General solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = 0$ is

A. $\tan^{-1} x + \tan^{-1} y = C$

B. $\sin^{-1} x - \cos^{-1} y = C$

C. $x\sqrt{1 - y^2} - y\sqrt{1 - x^2} = C$

D. $\sin^{-1} x + \sin^{-1} y = C$

E. $\cos^{-1} x + \cos^{-1} y = C$

(where C is arbitrary constant)

Choose the correct answer from the options given below

The solution of the differential equation $xdy - ydx = 0$ represents

The solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x+2}{3-y}, (y \neq 3), y(0) = 0$

represents

The solution of the initial value problem $x\frac{dy}{dx} = 2, y(1) = 2$ is

In the context of differential equation

Match List I with List II

Choose the correct answer from the options given below:

The general solution of the differential equation $\frac{dy}{dx} = \frac{3e^{2x}+3e^{4x}}{e^x+e^{-x}}$ is

The order and the degree of the differential equation $\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 = 2x^2 \log\left(\frac{d^2y}{dx^2}\right)$ are respectively: