The solution of the differential equation $\frac{dy}{dx} = \frac{y}{y^2 - x}$ is

The solution of the equation $(x + 2y^3)\frac{dy}{dx} - y = 0$ is

If a curve passes through the origin and the slope of the tangent to it at any point $(x, y)$ is $\frac{x^{2}-4x+y+8}{x-2}$,then this curve also passes through the point

Match the differential equations in List $I$ to their integrating factors in List $II$.

List $I$ (Differential Equation)	List $II$ (Integrating Factor)
$(P)$ $(x^3+1)\frac{dy}{dx}+x^2y=3x^2$	$(1)$ $x^3$
$(Q)$ $x^2\frac{dy}{dx}+3xy=x^6$	$(2)$ $(x^3+1)^2$
$(R)$ $(x^3+1)^2\frac{dy}{dx}+6x^2(x^3+1)y=x^2$	$(3)$ $(x^2+1)^2$
$(S)$ $(x^2+1)\frac{dy}{dx}+4xy=\ln x$	$(4)$ $x^2+1$
	$(5)$ $(x^3+1)^{1/3}$
	$(6)$ $(x^3+1)^{1/2}$

The correct match is:

Let $y=y(x)$ be the solution of the differential equation $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$,where $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{4}) = \frac{-1}{\ln 2}$,then $y(\frac{\pi}{6})$ is:

The integrating factor of the differential equation $x \cdot \frac{dy}{dx} + 2y = x^2$ is $(x \neq 0)$.