The solution of $\frac{dy}{dx} = \frac{x+y}{x-y}$ is

Solve the differential equation given below:
$\frac{x dy}{dx} = y + \sqrt{x^2 + y^2}$

The substitution $y = z^{\alpha}$ transforms the differential equation $(x^2y^2 - 1)dy + 2xy^3dx = 0$ into a homogeneous differential equation for

$I: y^{\prime}=\frac{y+x}{x} ; \quad II: y^{\prime}=\frac{x^2+y}{x^3} ; \quad III: y^{\prime}=\frac{2xy}{y^2-x^2}$
$S1$: 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.

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:

One ticket is selected at random from $50$ tickets numbered $\{00, 01, 02, \ldots, 49\}$. Then the probability that the sum of the digits on the selected ticket is $8$,given that the product of these digits is zero,is