Solve $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$

Let $y = y(x)$ be the solution curve of the differential equation $(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$ with the condition $y(0) = 0$. If the curve $y = y(x)$ passes through the point $(\alpha, -\frac{1}{2})$, then a value of $\alpha$ is:

The solution of $\frac{dy}{dx} = \sin(x + y) + \cos(x + y)$ is

Let $x = x(y)$ be the solution of the differential equation $y = (x - y \frac{dx}{dy}) \sin(\frac{x}{y})$,$y > 0$ and $x(1) = \frac{\pi}{2}$. Then $\cos(x(2))$ is equal to:

The curve passing through the point $(1,2)$ given that the slope of the tangent at any point $(x, y)$ is $\frac{3x}{y}$ represents

General solution of the differential equation $\sin^3 x \frac{dx}{dy} = \sin y$ is given by