Question

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

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

Hint:

For homogeneous differential equations of the form \( \frac{dy}{dx} = f\!\left(\frac{y}{x}\right) \), use the substitution \(y = vx\). This converts the equation into a separable form, which often leads to solutions involving logarithmic and inverse trigonometric functions.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The given differential equation is a homogeneous equation because the degrees of terms in the numerator and denominator are the same.
Step 2: Key Formula or Approach:
Substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$.
Step 3: Detailed Explanation:
Substituting $y=vx$ in the equation:
\[ v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v} \]
\[ x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v} \]
Separating variables:
\[ \frac{1 - v}{1 + v^2} dv = \frac{dx}{x} \]
\[ \int \left( \frac{1}{1 + v^2} - \frac{v}{1 + v^2} \right) dv = \int \frac{dx}{x} \]
Integrating:
\[ \tan^{-1} v - \frac{1}{2} \log(1 + v^2) = \log x + C \]
\[ \tan^{-1}\left(\frac{y}{x}\right) = \log x + \log\sqrt{1 + (y/x)^2} + C \]
\[ \tan^{-1}\left(\frac{y}{x}\right) = \log\left(x \cdot \sqrt{\frac{x^2 + y^2}{x^2}}\right) + C \]
\[ \tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2 + y^2} + C \]
Step 4: Final Answer:
The general solution is $\tan^{-1}\left(\frac{y}{x}\right) = \log\sqrt{x^2+y^2} + C$.