Let $y=y(x)$ be the solution of the differential equation:$\frac{dy}{dx}+\left(\frac{6{x}^{2}+\left(3{x}^{2}+2{x}^{3}+4\right){e}^{-2x}}{\left({x}^{3}+2\right)\left(2+{e}^{-2x}\right)}\right)y=2+{e}^{-2x},x\in \left(-1,2\right)$, satisfying $y(0)=\frac{3}{2}$. If $y(1)=\alpha \left(2+{\mathrm{e}}^{-2}\right)$, then $\alpha$ is equal to: [JEE Main 2026, 4 Apr (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: d) $\frac{13}{12}$. $\frac{d y}{d x}+P(x) y=Q(x)$ $Q(x)=2+e^{-2 x}$ $P(x)=\frac{3 x^2\left(2+e^{-2 x}\right)+2 e^{-2 x}\left(x^3+2\right)}{\left(x^3+2\right)\left(2+e^{-2 x}\right)}$ $=\frac{3 x^2}{x^3+2}+\frac{2 e^{-2 x}}{2+e^{-2 x}}$ Integrating factor $=e^{\int P(x) d x}=e^{\int \frac{3 x^2}{x^3+2} d x+\int \frac{2 e^{-2 x}}{2+e^{-2 x}} d x}=\frac{x^3+2}{2+e^{-2 x}}$ $\Rightarrow \frac{y \cdot\left(x^3+2\right)}{2+e^{-2 x}}=\int\left(x^3+2\right) d x+C=\frac{x^4}{4}+2 x+C$ $\because y(0)=\frac{3}{2} \Rightarrow C=1$ $\Rightarrow \frac{y \cdot\left(x^3+2\right)}{2+e^{-2 x}}=\frac{x^4}{4}+2 x+1$ Now, $\Rightarrow \frac{y(1) \cdot 3}{2+e^{-2}}=\frac{13}{4}$ $\Rightarrow y(1)=\frac{13}{12}\left(2+e^{-2}\right)$ $ \Rightarrow \alpha=\frac{13}{12}$

Let \(y=y(x)\) be the solution of the differential equation:\(\frac{dy}{dx}+\left(\frac{6{x}^{2}+\left(3{x}^{2}+2{x}^{3}…

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