Step 1: Switch to polar form:
The terms $x\,dy - y\,dx$ and $x\,dx + y\,dy$ hint at polar coordinates. Let $x = r\cos\theta$ and $y = r\sin\theta$. Then $x\,dy - y\,dx = r^2\,d\theta$ and $x\,dx + y\,dy = r\,dr$.
Step 2: Rewrite the equation:
Cross-multiply $\frac{dy}{dx} = \frac{x+y}{x-y}$ to get $(x-y)\,dy = (x+y)\,dx$. Expanding and regrouping:
\[ x\,dy - y\,dx = x\,dx + y\,dy \]
Step 3: Solve in r and theta:
Using the polar forms, $r^2\,d\theta = r\,dr$, so $d\theta = \frac{dr}{r}$. Integrating gives $\theta = \ln r + C$. Then $e^{\theta} = e^{C} r$, which we write as $e^{\theta} = c\,r$ after renaming the constant.
Step 4: Convert back and compare:
Here $r = \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}\frac{y}{x}$. So the solution is $c\sqrt{x^2+y^2} = e^{\tan^{-1}(y/x)}$, which is option (B) exactly. Options (C) and (D) have the wrong quantity under the root, since $r^2 = x^2+y^2$ always. Option (A) puts $y^2/x$ in the angle, but the angle is $y/x$.
Final Answer:
In polar form $e^{\theta} = c\,r$, which is option (B).
\[ \boxed{c(x^2+y^2)^{\frac{1}{2}} = e^{\tan^{-1}(y/x)}} \]