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:
The given expression $\frac{dy}{dx} = \frac{x+y}{x-y}$ is a first-order homogeneous differential equation because the total degree of each individual term in the numerator and denominator is exactly equal to 1. We need to find its general algebraic implicit curve solution.

Step 2: Key Formula or Approach:
For homogeneous differential equations, we use the standard substitution: $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ This transforms the differential expression into a standard variable-separable layout form in terms of $v$ and $x$.

Step 3: Detailed Explanation:
Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into our original equation: $$v + x\frac{dv}{dx} = \frac{x + vx}{x - vx}$$ Factor out $x$ from both the top and bottom on the right side: $$v + x\frac{dv}{dx} = \frac{1+v}{1-v}$$ Isolate the $x\frac{dv}{dx}$ term by subtracting $v$ from both sides: $$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1+v - v(1-v)}{1-v} = \frac{1+v-v+v^2}{1-v} = \frac{1+v^2}{1-v}$$ Now, separate the variables $v$ and $x$: $$\frac{1-v}{1+v^2} \, dv = \frac{1}{x} \, dx$$ Split the left-hand side fraction into two independent integrateable terms: $$\left( \frac{1}{1+v^2} - \frac{v}{1+v^2} \right) dv = \frac{1}{x} \, dx$$ Integrate both sides. To integrate the second term, multiply and divide by 2 so that the numerator becomes the exact derivative of the denominator: $$\int \frac{1}{1+v^2} \, dv - \frac{1}{2}\int \frac{2v}{1+v^2} \, dv = \int \frac{1}{x} \, dx$$ $$\tan^{-1}(v) - \frac{1}{2}\ln|1+v^2| = \ln|x| + c'$$ Substitute back $v = \frac{y}{x}$: $$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln\left|1+\frac{y^2}{x^2}\right| = \ln|x| + c'$$ $$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln\left|\frac{x^2+y^2}{x^2}\right| = \ln|x| + c'$$ Expand the logarithm log-division term $\ln\left|\frac{A}{B}\right| = \ln A - \ln B$: $$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\left(\ln|x^2+y^2| - \ln|x^2|\right) = \ln|x| + c'$$ Since $\ln|x^2| = 2\ln|x|$: $$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln|x^2+y^2| + \ln|x| = \ln|x| + c'$$ The $\ln|x|$ terms cancel out perfectly on both sides, leaving: $$\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln|x^2+y^2| = c$$

Step 4: Final Answer:
The general solution is $\tan^{-1}\left(\frac{y}{x}\right) - \frac{1}{2}\ln|x^2+y^2| = c$, which corresponds to option (C).