Differential Equations – Linear Differential Equation – JEE Main 28 Jan 2026 Shift 2

Solution:

Rewrite the equation in linear form $\frac{dy}{dx} + Py = Q$:
$$x\frac{dy}{dx} – y = x^2 \cot x \Rightarrow \frac{dy}{dx} – \frac{1}{x}y = x \cot x$$
Integrating Factor (I.F.):
$$I.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$$
Solution is given by $y(I.F.) = \int Q(I.F.) dx + C$:
$$y \cdot \frac{1}{x} = \int (x \cot x) \cdot \frac{1}{x} dx$$
$$\frac{y}{x} = \int \cot x dx = \ln|\sin x| + C$$
$$y = x(\ln|\sin x| + C)$$

Given condition $y(\frac{\pi}{2}) = \frac{\pi}{2}$:
$$\frac{\pi}{2} = \frac{\pi}{2}(\ln(\sin\frac{\pi}{2}) + C)$$
$$1 = \ln(1) + C \Rightarrow C = 1$$
Function is $y = x(\ln(\sin x) + 1)$.

Find $6y(\frac{\pi}{6}) – 8y(\frac{\pi}{4})$:
$$y(\frac{\pi}{6}) = \frac{\pi}{6}(\ln(\frac{1}{2}) + 1) = \frac{\pi}{6}(1 – \ln 2)$$
$$y(\frac{\pi}{4}) = \frac{\pi}{4}(\ln(\frac{1}{\sqrt{2}}) + 1) = \frac{\pi}{4}(1 – \frac{1}{2}\ln 2)$$

Substitute into the expression:
$$6 \left[ \frac{\pi}{6}(1 – \ln 2) \right] – 8 \left[ \frac{\pi}{4}(1 – \frac{1}{2}\ln 2) \right]$$
$$= \pi(1 – \ln 2) – 2\pi(1 – \frac{1}{2}\ln 2)$$
$$= \pi – \pi\ln 2 – 2\pi + \pi\ln 2$$
$$= -\pi$$

Ans. (3)