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Differential Equations

Multiple Choice Questions

91.

The solution of the differential equation $\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$ is

$\csc (x + y) + \tan (x + y) = x + c$
$x + \csc (x + y) = c$
$x + \tan (x + y) = c$
$x + \sec (x + y) = c$

92.

The differential equation of all straight lines touching the circle x² + y² = a² is

${(y - \frac{dy}{dx})}^{2} = a^{2} [1 + {(\frac{dy}{dx})}^{2}]$
${(y - x \frac{dy}{dx})}^{2} = a^{2} [1 + {(\frac{dy}{dx})}^{2}]$
$(y - x \frac{dy}{dx}) = a^{2} [1 + \frac{dy}{dx}]$
$(y - \frac{dy}{dx}) = a^{2} [1 - \frac{dy}{dx}]$

93.

The differential equation $|\frac{dy}{dx}| + |y| + 3 = 0$ admits

infinite number of solutions
no solutions
a unique solution
many solutions

94.

Solution of the differential equation $xdy - ydx - \sqrt{x^{2} + y^{2}} dx = 0$

$y - \sqrt{x^{2} + y^{2}} = {cx}^{2}$
$y + \sqrt{x^{2} + y^{2}} = {cx}^{2}$
$y + \sqrt{x^{2} + y^{2}} = {cy}^{2}$
$x - \sqrt{x^{2} + y^{2}} = {cy}^{2}$

95.

If $xsin (\frac{x}{y}) dy = [ysin (\frac{y}{x}) - x] dx$ and $y (1) = \frac{π}{2}$ , then the value of $\cos (\frac{y}{x})$ is equal to :

x
$\frac{1}{x}$
$\log (x)$
e^x

96.

The differential equation of the system of all circles of radius r in the xy plane is :

${[1 + {(\frac{dy}{dx})}^{3}]}^{2} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{2}$
${[1 + {(\frac{dy}{dx})}^{3}]}^{2} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{3}$
${[1 + {(\frac{dy}{dx})}^{2}]}^{3} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{2}$
${[1 + {(\frac{dy}{dx})}^{2}]}^{3} = r^{2} {(\frac{d^{2} y}{{dx}^{2}})}^{3}$

97.

The general solution of the differential equation

$\frac{d^{2} y}{{dx}^{2}} + 2 \frac{d y}{d x} + y = 2 e^{3 x}$ is given by

$y = (c_{1} + c_{2} x) e^{x} + \frac{e^{3 x}}{8}$
$y = (c_{1} + c_{2} x) e^{- x} + \frac{e^{- 3 x}}{8}$
$y = (c_{1} + c_{2} x) e^{- x} + \frac{e^{3 x}}{8}$
$y = (c_{1} + c_{2} x) e^{x} + \frac{e^{- 3 x}}{8}$

98.

The solution of the differential ydx + (x - y³)dy = 0 is:

xy = $\frac{1}{3} y^{3} + c$
xy = y⁴ + c
y⁴ = 4xy + c
4y = y³ + c

99.
If the distances covered by a particle in time t is proportional to the cube root of its velocity, then the acceleration is

a constant

$\propto s^{3}$

$\propto \frac{1}{s^{3}}$

$\propto s^{5}$

$\propto s^{5}$

$∵ s^{3} \propto v \Rightarrow \frac{dS}{dt} = {ks}^{3} . . . (i) \Rightarrow \frac{d^{2} S}{{dt}^{2}} = 3 {ks}^{2} \frac{dS}{dt} \Rightarrow \frac{d^{2} S}{{dt}^{2}} = 3 {ks}^{2} ({ks}^{3}) [∵ using Eq . (i)] \Rightarrow \frac{d^{2} S}{{dt}^{2}} = 3 k^{2} s^{2}$

Thus, From above, it is clear that acceleration of a particle is proportional to s⁵.

100.

The solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \frac{ϕ (\frac{y}{x})}{ϕ' (\frac{y}{x})}$ is

$xϕ (\frac{y}{x}) = k$
$ϕ (\frac{y}{x}) = kx$
$yϕ (\frac{y}{x}) = k$
$ϕ (\frac{y}{x}) = ky$

Previous Year Papers

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Multiple Choice Questions

99. If the distances covered by a particle in time t is proportional to the cube root of its velocity, then the acceleration is a constant ∝ s3 ∝ 1s3 ∝ s5

99.
If the distances covered by a particle in time t is proportional to the cube root of its velocity, then the acceleration is

a constant

$\propto s^{3}$

$\propto \frac{1}{s^{3}}$

$\propto s^{5}$