The general solution of the differential equation $(x-y-1) dy = (x+y+1) dx$ is

$I: y^{\prime}=\frac{y+x}{x} ; \quad II: y^{\prime}=\frac{x^2+y}{x^3} ; \quad III: y^{\prime}=\frac{2xy}{y^2-x^2}$
$S1$: Differential equations given by $I$ and $II$ are homogeneous differential equations.
$S2$: Differential equations given by $II$ and $III$ are homogeneous differential equations.
$S3$: Differential equations given by $I$ and $III$ are homogeneous differential equations.

The general solution of the differential equation $(x^3-y^3) dx = (x^2y - xy^2) dy$ is

$A$ die is thrown twice. Let $A$ be the event of getting a prime number when the die is thrown first time and $B$ be the event of getting an even number when the die is thrown second time. Then $P(A / \overline{B})=$

Let $f(x) = \sqrt{\lim_{r \rightarrow x} \left\{ \frac{2r^2 \left[(f(r))^2 - f(x)f(r)\right]}{r^2 - x^2} - r^3 e^{\frac{f(r)}{r}} \right\}}$ be differentiable in $(-\infty, 0) \cup (0, \infty)$ and $f(1) = 1$. Then the value of $ea$,such that $f(a) = 0$,is equal to:

The general solution of $\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$ is