Let \(y=y(x)\) be the solution of the differential equation \[ \sec ^{2} x d x+\left(e^{2 y} \tan ^{2} x+\tan x\right) d y=0, \] \(0<x<\frac{\pi}{2}\), \(y\left(\frac{\pi}{4}\right)=0\). If \(y\left(\frac{\pi}{6}\right)=\alpha\), Then \(\mathrm{e}^{8 \alpha}\) is equal to ______.

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

sec^{2} x d x + (e^{2 y} tan^{2} x + tan x) d y = 0,

0 < x < \frac{π}{2}

y (\frac{π}{4}) = 0

. If

y (\frac{π}{6}) = α

,

Then

e^{8 α}

is equal to ______.

Answer: 9

sec^{2} x d x + (e^{2 y} tan^{2} x + tan x) d y = 0

. Let

t = tan x

, so

d t = sec^{2} x d x

, and the equation becomes

d t + (e^{2 y} t^{2} + t) d y = 0 \Rightarrow \frac{d t}{d y} + t = - e^{2 y} t^{2}

,

a Bernoulli equation in

t

. Divide by

t^{2}

and set

u = \frac{1}{t}

, so

\frac{d u}{d y} = - \frac{1}{t ^{2}} \frac{d t}{d y}

- \frac{d u}{d y} + u = - e^{2 y} \Rightarrow \frac{d u}{d y} - u = e^{2 y}

.

This is linear with integrating factor

e^{- y}

\frac{d}{d y} (u e^{- y}) = e^{2 y} \cdot e^{- y} = e^{y}

u e^{- y} = e^{y} + C \Rightarrow u = e^{2 y} + C e^{y}

, i.e.

cot x = e^{2 y} + C e^{y}

(since

u = 1/ t = cot x

).

Using

y (\frac{π}{4}) = 0

cot \frac{π}{4} = 1 = e^{0} + C e^{0} = 1 + C \Rightarrow C = 0

.

So

cot x = e^{2 y}

.

At

x = \frac{π}{6}

y = α

cot \frac{π}{6} = 3 = e^{2 α}

e^{8 α} = (e^{2 α})^{4} = (3)^{4} = 9

JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Integer answer