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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

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

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

- 4y

$\cos^{- 1} (\frac{y}{b}) = 2 \log (\frac{x}{2}), where x > 0 On differentiating w r t x we get - \frac{1}{\sqrt{1 - \frac{y^{2}}{b^{2}}}} . \frac{1}{b} \frac{dy}{dx} = 2 . \frac{1}{(\frac{x}{2})} \frac{1}{2} \Rightarrow - \frac{1}{\sqrt{b^{2} - y^{2}}} . \frac{dy}{dx} = \frac{2}{x} \Rightarrow x \frac{dy}{dx} = - 2 \sqrt{b^{2} - y^{2}} . . . (i) Again, differentiating w r t . x we get, x^{2} \frac{d^{2} y}{{dx}^{2}} + x \frac{dy}{dx} = - 2 . \frac{1}{2} {(b^{2} - y^{2})}^{- \frac{1}{2}} (- 2 y) \frac{dy}{dx} \Rightarrow x^{2} \frac{d^{2} y}{{dx}^{2}} + x \frac{dy}{dx} = \frac{2 y}{- \frac{x}{2} . \frac{dy}{dx}} . \frac{dy}{dx} (from eq (i)) \Rightarrow x^{2} \frac{d^{2} y}{{dx}^{2}} + x \frac{dy}{dx} = - 4 y$

Previous Year Papers

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Multiple Choice Questions

370. If cos-1yb = 2logx2, where x > 0, thenx2d2ydx2 + xdydx = ? 4y - 4y 0 - 8y

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