The solution of the differential equation \(\displaystyle\frac{dy}{dx}=\frac{x + y}{x-y}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\tan^{-1}\displaystyle\frac{y}{x}= \frac{1}{2}\log_e(x^2+ y^2)+C\): C is an arbitrary constant
The right side is a function of \(\frac{y}{x}\) alone, so this is a homogeneous differential equation. Let
\(y = vx\)
\( \displaystyle \frac{dy}{dx} = v + x\frac{dv}{dx} \)
Substituting:
\( \displaystyle v + x\frac{dv}{dx} = \frac{1 + v}{1 - v} \)
\( \displaystyle x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v^2}{1 - v} \)
Separating the variables:
\( \displaystyle \frac{1 - v}{1 + v^2}\,dv = \frac{dx}{x} \)
\( \displaystyle \int\frac{dv}{1 + v^2} - \int\frac{v}{1 + v^2}\,dv = \int\frac{dx}{x} \)
\( \displaystyle \tan^{-1}v - \frac{1}{2}\log_e(1 + v^2) = \log_e|x| + C \)
Putting back \(v = \frac{y}{x}\):
\( \displaystyle \frac{1}{2}\log_e\left(1 + \frac{y^2}{x^2}\right) = \frac{1}{2}\log_e(x^2 + y^2) - \log_e|x| \)
\( \displaystyle \tan^{-1}\frac{y}{x} - \frac{1}{2}\log_e(x^2 + y^2) + \log_e|x| = \log_e|x| + C \)
The \(\log_e|x|\) terms cancel:
\( \displaystyle \tan^{-1}\frac{y}{x} = \frac{1}{2}\log_e(x^2 + y^2) + C \)
NCERT (2026-27 edition):
Homogeneous Differential Equations
