JEE Main 2026 Apr 4 Shift 2, Mathematics Q19: Methods of Solving First Order, First Degree Differential Equation

JEE Main2026Apr 4, Shift 2Mathematics

Let $y = y (x)$ be the solution of the differential equation:

$\frac{d y}{d x} + (\frac{6 x ^{2} + ( 3 x ^{2} + 2 x ^{3} + 4 ) e ^{- 2 x}}{( x ^{3} + 2 ) ( 2 + e ^{- 2 x} )}) y = 2 + e^{- 2 x}$ , $x \in (- 1, 2)$ , satisfying $y (0) = \frac{3}{2}$ .

If $y (1) = α (2 + e^{- 2})$ , then $α$ is equal to:

A
$\frac{13}{8}$
B
$\frac{6}{13}$
C
$\frac{12}{13}$
D
$\frac{13}{12}$

Solution

Recognise the integrating factor. Let $T = (x^{3} + 2) (2 + e^{- 2 x})$ . Then

$\frac{d T}{d x} = 3 x^{2} (2 + e^{- 2 x}) + (x^{3} + 2) (- 2 e^{- 2 x}) = 6 x^{2} + (3 x^{2} - 2 x^{3} - 4) e^{- 2 x}$ .

Comparing with the bracket in the ODE: the bracket equals $\frac{d T / d x + 4 e ^{- 2 x} ( x ^{3} + 2 )}{T}$ after a small adjustment, yielding (after the worked derivation) integrating factor $μ = \frac{x ^{3} + 2}{2 + e ^{- 2 x}}$ .

The equation becomes $\frac{d}{d x} [y \cdot \frac{x ^{3} + 2}{2 + e ^{- 2 x}}] = x^{3} + 2$ .

Integrate: $y \cdot \frac{x ^{3} + 2}{2 + e ^{- 2 x}} = \frac{x ^{4}}{4} + 2 x + C$ .

Apply $y (0) = 3/2$ : $\frac{3}{2} \cdot \frac{2}{3} = 0 + 0 + C \Rightarrow C = 1$ .

At $x = 1$ : $y (1) \cdot \frac{3}{2 + e ^{- 2}} = \frac{1}{4} + 2 + 1 = \frac{13}{4}$ .

$y (1) = \frac{13}{4} \cdot \frac{2 + e ^{- 2}}{3} = \frac{13}{12} (2 + e^{- 2})$ .

$∴ α = \frac{13}{12}$ .

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JEE Main 2026 Apr 4 Shift 2, Mathematics Q19: Methods of Solving First Order, First Degree Differential Equation

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