Let $y=y(x)$ be the solution of the differential equation : $$\frac{d y}{d x}+\left(\frac{6 x^{2}+\left(3 x^{2}+2 x^{3}+4\right) e^{-2 x}}{\left(x^{3}+2\right)\left(2+e^{-2 x}\right)}\right) y=2+e^{-2 x}$$ $\mathrm{x} \in(-1,2)$, satisfying $\mathrm{y}(0)=\frac{3}{2}$. If $y(1)=\alpha\left(2+e^{-2}\right)$, than $\alpha$ is equal to :

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation :

\frac{d y}{d x} + (\frac{6 x ^{2} + ( 3 x ^{2} + 2 x ^{3} + 4 ) e ^{- 2 x}}{( x ^{3} + 2 ) ( 2 + e ^{- 2 x} )}) y = 2 + e^{- 2 x}

x \in (- 1, 2)

, satisfying

y (0) = \frac{3}{2}

.
If

y (1) = α (2 + e^{- 2})

, than

α

is equal to :

Answer: (D)

The coefficient of

y

is the logarithmic derivative of

\frac{x ^{3} + 2}{2 + e ^{- 2 x}}

, since

\frac{d}{d x} ln \frac{x ^{3} + 2}{2 + e ^{- 2 x}} = \frac{3 x ^{2}}{x ^{3} + 2} + \frac{2 e ^{- 2 x}}{2 + e ^{- 2 x}} = \frac{6 x ^{2} + ( 3 x ^{2} + 2 x ^{3} + 4 ) e ^{- 2 x}}{( x ^{3} + 2 ) ( 2 + e ^{- 2 x} )}

. So the integrating factor is

μ = \frac{x ^{3} + 2}{2 + e ^{- 2 x}}

, and the equation becomes

\frac{d}{d x} (μ y) = μ (2 + e^{- 2 x}) = x^{3} + 2

. Integrating,

μ y = \frac{x ^{4}}{4} + 2 x + C

. At

x = 0

μ (0) = \frac{2}{3}

and

y (0) = \frac{3}{2}

, so

μ y = 1 = C

. At

x = 1

μ (1) = \frac{3}{2 + e ^{- 2}}

and

μ y = \frac{1}{4} + 2 + 1 = \frac{13}{4}

, giving

y (1) = \frac{13}{4} \cdot \frac{2 + e ^{- 2}}{3} = \frac{13}{12} (2 + e^{- 2})

. Hence

α = \frac{13}{12}

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option