JEE Main 2026 — Mathematics Calculus

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

$P(x) = \dfrac{6x^2 + (3x^2 + 2x^3 + 4)e^{-2x}}{(x^3+2)(2+e^{-2x})}$P(x)=(x3+2)(2+e−2x)6x2+(3x2+2x3+4)e−2x​ and $Q(x) = 2 + e^{-2x}$Q(x)=2+e−2x.

We can rewrite $P(x)$P(x) as:

$$P(x) = \dfrac{3x^2(2+e^{-2x}) + 2e^{-2x}(x^3+2)}{(x^3+2)(2+e^{-2x})} = \dfrac{3x^2}{x^3+2} + \dfrac{2e^{-2x}}{2+e^{-2x}}$$P(x)=(x3+2)(2+e−2x)3x2(2+e−2x)+2e−2x(x3+2)​=x3+23x2​+2+e−2x2e−2x​

The integrating factor (IF) is given by:

$$IF = e^{\int P(x) dx} = e^{\int \left( \dfrac{3x^2}{x^3+2} + \dfrac{2e^{-2x}}{2+e^{-2x}} \right) dx}$$IF=e∫P(x)dx=e∫(x3+23x2​+2+e−2x2e−2x​)dx

$$IF = e^{\ln(x^3+2) - \ln(2+e^{-2x})} = \dfrac{x^3+2}{2+e^{-2x}}$$IF=eln(x3+2)−ln(2+e−2x)=2+e−2xx3+2​

The general solution of the differential equation is:

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

Substituting the values, we get:

$$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \int (2+e^{-2x}) \left( \dfrac{x^3+2}{2+e^{-2x}} \right) dx + C$$y(2+e−2xx3+2​)=∫(2+e−2x)(2+e−2xx3+2​)dx+C

$$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \int (x^3+2) dx + C$$y(2+e−2xx3+2​)=∫(x3+2)dx+C

$$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \dfrac{x^4}{4} + 2x + C$$y(2+e−2xx3+2​)=4x4​+2x+C

Given $y(0) = \dfrac{3}{2}$y(0)=23​, substituting $x = 0$x=0:

$$\dfrac{3}{2} \left( \dfrac{0+2}{2+e^0} \right) = 0 + 0 + C$$23​(2+e00+2​)=0+0+C

$$\dfrac{3}{2} \left( \dfrac{2}{3} \right) = C \Rightarrow C = 1$$23​(32​)=C⇒C=1

Thus, the particular solution is:

$$y \left( \dfrac{x^3+2}{2+e^{-2x}} \right) = \dfrac{x^4}{4} + 2x + 1$$y(2+e−2xx3+2​)=4x4​+2x+1

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

$$y(1) \left( \dfrac{1+2}{2+e^{-2}} \right) = \dfrac{1}{4} + 2 + 1$$y(1)(2+e−21+2​)=41​+2+1

$$y(1) \left( \dfrac{3}{2+e^{-2}} \right) = \dfrac{13}{4}$$y(1)(2+e−23​)=413​

$$y(1) = \dfrac{13}{12}(2+e^{-2})$$y(1)=1213​(2+e−2)

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

$$\alpha = \dfrac{13}{12}$$α=1213​