JEE Main 2026 — Differential Equations Question with Solution

The given differential equation is a linear differential equation of the form $\frac{dy}{dx}+P(x)y=Q(x)$, where

$P(x)=\frac{6x^2+(3x^2+2x^3+4)e^{-2x}}{(x^3+2)(2+e^{-2x})}$ and $Q(x)=2+e^{-2x}$.

We can rewrite $P(x)$ as:

$P(x)=\frac{3x^2(2+e^{-2x})+2e^{-2x}(x^3+2)}{(x^3+2)(2+e^{-2x})}=\frac{3x^2}{x^3+2}+\frac{2e^{-2x}}{2+e^{-2x}}$

The integrating factor (IF) is given by:

$IF=e^{\int P(x)dx}=e^{\int \left(\frac{3x^2}{x^3+2}+\frac{2e^{-2x}}{2+e^{-2x}}\right)dx}$

$IF=e^{\ln (x^3+2)-\ln (2+e^{-2x})}=\frac{x^3+2}{2+e^{-2x}}$

The general solution of the differential equation is:

$y\cdot (IF)=\int Q(x)\cdot (IF)dx+C$

Substituting the values, we get:

$y\left(\frac{x^3+2}{2+e^{-2x}}\right)=\int (2+e^{-2x})\left(\frac{x^3+2}{2+e^{-2x}}\right)dx+C$

$y\left(\frac{x^3+2}{2+e^{-2x}}\right)=\int (x^3+2)dx+C$

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

Given $y(0)=\frac{3}{2}$, substituting $x=0$:

$\frac{3}{2}\left(\frac{0+2}{2+e^0}\right)=0+0+C$

$\frac{3}{2}\left(\frac{2}{3}\right)=C\Rightarrow C=1$

Thus, the particular solution is:

$y\left(\frac{x^3+2}{2+e^{-2x}}\right)=\frac{x^4}{4}+2x+1$

To find $y(1)$, substitute $x=1$:

$y(1)\left(\frac{1+2}{2+e^{-2}}\right)=\frac{1}{4}+2+1$

$y(1)\left(\frac{3}{2+e^{-2}}\right)=\frac{13}{4}$

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

Comparing this with $y(1)=\alpha (2+e^{-2})$, we get:

$\alpha =\frac{13}{12}$