The solution of d2xdy2 - x = k wher

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 Multiple Choice QuestionsMultiple Choice Questions

71.

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

  • 1/2

  • 2

  • 1/2π

  • None


72.

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


73.

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


74.

The solution of the equation dydx = cosx - y is

  • x + cotx - y2 = C

  • y + cotx - y2 = C

  • x + tanx - y2 = C

  • None of these


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75.

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


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76.

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)


C.

x = k(e- y - 1)

Rewriting given differential equation as,

(D2 - 1)x = k                 ...(i)where,   D = ddy

Its auxiliary equation is m2 - 1 = 0, so that

           m = 1, - 1

Hence, CF = C1ey + C2e- y

where, C1, C2 are arbitrary constants.

Now, PI = 1D2 - 1k              = k . 1D2 - 1e0 . y              = k . 102 - 1e0 . y = - k

So, solution of Eq. (i) is

x = C1ey + C2e- y - k           ...(ii)

Given that x = 0, when y = 0

Hence, Eq. (ii) gives

0  = C1 + C2 - k

 C1 + C2 = k                   ...(iii)

Multiplying both sides of Eq. (ii) by e- y , we get

x . e- y = C1 + C2e- 2y - ke- y   ...(iv)

Given that x ➔ m when y ➔ 00, m being a finite quantity

So, Eq. (iv) becomes

x × 0 = C1 + C2 × 0 - k × 0     C1 = 0                                       ...(v)

From Eqs. (iv) and (v), we get

C1 = 0 and C2 = k

Hence, Eq. (ii) becomes

x = ke- y - k

   = k(e- y - 1)


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77.

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

  • circles

  • parabolas

  • ellipses

  • hyperbolas


78.

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

  • hyperbola

  • circle

  • ellipse

  • parabola


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79.

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

  • k . ex22

  • k . ey22

  • k . ex2

  • k . exy2


80.

The differential equation of the rectangular hyperbola, where axes are the asymptotes of the hyperbola, is

  • ydudx = x

  • xdydx = - y

  • xdydx = y

  • xdy + ydx = c


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