Question

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

• A. $\tan^{-1}\left(\frac{x}{y}\right) + \frac{1}{2}\ln|x^2+y^2| = c$
• B. $\tan^{-1}\left(\frac{y}{x}\right) + \frac{1}{2}\ln|x^2+y^2| = c$
• C. $\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln|x^2+y^2| = c$
• D. $\tan^{-1}\left(\frac{x}{y}\right) - \frac{1}{2}\ln|x^2+y^2| = c$

Hint:

Converting the problem to polar coordinates ($x = r\cos\theta, y = r\sin\theta$) is a great shortcut. The homogeneous direction equation simplifies directly to $\frac{dr}{r} = d\theta$, which integrates to $\theta - \ln r = c$. Since $\theta = \tan^{-1}(y/x)$ and $r = \sqrt{x^2+y^2}$, this yields option (C) immediately!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Solve the homogeneous differential equation dy/dx = (x+y)/(x–y).

Step 2: Key Formula or Approach:
Substitute y = vx, then dy/dx = v + x(dv/dx), separate variables.

Step 3: Detailed Explanation:
v + x(dv/dx) = (1+v)/(1–v) → x(dv/dx) = (1+v²)/(1–v). Separating: (1–v)/(1+v²) dv = dx/x. Integrating: tan⁻¹v – ½ln(1+v²) = ln|x| + c. Back-substituting v=y/x and simplifying yields tan⁻¹(y/x) – ½ln(x²+y²) = c.

Step 4: Final Answer:
The general solution is tan⁻¹(y/x) – ½ln(x²+y²) = c, matching option (C).