The particular solution of the differential equation \(\left\{x\sin^2\left(\displaystyle\frac{y}{x}\right) - y\right\}dx + x\,dy = 0\), \(y = \displaystyle\frac{\pi}{4}\) when \(x = 1\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x = e^{\cot\frac{y}{x} - 1}\)
Dividing the equation by \(x\,dx\):
\(\displaystyle \frac{dy}{dx} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right)\)
The right side depends only on \(\frac{y}{x}\), so the equation is homogeneous. Let
\(y = vx\)
\(\displaystyle \frac{dy}{dx} = v + x\frac{dv}{dx}\)
Then
\(\displaystyle v + x\frac{dv}{dx} = v - \sin^2 v\)
\(\displaystyle x\frac{dv}{dx} = -\sin^2 v\)
Separating the variables:
\(\displaystyle \mathrm{cosec}^2 v\,dv = -\frac{dx}{x}\)
Integrating both sides:
\(-\cot v = -\log|x| - C\)
\(\displaystyle \cot\frac{y}{x} = \log|x| + C\)
At \(x = 1\), \(y = \frac{\pi}{4}\): \(\cot\frac{\pi}{4} = 1\) and \(\log 1 = 0\), so \(C = 1\).
\(\displaystyle \cot\frac{y}{x} = \log|x| + 1\)
\(\displaystyle \log|x| = \cot\frac{y}{x} - 1\)
Near the given point x is positive, so
\(\displaystyle x = e^{\cot\frac{y}{x} - 1}\)
NCERT (2026-27 edition):
Homogeneous Differential Equations
