12y + 3 = C

  • 2y + 12x - 3 = C

  • 2x + 32y - 1 = C

  • 2x - 12y - 1 = C


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669.

The solution of the differential equation dydx = xlogx2 + xsiny + ycosy will be

  • ysin(y) = x2log(x) + C

  • ysin(y) = x2 + C

  • ysin(y) = x2 + lo(x) + C

  • ysin(y) = xlog(x) + C


670.

The order and degree of the differential equation y'4 = 1 - y''' are respectively.

  • 3, 4

  • 1, 2

  • 3, 2

  • 3, 1


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