JEE Main 2026 · 4 Apr · Shift 2 · Mathematics Q19

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MathematicsSingle correct+4 / -1Question 19 of 75

Let $y=y(x)$ be the solution of the differential equation:
$\frac{dy}{dx}+\left(\frac{6x^2+\left(3x^2+2x^3+4\right)e^{-2x}}{\left(x^3+2\right)\left(2+e^{-2x}\right)}\right)y=2+e^{-2x},$
$x\in(-1,2)$ , satisfying $y(0)=\dfrac32$ . If $y(1)=\alpha\left(2+e^{-2}\right)$ , then $\alpha$ is equal to:

\dfrac{13}8

\dfrac6{13}

\dfrac{12}{13}

\dfrac{13}{12}

Answer and worked solution

Official answer: D

Integrating factor. The coefficient is $\dfrac d{dx}\ln\dfrac{x^3+2}{2+e^{-2x}}$ , so the integrating factor is $\dfrac{x^3+2}{2+e^{-2x}}$ .

Integrate. $\dfrac d{dx}\left[y\cdot\text{IF}\right]=x^3+2$ , so $y\cdot\text{IF}=\frac{x^4}4+2x+C$ .

Constant. At $x=0$ : $\frac32\cdot\frac23=1=C$ .

At $x=1$ : $y\cdot\frac3{2+e^{-2}}=\frac{13}4$ , so $y(1)=\frac{13}{12}(2+e^{-2})$ and $\alpha=\frac{13}{12}$ .

(D)

Topics:Linear first-order equations

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JEE Main 2026 · 4 Apr · Shift 2, Mathematics question 19