The integrating factor of the differential equation $(1 - y^2)\frac{dx}{dy} + yx = ay$ for $(-1 < y < 1)$ is

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

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:

If $y=y(x)$ is the solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ satisfying $y(1) = 1$,then the value of $y\left(\frac{1}{2}\right)$ is

Let $y=y(x)$ be the solution of the differential equation $x(1-x^{2}) \frac{dy}{dx}+(3x^{2}y-y-4x^{3})=0, x>1$ with $y(2)=-2$. Then $y(3)$ is equal to

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x$,with the initial condition $y(0) = \frac{1}{3} + e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to: