The differential equation of the family of circles touching Y-axi

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 Multiple Choice QuestionsMultiple Choice Questions

191.

The order and degree of the differential equation d2ydx2 = 1 - dydx43 are respectively

  • 2, 3

  • 3, 2

  • 2, 4

  • 2, 2


192.

Form the differential equation of all family of lines y = mx ± 4m eliminating the arbitrary m constant 'm' is

  • d2ydx2 = 0

  • xdydx - ydydx +4 = 0

  • xdydx2 + ydydx +4 = 0

  • dydx = 0


193.

The differential equation of family of circles whose centre lies on x-axis, is

  • d2ydx2 + dydx2 + 1 = 0

  • yd2ydx2 + dydx2 - 1 = 0

  • yd2ydx2 - dydx2 - 1 = 0

  • yd2ydx2 + dydx2 + 1 = 0


194.

The solution of the differential equation y1 + logxdydx - xlogx = 0 is

  • x log(x) = y + c

  • x log(x) = yc

  • y(1 + log(x) = c

  • log(x) - y = c


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195.

The order of the differential equation whose solution is aex + be2x + ce3x + d = 0, is

  • 1

  • 2

  • 3

  • 4


196.

The differential equation of all circles which pass through the origin and whose centres lie on y-axis is

  • x2 - y2dydx - 2xy = 0

  • x2 - y2dydx + 2xy = 0

  • x2 - y2dydx - xy = 0

  • x2 - y2dydx + xy = 0


197.

If m and n are order and degree of the equation

d2ydx25 + 4 . d2ydx23d3ydx3 + d3ydx3 = x2 - 1, then

 

  • m = 3, n = 3

  • m = 3, n = 2

  • m = 3, n = 5

  • m = 3, n = 1


198.

The integrating factor of the differential equation dydxxlogx + y = 2logx is iven by

  • ex

  • log(x)

  • log(log(x))

  • x


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199.

The differential equation whose solution is (x - h)2 + (y - k)2 = a2 (a is a constant), is

  • 1+dydx23 = a2d2ydx2

  • 1+dydx23 = a2d2ydx22

  • 1+dydx3 = a2d2ydx22

  • None of these


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200.

The differential equation of the family of circles touching Y-axis at the origin is

  • x2 + y2dydx - 2xy = 0

  • x2 - y2 + 2xydydx = 0

  • x2 - y2dydx - 2xy = 0

  • x2 + y2dydx + 2xy = 0


B.

x2 - y2 + 2xydydx = 0

Let centre of circle on X-axis be (h, 0),

The radius of circle will be h.

 The equation of circle having centre (h, 0) and radius h is

     x - h2 + y - 02 = h2 x2 + h2 - 2hx + y2 = h2          x2 - 2hx + y2 = 0     ...(i)On differentiating both sides w.r.tx, we get2x - 2h +2ydydx = 0                     h = x + ydydxOn putting h = x + ydydx in Eq. (i), we getx2 - 2x + ydydxx + y2 = 0 - x2 + y2 - 2xydydx = 0    x2 - y2 + 2xydydx = 0


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