Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $x \frac{d y}{d x}-y=x^{2} \cot x, x \in(0, \pi)$. If $y\left(\frac{\pi}{2}\right)=\frac{\pi}{2}$, then $6 y\left(\frac{\pi}{6}\right)-8 y\left(\frac{\pi}{4}\right)$ is equal to :

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

x \frac{d y}{d x} - y = x^{2} cot x, x \in (0, π)

.
If

y (\frac{π}{2}) = \frac{π}{2}

, then

6 y (\frac{π}{6}) - 8 y (\frac{π}{4})

is equal to :

Answer: (C)

Dividing

x \frac{d y}{d x} - y = x^{2} cot x

x^{2}

gives

\frac{d}{d x} (\frac{y}{x}) = cot x

, so

\frac{y}{x} = ln (sin x) + C

. The condition

y (\frac{π}{2}) = \frac{π}{2}

gives

1 = ln 1 + C

, so

C = 1

and

y = x (ln (sin x) + 1)

. Then

y (\frac{π}{6}) = \frac{π}{6} (1 - ln 2), y (\frac{π}{4}) = \frac{π}{4} (1 - \frac{1}{2} ln 2),

and

6 y (\frac{π}{6}) - 8 y (\frac{π}{4}) = π (1 - ln 2) - 2 π (1 - \frac{1}{2} ln 2) = π - π ln 2 - 2 π + π ln 2 = - π .

The required value is

- π

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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