If $\frac{dy}{dx} = \frac{xy + y}{xy + x}$,then the solution of the differential equation is:

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

Let $y=y(x)$ be a solution of the differential equation,$\sqrt{1-x^{2}} \frac{dy}{dx}+\sqrt{1-y^{2}}=0, |x| < 1$. If $y\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$,then $y\left(\frac{-1}{\sqrt{2}}\right)$ is equal to

General solution of the differential equation $(y^3+y)(x^2+1) dy = (xy^4+2y^2x) dx$ is (where $C$ is a constant of integration.)

If $y(x)$ is the solution of the differential equation $x dy - (y^2 - 4y) dx = 0$ for $x > 0$ and $y(1) = 2$,and the slope of the curve $y = y(x)$ is never zero,then the value of $10y(\sqrt{2})$ is . . . .

The equation of the curve passing through $\left(\frac{\pi}{6}, 0\right)$ and satisfying the differential equation $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$ is: