JEE Main 2026 — Mathematics Calculus

The differential equation $x\frac{dy}{dx} - y = x^2 \cot x$xdxdy​−y=x2cotx can be rewritten as $\frac{dy}{dx} - \frac{y}{x} = x \cot x$dxdy​−xy​=xcotx.

The integrating factor is $\mu(x) = \frac{1}{x}$μ(x)=x1​. Multiplying by $\mu(x)$μ(x) gives $\frac{d}{dx}\left(\frac{y}{x}\right) = \cot x$dxd​(xy​)=cotx.

Integrating: $\frac{y}{x} = \ln|\sin x| + C$xy​=ln∣sinx∣+C, so $y = x\ln(\sin x) + Cx$y=xln(sinx)+Cx.

Using $y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$y(2π​)=2π​: we get $\frac{\pi}{2} = 0 + C\frac{\pi}{2}$2π​=0+C2π​, thus $C = 1$C=1.

Therefore $y = x(1 + \ln(\sin x))$y=x(1+ln(sinx)).

Now: $y\left(\frac{\pi}{6}\right) = \frac{\pi}{6}(1 - \ln 2)$y(6π​)=6π​(1−ln2) and $y\left(\frac{\pi}{4}\right) = \frac{\pi}{4}\left(1 - \frac{\ln 2}{2}\right)$y(4π​)=4π​(1−2ln2​).

Thus $6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right) = \pi(1 - \ln 2) - 2\pi\left(1 - \frac{\ln 2}{2}\right) = -\pi$6y(6π​)−8y(4π​)=π(1−ln2)−2π(1−2ln2​)=−π.