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Differential Equations

Multiple Choice Questions

361.

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

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

362.

$The differential equation of the family y = {ae}^{x} + {bxe}^{x} + {cx}^{2} e^{x} of curves, where a, b, c are arbitrary constants, is$

$y''' + 3 y'' + 3' + y = 0$
$y''' + 3 y'' - 3' - y = 0$
$y''' - 3 y'' - 3' + y = 0$
$y''' - 3 y'' + 3' - y = 0$

363.

$The solution of \tan (y) \frac{dy}{dx} = \sin (x + y) + \sin (x - y) is$

sec(y) = 2cos(x) + c
sec(y) = - 2cos(x) + c
tan(y) = - 2cos(x) + c
sec²(y) = - 2cos(x) + c

364.
A family of curves has the differential equation $xy \frac{dy}{dx} = 2 y^{2} - x^{2}$ . Then, the family of curves is

$y^{2} = {cx}^{2} + x^{3}$

$y^{2} = {cx}^{4} + x^{3}$

$y^{2} = x + {cx}^{4}$

$y^{2} = x^{2} + {cx}^{4}$

$y^{2} = x^{2} + {cx}^{4}$

$xy \frac{dy}{dx} = 2 y^{2} - x^{2} \frac{dy}{dx} = \frac{2 y}{x} - \frac{x}{y} \frac{dy}{dx} - \frac{2 y}{x} = - \frac{x}{y} \Rightarrow y \frac{dy}{dx} - \frac{2 y^{2}}{x} = - x . . . (i) Put v = y^{2} . . . (ii) \Rightarrow (\frac{dv}{dx} = 2 y \frac{dy}{dx}) \Rightarrow (\frac{1}{2} \frac{dv}{dx} = y \frac{dy}{dx}) From eq . (i) \frac{1}{2} \frac{dv}{dx} - \frac{2 v}{x} = - x \frac{dv}{dx} - \frac{4 v}{x} = - 2 x IF = e^{\int pdx} = e^{- \frac{4}{x} dx} = e^{- 4 \log (x)} = e^{\log (x^{- 4})} = x^{- 4} Complete solution is x^{- 4} v = \int (- 2 x) x^{- 4} dx + c \frac{v}{x^{4}} = - 2 \int \frac{dx}{x^{3}} + c \Rightarrow \frac{v}{x^{4}} = \frac{1}{x^{2}} + c v = x^{2} + {cx}^{4} y^{2} = {cx}^{4} + x^{2} (from eq (ii)) which is the required family of curves .$

365.

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

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

366.

If y = y(x) is the solution of the differential equation $(\frac{2 + \sin (x)}{y + 1}) \frac{dy}{dx} + \cos (x) = 0$ then y( $\frac{π}{2}$ )is equal to

$\frac{1}{3}$
$\frac{2}{3}$
1
$\frac{4}{3}$

367.

$If u = f (r), where r^{2} = x^{2} + y^{2}, then (\frac{\partial u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}}) = ?$

f''(r)
f''(r) +f'(r)
f''(r) + $\frac{1}{r}$ f'(r)
f''(r) + rf'(r)

368.

$If \frac{dy}{dx} + 2 xtan (x - y) = 1, then \sin (x - y) = ?$

${Ae}^{- x^{2}}$
Ae^2x
${Ae}^{x^{2}}$
Ae^{- 2x}

369.

$An integrating factor of the differential equation (1 - x^{2}) \frac{dy}{dx} + xy = \frac{x^{4}}{(1 + x^{5})} {(\sqrt{1 - x^{2}})}^{3} is$

$\sqrt{1 - x^{2}}$
$\frac{x}{\sqrt{1 - x^{2}}}$
$\frac{x^{2}}{\sqrt{1 - x^{2}}}$
$\frac{1}{\sqrt{1 - x^{2}}}$

370.

$If \cos^{- 1} (\frac{y}{b}) = 2 \log (\frac{x}{2}), where x > 0, then x^{2} \frac{d^{2} y}{{dx}^{2}} + x \frac{dy}{dx} = ?$

4y
- 4y
0
- 8y

Previous Year Papers

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Multiple Choice Questions

364. A family of curves has the differential equation xydydx = 2y2 - x2. Then, the family of curves is y2 = cx2 + x3 y2 = cx4 + x3 y2 = x + cx4 y2 = x2 + cx4

364.
A family of curves has the differential equation $xy \frac{dy}{dx} = 2 y^{2} - x^{2}$ . Then, the family of curves is

$y^{2} = {cx}^{2} + x^{3}$

$y^{2} = {cx}^{4} + x^{3}$

$y^{2} = x + {cx}^{4}$

$y^{2} = x^{2} + {cx}^{4}$