Let a curve $y=y(x)$ be given by the solution of the differential equation $\cos \left(\frac{1}{2} \cos ^{-1}\left(e^{-x}\right)\right) d x=\sqrt{e^{2 x}-1} d y$ If it intersects $y$-axis at $y=-1$, and the intersection point of the curve with $x$-axis is $(\alpha, 0)$, then $\mathrm{e}^{\alpha}$ is equal to

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Differential Equations

Let a curve

y = y (x)

be given by the solution of the differential equation

cos (\frac{1}{2} cos^{- 1} (e^{- x})) d x = e^{2 x} - 1 d y

If it intersects

y

-axis at

y = - 1

, and the intersection point of the curve with

x

-axis is

(α, 0)

, then

e^{α}

is equal to

Answer: 2

With

cos (\frac{1}{2} cos^{- 1} t) = \frac{1 + t}{2}

and

t = e^{- x}

, the equation reads

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

Since

e^{2 x} - 1 = e^{x} (1 - e^{- x}) (1 + e^{- x})

, the factor

1 + e^{- x}

cancels and

d y = \frac{e ^{- x} d x}{2 1 - e ^{- x}}

Putting

v = 1 - e^{- x}

, so that

d v = e^{- x} d x

, integration gives

y = 2 v + c = 2 (1 - e^{- x}) + c

The curve meets the

y

-axis at

y = - 1

, where

x = 0

and

v = 0

, so

c = - 1

. Setting

y = 0

then gives

2 (1 - e^{- α}) = 1

, so

e^{- α} = \frac{1}{2}

and

e^{α} = 2

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Integer answer