Let y=y(x) be the solution of the differential equation sec x frac dy dx-2 y=2+3 sin x, x in (-(pi / 2), (pi / 2)), y(0)=-(7 / 4). Then y((pi / 6)) is equal to:

(d y / d x)-2 y cos x=2 cos x+3 sin x * cos x I.F. = e^-2 sin x e^-2 sin x * y=integral e^-2 sin x(3 sin x cos x+2 cos x) d x y . e^-2 sin x=e^-2 sin x(-(3 / 2) sin x-(7 / 4))+C => y=-(3 / 2) sin x-(7 / 4)+ C . e^2 sin x Because y(0)=-(7 / 4) => C=0 y((pi / 6))=(-3 / 2) * (1 / 2)-(7 / 4)=(-5 / 2)

Let y y x be the solution of the differential equation ... | null (Mathematics)

JEE Mains 2026 21st January Shift 2

Mathematics

Let

y=y(x)

be the solution of the differential equation

\sec x \frac{\mathrm{dy}}{\mathrm{dx}}-2 \mathrm{y}=2+3 \sin \mathrm{x}, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

y(0)=-\frac{7}{4}

. Then

y\left(\frac{\pi}{6}\right)

is equal to:

-\frac{5}{2}

-\frac{5}{4}

-3 \sqrt{3}-7

-3 \sqrt{2}-7

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Correct Answer:

Explanation

\frac{d y}{d x}-2 y \cos x=2 \cos x+3 \sin x \cdot \cos x\\

I.F.

=\mathrm{e}^{-2 \sin \mathrm{x}}\\

e^{-2 \sin x} \cdot y=\int e^{-2 \sin x}(3 \sin x \cos x+2 \cos x) d x\\

y . e^{-2 \sin x}=e^{-2 \sin x}\left(-\frac{3}{2} \sin x-\frac{7}{4}\right)+C\\

\Rightarrow \mathrm{y}=-\frac{3}{2} \sin \mathrm{x}-\frac{7}{4}+\mathrm{C} . \mathrm{e}^{2 \sin \mathrm{x}}\\

\because y(0)=-\frac{7}{4} \Rightarrow C=0\\

y\left(\frac{\pi}{6}\right)=\frac{-3}{2} \cdot \frac{1}{2}-\frac{7}{4}=\frac{-5}{2}

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