The general solution of $\frac{dy}{dx} = \frac{x+y+1}{y-x+1}$ is

$A$ function $y = f(x)$ satisfies the differential equation $\frac{dy}{dx} - y = \cos x - \sin x$ with the initial condition that $y$ is bounded when $x \rightarrow \infty$. The area enclosed by $y = f(x)$,$y = \cos x$,and the $y$-axis is

The solution of the differential equation $e^{-x}(y+1) dy + (\cos^2 x - \sin 2x) y dx = 0$ at $x=0, y=1$ is

Let $S = (0, 2 \pi) - \left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. Let $y = y(x)$,$x \in S$,be the solution curve of the differential equation $\frac{dy}{dx} = \frac{1}{1 + \sin 2x}$ with $y\left(\frac{\pi}{4}\right) = \frac{1}{2}$. If the sum of abscissas of all the points of intersection of the curve $y = y(x)$ with the curve $y = \sqrt{2} \sin x$ is $\frac{k \pi}{12}$,then $k$ is equal to:

$A$ curve $y = f(x)$ passing through the point $\left(1, \frac{1}{\sqrt{e}}\right)$ satisfies the differential equation $\frac{dy}{dx} + x e^{-\frac{x^2}{2}} = 0.$ Then which of the following does not hold good?

The differential equation $\frac{dx}{dy} = \frac{3y}{2x}$ represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity: