Problem Statement
Find the particular solution of the differential equation
$$x \frac{dy}{dx} - y + x \sin\left(\frac{y}{x}\right) = 0$$
given that $y = \frac{\pi}{2}$ when $x = 1$.
Verified Solution & Marking Scheme
Express DE in standard homogeneous form
$$x \frac{dy}{dx} = y - x \sin\left(\frac{y}{x}\right) \implies \frac{dy}{dx} = \frac{y}{x} - \sin\left(\frac{y}{x}\right)$$
Substitute y = vx
Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the DE:
$$v + x\frac{dv}{dx} = v - \sin v \implies x\frac{dv}{dx} = -\sin v$$
Separate variables and integrate
$$-\frac{dv}{\sin v} = \frac{dx}{x} \implies -\csc v \, dv = \frac{dx}{x}$$
Integrating both sides:
$$-\ln|\csc v - \cot v| = \ln|x| + C_1 \implies \ln|x| + \ln|\csc v - \cot v| = -C_1$$
$$\ln|x(\csc v - \cot v)| = C \implies x(\csc v - \cot v) = C$$
Using $\csc v - \cot v = \frac{1 - \cos v}{\sin v} = \tan(v/2)$:
$$x \tan\left(\frac{y}{2x}\right) = C$$
Apply initial condition y = π/2 at x = 1
$$1 \cdot \tan\left( \frac{\pi/2}{2(1)} \right) = C \implies \tan\left(\frac{\pi}{4}\right) = C \implies C = 1$$
Therefore, the particular solution is:
$$x \tan\left(\frac{y}{2x}\right) = 1 \quad \text{or} \quad \tan\left(\frac{y}{2x}\right) = \frac{1}{x} \implies y = 2x \tan^{-1}\left(\frac{1}{x}\right)$$