JEE Main 2026 — Mathematics Calculus

The differential equation is $\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x$secxdxdy​−2y=2+3sinx, or $\frac{dy}{dx} - 2\cos x \cdot y = (2 + 3\sin x)\cos x$dxdy​−2cosx⋅y=(2+3sinx)cosx.

This is a linear ODE with integrating factor $\mu(x) = e^{-2\sin x}$μ(x)=e−2sinx.

Multiplying both sides and integrating: $\frac{d}{dx}[ye^{-2\sin x}] = (2 + 3\sin x)\cos x \cdot e^{-2\sin x}$dxd​[ye−2sinx]=(2+3sinx)cosx⋅e−2sinx

Computing the integral: $\displaystyle \int 2\cos x \cdot e^{-2\sin x} dx = -e^{-2\sin x}$∫2cosx⋅e−2sinxdx=−e−2sinx

For the second term, using integration by parts: $\displaystyle \int 3\sin x \cos x \cdot e^{-2\sin x} dx = \frac{3}{4}e^{-2\sin x}(-2\sin x - 1)$∫3sinxcosx⋅e−2sinxdx=43​e−2sinx(−2sinx−1)

Thus: $ye^{-2\sin x} = -e^{-2\sin x} + \frac{3}{4}e^{-2\sin x}(-2\sin x - 1) + C$ye−2sinx=−e−2sinx+43​e−2sinx(−2sinx−1)+C

Simplifying: $y = -\frac{7}{4} - \frac{3}{2}\sin x + Ce^{2\sin x}$y=−47​−23​sinx+Ce2sinx

Using $y(0) = -\frac{7}{4}$y(0)=−47​ gives $C = 0$C=0.

Therefore: $y(\frac{\pi}{6}) = -\frac{7}{4} - \frac{3}{2} \cdot \frac{1}{2} = -\frac{5}{2}$y(6π​)=−47​−23​⋅21​=−25​