Differential Equation – Homogeneous – JEE Main 2025 Shift 2

Solution:

The given differential equation is:
$$y = \left(x – y \frac{dx}{dy}\right) \sin\left(\frac{x}{y}\right)$$

Multiplying by $dy$, we can rearrange the terms:
$$y \, dy = (x \, dy – y \, dx) \sin\left(\frac{x}{y}\right)$$
$$y \, dy = -(y \, dx – x \, dy) \sin\left(\frac{x}{y}\right)$$

Dividing both sides by $y^2$:
$$\frac{dy}{y} = – \left( \frac{y \, dx – x \, dy}{y^2} \right) \sin\left(\frac{x}{y}\right)$$

We know that $d\left(\frac{x}{y}\right) = \frac{y \, dx – x \, dy}{y^2}$. Substituting this into the equation:
$$\frac{dy}{y} = – \sin\left(\frac{x}{y}\right) d\left(\frac{x}{y}\right)$$

Integrating both sides:
$$\int \frac{dy}{y} = \int – \sin\left(\frac{x}{y}\right) d\left(\frac{x}{y}\right)$$
$$\ln y = \cos\left(\frac{x}{y}\right) + C$$

Applying the initial condition $x(1) = \frac{\pi}{2}$ (i.e., when $y=1, x=\frac{\pi}{2}$):
$$\ln 1 = \cos\left(\frac{\pi/2}{1}\right) + C$$
$$0 = \cos\left(\frac{\pi}{2}\right) + C$$
$$0 = 0 + C \Rightarrow C = 0$$

So, the particular solution is:
$$\ln y = \cos\left(\frac{x}{y}\right)$$

We need to find $\cos(x(2))$. Let $x_2 = x(2)$ be the value of $x$ when $y=2$.
Substitute $y=2$ into the solution:
$$\ln 2 = \cos\left(\frac{x_2}{2}\right)$$

Using the double angle identity $\cos \theta = 2\cos^2\left(\frac{\theta}{2}\right) – 1$:
$$\cos(x_2) = 2\cos^2\left(\frac{x_2}{2}\right) – 1$$
Substitute $\cos\left(\frac{x_2}{2}\right) = \ln 2$:
$$\cos(x_2) = 2(\ln 2)^2 – 1$$

Ans. (2)