Solve the differential equation $(x \, dy - y \, dx) y \sin \left( \frac{y}{x} \right) = (y \, dx + x \, dy) x \cos \left( \frac{y}{x} \right)$.

$\sec \left( \frac{y}{x} \right) = C \, xy$

The correct answer is Option (2) → $\sec \left( \frac{y}{x} \right) = C \, xy$ ##

The given differential equation can be written as

$\left[ xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right) \right] dy = \left[ xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right) \right] dx$

or $\quad \frac{dy}{dx} = \frac{xy \cos \left( \frac{y}{x} \right) + y^2 \sin \left( \frac{y}{x} \right)}{xy \sin \left( \frac{y}{x} \right) - x^2 \cos \left( \frac{y}{x} \right)}$

Dividing numerator and denominator on RHS by $x^2$, we get

$\frac{dy}{dx} = \frac{\frac{y}{x} \cos \left( \frac{y}{x} \right) + \left( \frac{y^2}{x^2} \right) \sin \left( \frac{y}{x} \right)}{\frac{y}{x} \sin \left( \frac{y}{x} \right) - \cos \left( \frac{y}{x} \right)} \quad \dots (1)$

Clearly, equation $(1)$ is a homogeneous differential equation of the form $\frac{dy}{dx} = g \left( \frac{y}{x} \right)$.

To solve it, we make the substitution

$y = vx \quad \dots (2)$

or $\quad \frac{dy}{dx} = v + x \frac{dv}{dx}$

or $\quad v + x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} \quad$ (using (1) and (2))

or $\quad x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v}$

or $\quad \left( \frac{v \sin v - \cos v}{v \cos v} \right) dv = \frac{2 \, dx}{x}$

Therefore $\quad \displaystyle\int \left( \frac{v \sin v - \cos v}{v \cos v} \right) dv = 2 \displaystyle\int \frac{1}{x} dx$

or $\quad \displaystyle\int \tan v \, dv - \displaystyle\int \frac{1}{v} dv = 2 \displaystyle\int \frac{1}{x} dx$

or $\quad \log | \sec v | - \log | v | = 2 \log | x | + \log | C_1 |$

or $\quad \log \left| \frac{\sec v}{v \, x^2} \right| = \log | C_1 |$

or $\quad \frac{\sec v}{v \, x^2} = \pm C_1 \quad \dots (3)$

Replacing $v$ by $\frac{y}{x}$ in equation $(3)$, we get

$\frac{\sec \left( \frac{y}{x} \right)}{\left( \frac{y}{x} \right) (x^2)} = C \text{ where, } C = \pm C_1$

or $\quad \sec \left( \frac{y}{x} \right) = C \, xy$

which is the general solution of the given differential equation.