The general solution of the differential equation \(x\displaystyle\frac{dy}{dx} = y + x\tan\left(\frac{y}{x}\right)\), \(x \gt 0\), is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\sin\left(\displaystyle\frac{y}{x}\right) = Cx\)
Dividing by x:
\( \displaystyle \frac{dy}{dx} = \frac{y}{x} + \tan\left(\frac{y}{x}\right) \)
The right side is a function of \(\frac{y}{x}\), so the equation is homogeneous. Let \(y = vx\), so \(\frac{dy}{dx} = v + x\frac{dv}{dx}\):
\( \displaystyle v + x\frac{dv}{dx} = v + \tan v \)
\( \displaystyle \cot v\,dv = \frac{dx}{x} \)
Integrating:
\( \log|\sin v| = \log x + \log|C| \)
\( \sin v = Cx \)
Putting back \(v = \frac{y}{x}\): \(\sin\left(\frac{y}{x}\right) = Cx\).
Why the other options are incorrect
Option (2) \(\cos\left(\frac{y}{x}\right) = Cx\) — Incorrect. Differentiating \(\cos v = Cx\) gives \(x\frac{dv}{dx} = -\cot v\), not \(\tan v\).
Option (3) \(\tan\left(\frac{y}{x}\right) = Cx\) — Incorrect. Differentiating \(\tan v = Cx\) gives \(x\frac{dv}{dx} = \sin v\cos v\), not \(\tan v\).
Option (4) \(\sin\left(\frac{y}{x}\right) = \frac{C}{x}\) — Incorrect. Differentiating \(\sin v = \frac{C}{x}\) gives \(x\frac{dv}{dx} = -\tan v\), the wrong sign.