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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

y⁴ = 4xy + c

$ydx + (x - y^{3}) dy = 0 Here, M = y and N = x - y^{3} . We have \frac{\partial M}{\partial y} = 1 and \frac{\partial N}{\partial x} = 1, i . e ., \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

Hence, the given equation is exact.

Integrating M, i.e., y w.r.t. x, treating y as a constant, we get xy.

Again, the only terms in N which do not contain x are - y³.

Integrating (- y³) w.r.t. y we obtain $- \frac{y^{4}}{4}$ .

Hence, the solution of the given differential equation is

$xy - \frac{y^{4}}{4} + c = 0 \Rightarrow y^{4} = 4 xy + c$

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
$\propto s^{3}$
$\propto \frac{1}{s^{3}}$
$\propto s^{5}$

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

98. The solution of the differential ydx + (x - y3)dy = 0 is: xy = 13y3 + c xy = y4 + c y4 = 4xy + c 4y = y3 + c

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