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

491.

Let F denotes the family of ellipses whose centre is at the origin and major axis is the y-axis. Then, equation of the family F is :

$\frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} (x \frac{dy}{dx} - y) = 0$
$xy \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} (x \frac{dy}{dx} - y) = 0$
$xy \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} (x \frac{dy}{dx} - y) = 0$
$\frac{d^{2} y}{{dx}^{2}} - \frac{dy}{dx} (x \frac{dy}{dx} - y) = 0$

492.

Solution of the equation $x {(\frac{dy}{dx})}^{2} + 2 \sqrt{xy} \frac{dy}{dx} + y = 0$ is :

x + y = a
$\sqrt{x} - \sqrt{y} = \sqrt{a}$
x² + y² = a²
$\sqrt{x} + \sqrt{y} = \sqrt{a}$

493.
The solution of differential equation (x + y )(dx - dy) = dx + dy is :

x - y = ke^{x - y}

x + y = ke^{x + y}

x + y = k(x - y)

x + y = ke^{x - y}

x + y = ke^{x - y}

(x + y)dx - (x + y)dy = dx + dy

$\Rightarrow (x + y - 1) dx = (x + y + 1) dy \Rightarrow \frac{dy}{dx} = (\frac{x + y - 1}{x + y + 1}) Let x + y = v and \frac{dy}{dx} = \frac{dv}{dx} - 1 ∴ \frac{dv}{dx} - 1 = \frac{v - 1}{v + 1} \Rightarrow \frac{dv}{dx} = \frac{v - 1 + v + 1}{v + 1} \Rightarrow \int (\frac{v + 1}{2 v}) dv = \int dx \Rightarrow \frac{1}{2} \int 1 dv + \frac{1}{2} \int \frac{1}{v} dv = \int 1 dx \Rightarrow \frac{1}{2} v + \frac{1}{2} \log (v) = x + \log (c) \Rightarrow \frac{1}{2} (x + y) + \frac{1}{2} \log (x + y) = x + \log (c) \Rightarrow \log (\frac{x + y}{c^{2}}) = x - y \Rightarrow x + y = {ke}^{x - y}$