Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Differential Equations

Multiple Choice Questions

211.

The solution of the differential equation $(x + 2 y^{3}) \frac{dy}{dx} = Y$ is

y³ + Cx = y
$\frac{{xy}^{4}}{2} + xy = Cy$
y³ + Cy = x
x + 2y³ = y + C

212.

The solution ofthe differential equation

$\frac{dy}{dx} = \frac{y (\log (y) - \log (x) + 1)}{x}$ is

x = ye^cy
y = xe^cy
x = ye^cx
None of these

213.

The solution of the differential equation $\frac{dy}{dx} + \sin (\frac{y + x}{2}) + \sin (\frac{y - x}{2}) = 0$ is

$\log (\tan (\frac{y}{2})) = C - 2 \sin (x)$
$\log (\tan (\frac{y}{4})) = C - 2 \sin (\frac{x}{2})$
$\log (\tan (\frac{y}{2} + \frac{π}{4})) = C - 2 \sin (x)$
$\log (\tan (\frac{y}{2} + \frac{π}{4})) = C - 2 \sin (\frac{x}{2})$

214.

The solution of differential equation $(x^{2} + y^{2}) - 2 xy \frac{dy}{dx} = 0$ is

$x^{2} + y^{2} = xC$
$x^{2} - y^{2} = xC$
x² + y² = C
x² - y² = C

215.

The solution of differential equation $\frac{dy}{dx} = \frac{x (2 \log (x) + 1)}{\sin (y) + ycos (y)}$ is

$ysin (y) = x^{2} \log (x) + C$
$y = x^{2} + \log (x) + C$
$ysin (y) = x^{2} + C$
None of these

216.

Solution of $2 ysin (x) \frac{dy}{dx} = 2 \sin (x) \cos (x) - y^{2} \cos (x), x = \frac{π}{2}$ , y = 1 is given by

y² = sin(x)
y = sin²(x)
y² = cos(x) + 1
None of these

217.

Solution of $x^{2} \frac{dy}{dx} - xy = 1 + \cos (\frac{y}{x})$ is

$\tan (\frac{y}{2 x}) = C - \frac{1}{2 x^{2}}$
$\tan (\frac{y}{x}) = C + \frac{1}{x}$
$\cos (\frac{y}{x}) = 1 + \frac{C}{x}$
$x^{2} = (C + x^{2}) \tan (\frac{y}{x})$

218.

The solution of $\frac{dy}{dx} = \cos (x) (2 - ycsc (x))$ where y = 2, when x = $\frac{π}{2}$ is

$y = \sin (x) + \csc (x)$
$y = \tan (\frac{x}{2}) + \cot (\frac{x}{2})$
$y = \frac{1}{\sqrt{2}} \sec (\frac{x}{2}) + \sqrt{2} \cos (\frac{x}{2})$
None of the above

219.
The solution of the equation $\sin^{- 1} (\frac{dy}{dx}) = x + y$ is

$\tan (x + y) + \sec (x + y) = x + C$

$\tan (x + y) - \sec (x + y) = x + C$

$\tan (x + y) - \sec (x + y) + x + C = 0$

None of the above

$\tan (x + y) - \sec (x + y) = x + C$

$Given, \sin^{- 1} (\frac{dy}{dx}) = x + y \frac{dy}{dx} = \sin (x + y) Now, put x + y = v and \frac{dy}{dx} = \frac{dv}{dx} - 1 ∴ \frac{dy}{dx} = \sin (x + y) reduces to \frac{dv}{1 + \sin (v)} = dx Now, on integrating both sides, we get \tan (v) - \sec (v) = x + C \tan (x + y) - \sec (x + y) = x + C$

220.

The solution of differential equation

$4 xy \frac{dy}{dx} = \frac{3 {(1 + x)}^{2} (1 + y^{2})}{(1 + x^{2})}$ is

$\log (1 + y) = \log (x) + 2 \tan (x) + C$
$\log (1 + y^{2}) = 3 \log (\frac{1}{x}) + 6 \tan^{- 1} (x) + C$
$\log (1 + y^{2}) = 3 \log (x) + 6 \tan^{- 1} (x) + C$
None of the above

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

219. The solution of the equation sin-1dydx = x + y is tanx + y + secx + y = x + C tanx + y - secx + y = x + C tanx + y - secx + y + x + C = 0 None of the above

219.
The solution of the equation $\sin^{- 1} (\frac{dy}{dx}) = x + y$ is

$\tan (x + y) + \sec (x + y) = x + C$

$\tan (x + y) - \sec (x + y) = x + C$

$\tan (x + y) - \sec (x + y) + x + C = 0$

None of the above