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Differential Equations

Multiple Choice Questions

71.

If T = 2 $π$ $\sqrt{\frac{1}{g}}$ , then relative errors in T and l are in the ratio

1/2
2
1/2 $π$
None

72.

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

y = (C₁ + C₂)sin(x + C₃) - C₄e^{x + C₅}, is

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 $\frac{d y}{d x} = \cos (x - y)$ is

$x + \cot (\frac{x - y}{2}) = C$
$y + \cot (\frac{x - y}{2}) = C$
$x + \tan (\frac{x - y}{2}) = C$
None of these

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

76.
The solution of $\frac{d^{2} x}{d y^{2}} - x = k$ 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(e^y + e^{- y} - 2)

x = k(e^{- y} - 1)

x = k(e^y - 1)

x = k(e^{- y} - 1)

Rewriting given differential equation as,

$(D^{2} - 1) x = k . . . (i) where, D = \frac{d}{dy}$

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

m = 1, - 1

Hence, CF = C₁e^y + C₂e^{- y}

where, C₁, C₂ are arbitrary constants.

$Now, PI = \frac{1}{D^{2} - 1} k = k . \frac{1}{D^{2} - 1} e^{0 . y} = k . \frac{1}{0^{2} - 1} e^{0 . y} = - k$

So, solution of Eq. (i) is

x = C₁e^y + C₂e^{- y} - k ...(ii)

Given that x = 0, when y = 0

Hence, Eq. (ii) gives

0 = C₁ + C₂ - k

$\Rightarrow C_{1} + C_{2} = k . . . (iii)$

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

$x \times 0 = C_{1} + C_{2} \times 0 - (k \times 0) C_{1} = 0 . . . (v)$

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

C₁ = 0 and C₂ = k

Hence, Eq. (ii) becomes

x = ke^{- y} - k

= k(e^{- y} - 1)

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 $\frac{d y}{d x} = \frac{x^{2} + y^{2} + 1}{2 xy}$ , satisfying y(1) = 0 is given by

hyperbola
circle
ellipse
parabola

79.

If $x . \frac{d y}{d x} + y = x . \frac{f (xy)}{f' (xy)}$ , then f(xy) is equal to

$k . e^{\frac{x^{2}}{2}}$
$k . e^{\frac{y^{2}}{2}}$
$k . e^{x^{2}}$
$k . e^{\frac{xy}{2}}$

80.

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

$y \frac{d u}{d x} = x$
$x \frac{d y}{d x} = - y$
$x \frac{d y}{d x} = y$
xdy + ydx = c

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

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)

76.
The solution of $\frac{d^{2} x}{d y^{2}} - x = k$ 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(e^y + e^{- y} - 2)

x = k(e^{- y} - 1)

x = k(e^y - 1)