JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Differential Equations

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)

Answer: (B)

Solution

Rewrite the DE as

\frac{d y}{d x} = \frac{y ^{3} ( e ^{5 x} + 1 )}{e ^{x} ( 1 + y ^{4} )}

. Separating variables:

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

. Integrating:

- \frac{1}{2 y ^{2}} + \frac{y ^{2}}{2} = \frac{e ^{4 x}}{4} - e^{- x} + C

. Using

y (0) = \frac{1}{2}

- 1 + \frac{1}{2} = \frac{1}{4} - 1 + C \Rightarrow C = 0

. At

x = lo g_{e} 2

- \frac{1}{2 y ^{2}} + \frac{y ^{2}}{2} = 4 - \frac{1}{2} = \frac{7}{2}

. Let

t = y^{2}

\frac{t ^{2} - 1}{2 t} = \frac{7}{2} \Rightarrow t^{2} - 7 t - 1 = 0 \Rightarrow t = \frac{7 + 53}{2}

. Thus

y = \frac{7 + 53}{2}

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Single correct option