The differential equation of the curve for which the initial ordi

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 Multiple Choice QuestionsMultiple Choice Questions

411.

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

  • 1

  • 2

  • 3

  • None


412.

If T = 2π1g, then relative errors in T and l are in the ratio

  • 1/2

  • 2

  • 1/2π

  • None


413.

The order of the differential equation whose general solution is given by

y = (C1 + C2)sin(x + C3) - C4ex + C5, is

  • 2

  • 3

  • 4

  • 5


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414.

The differential equation of the curve for which the initial ordinate of any tangent is equal to the corresponding subnormal

  • is linear

  • is homogeneous of second degree

  • has separable variables

  • is of second order


A.

is linear

If y = f(x) is the curve, then

Y - y = dydxX - x

is the equation of the tangent at (x, y).

Putting X = 0, then the initial ordinate of the tangent is y - xdydx

 Subnormal at this point is given by ydydx

ydydx = y - xdydx or yx +y = dydx

which is a homogeneous equation and by rewriting it as

dxdy = x + yy = xy + 1 or dxdy - xy = 1

We see that it is also a linear equation.


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415.

The solution of the equation dydx = cosx - y is

  • x + cotx - y2 = C

  • y + cotx - y2 = C

  • x + tanx - y2 = C

  • None of these


416.

The differential equation of all parabolas each of which has a latusrectum 4a and whose axes are parallel to the Y-axis is

  • of order 1 and degree 2

  • of order 2 and degree 3

  • of order 2 and degree 1

  • of order 2 and degree 2


417.

The solution of d2xdy2 - x = k where k is a non-zero constant, vanishes when y = 0 and tends of finite limit as y tends to infinity, is 

  • x = k(1 + e- y)

  • x = k(ey + e- y - 2)

  • x = k(e- y - 1)

  • x = k(ey - 1)


418.

The differential equation (3x + 4y + 1)dx + (4x + 5y + 1)dy = 0 represents a family of

  • circles

  • parabolas

  • ellipses

  • hyperbolas


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419.

The solution of dydx = x2 + y2 + 12xy, satisfying y(1) = 0 is given by

  • hyperbola

  • circle

  • ellipse

  • parabola


420.

If x . dydx + y = x . fxyf'xy, then f(xy) is equal to

  • k . ex22

  • k . ey22

  • k . ex2

  • k . exy2


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