JEE Advanced 2026 — Paper 2 — MATH-I solution

JEE Advanced2026Paper 2MATH-I

Let $y : (- \infty, \infty) \to (0, \infty)$ be the solution of the differential equation

$\frac{d y}{d x} = \frac{e ^{5 x} y ^{3} + y ^{3}}{e ^{x} + e ^{x} y ^{4}}$

satisfying $y (0) = \frac{1}{2}$ . Then the value of $y (lo g_{e} 2)$ is

Solution

The right side factors as $\frac{y ^{3} ( e ^{5 x} + 1 )}{e ^{x} ( 1 + y ^{4} )}$ , allowing separation:

$\int \frac{1 + y ^{4}}{y ^{3}} d y = \int \frac{e ^{5 x} + 1}{e ^{x}} d x$ .

Integrating both sides: $- \frac{1}{2 y ^{2}} + \frac{y ^{2}}{2} = \frac{e ^{4 x}}{4} - e^{- x} + C$ .

Using $y (0) = \frac{1}{2}$ : LHS $= - 1 + \frac{1}{4} = - \frac{3}{4}$ , RHS $= \frac{1}{4} - 1 + C = - \frac{3}{4} + C$ , so $C = 0$ .

At $x = lo g_{e} 2$ : $- \frac{1}{2 y ^{2}} + \frac{y ^{2}}{2} = \frac{16}{4} - \frac{1}{2} = \frac{7}{2}$ .

Let $t = y^{2}$ : $\frac{t ^{2} - 1}{2 t} = \frac{7}{2}$ , so $t^{2} - 7 t - 1 = 0$ , giving $t = \frac{7 + 53}{2}$ (positive root).

$∴ y = \frac{7 + 53}{2}$ .

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