JEE Main 2026 (Online) 4th April Evening ShiftQuestion 1
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 $y(0)=\frac{3}{2}$. If $y(1)=\alpha(2+e^{-2})$, then $\alpha$ is equal to:
Differential Equation
Solution
Concept. The coefficient of $y$ splits into two logarithmic derivatives, so the IF is a simple quotient. Formulas used. (1) $\dfrac{6x^2+(3x^2+2x^3+4)e^{-2x}}{(x^3+2)(2+e^{-2x})}=\dfrac{3x^2}{x^3+2}+\dfrac{2e^{-2x}}{2+e^{-2x}}$(2) $\int\frac{f'}{f}\,dx=\ln|f|$Step 1 - IF. By (1) and (2), $\int P\,dx=\ln(x^3+2)-\ln(2+e^{-2x})$, so $$\text{IF}=\frac{x^3+2}{2+e^{-2x}}.$$ Step 2 - Solve. The right side times IF is $x^3+2$: $$y\cdot\frac{x^3+2}{2+e^{-2x}}=\int(x^3+2)\,dx=\frac{x^4}{4}+2x+C.$$ Step 3 - Constant. At $x=0$: $\frac32\cdot\frac23=1=C$. Step 4 - At $x=1$. $$y\cdot\frac{3}{2+e^{-2}}=\frac14+2+1=\frac{13}{4}\ \Longrightarrow\ y=\frac{13}{12}\left(2+e^{-2}\right).$$ So $\alpha=\frac{13}{12}$. Answer: (D)