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

551.

The solution of the differential equation $\frac{dy}{dx} = 2 e^{x - y} + x^{2} e^{- y}$ is

$e^{y} = 2 e^{x} + \frac{x^{3}}{3} + C$
$e^{- y} = 2 e^{x} + \frac{x^{- 3}}{3} + C$
$e^{- y} = 2 e^{x} + \frac{x^{3}}{3} + C$
$e^{y} = 2 e^{- x} + \frac{x^{3}}{3} + C$

552.

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

553.

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

554.

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

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

555.

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

556.

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

557.

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

558.

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

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

559.
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

$y = \sin (x) + \csc (x)$

$Given, \frac{dy}{dx} = \cos (x) (2 - ycsc (x)) \frac{dy}{dx} = 2 \cos (x) - ycot (x) \Rightarrow \frac{dy}{dx} + ycot (x) = 2 \cos (x) IF = e^{\int \cot (x) dx} = \sin (x) ysin (x) = \int 2 \cos (x) . \sin (x) dx + C ysin (x) = \sin^{2} (x) + C at y = 2 and x = \frac{π}{2} \Rightarrow C = 1 ∴ ysin (x) = \sin^{2} (x) + 1 [∵ from Eq . (i)] \Rightarrow y = \sin (x) + \csc (x)$