$\frac{dy}{dx}=\frac{y+x}{y-x}$

$\begin{aligned} \frac {dy}{dx} & = \frac {y+x}{y-x} & \small \color{#3D99F6} \text{Let }y = vx \implies \frac {dy}{dx} = v + x \frac {dv}{dx} \\ v + x \frac {dv}{dx} & = \frac {vx+x}{vx-x} \\ v + x \frac {dv}{dx} & = \frac {v+1}{v-1} \\ x \frac {dv}{dx} & = \frac {v+1}{v-1} - v \\ & = \frac {1+2v-v^2}{v-1} \\ \int \frac {v-1}{1+2v-v^2} dv & = \int \frac 1x dx \\ - \frac 12 \int \frac {2-2v}{1+2v-v^2} dv & = \int \frac 1x dx \\ - \frac 12 \ln \left(1+2v-v^2\right) & = \ln x + \color{#3D99F6} C & \small \color{#3D99F6} \text{where }C \text{ is the constant of integration.} \\ \ln \left(1 + \frac {2y}x - \frac {y^2}{x^2} \right) & = - 2\ln x + c_1 \\ \ln \left(1 + \frac {2y}x - \frac {y^2}{x^2} \right) + \ln x^2 & = c_1 \\ \ln \left(x^2 + 2xy - y^2 \right) & = c_1 \\ x^2 + 2xy - y^2 & = c_2 & \small \color{#3D99F6} \text{Put } x=1 \text{ and }y = 0, 2 \\ \implies 1 & = c_2 \\ \implies x^2 + 2xy - y^2 - 1 & = 0 & \small \color{#3D99F6} \text{Put } x=5 \\ 25 + 10y - y^2 - 1& = 0 \\ y^2 - 10y \color{#3D99F6} - 24 & = 0 \end{aligned}$

By Vieta's formula , the products of roots of $y(5)$ is $\boxed{\color{#3D99F6}-24}$ .

Start from the original equation $\frac{dy}{dx}=\frac{y+x}{y-x}$ .

Multiply both sides by $(y-x)\ dx$ to get $\displaystyle\int(y-x)\ dy=\displaystyle\int(y+x)\ dx$ .

Apply the linearity of integration: $\displaystyle\int y\ dy-\displaystyle\int x\ dy=\displaystyle\int y\ dx+\displaystyle\int x\ dx$ .

This can be simplified to $\frac{1}{2}y^2-\displaystyle\int x\ dy=\displaystyle\int y\ dx+\frac{1}{2}x^2$ .

Manipulate to $\frac{1}{2}y^2-\frac{1}{2}x^2=\displaystyle\int y\ dx+\displaystyle\int x\ dy$ . Note that the right-hand side is equal to $xy+C$ (integration by parts).

Finally, use the quadratic equation to find $y=x±\sqrt{2x^2+C}$ . Using the initial conditions, we find $C=-1$ .

Thus, the two possible values for $y(5)$ is $12$ and $-2$ . The final answer is $-24$ .