Let $y=y(x)$ be the solution of the differential equation $(x-x^{3}) dy=(y+yx^{2}-3x^{4}) dx, x>2$. If $y(3)=3$,then $y(4)$ is equal to :

Consider the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If the value of $y$ is $1$ when $x = 1$,then the value of $x$ for which $y = 2$ is:

The solution of $(1+x^2) \frac{dy}{dx} + 2xy - 4x^2 = 0$ 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 $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$ such that $y(0)=0,$ then $y(1)$ is equal to

Let $y = y_1(x)$ and $y = y_2(x)$ be the solution curves of the differential equation $\frac{dy}{dx} = y + 7$ with initial conditions $y_1(0) = 0$ and $y_2(0) = 1$ respectively. Then the curves $y = y_1(x)$ and $y = y_2(x)$ intersect at