If $$\frac{d y}{d x}=\frac{2^{x+y}-2^{x}}{2^{y}}, y(0)=1$$, then $$y(1)$$ is equal to:

(B) $$\log _{2}\left(1+\mathrm{e}^{2}\right)$$

(D) $$\log _{2}(2 \mathrm{e})$$

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Differential Equations

\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1

, then

y (1)

is equal to:

Answer: (A)

Rewrite the differential equation:

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

. Separating variables:

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

. Let

u = 2^{y} - 1

, then

d u = 2^{y} ln 2 d y

, so

\frac{2 ^{y}}{2 ^{y} - 1} d y = \frac{u + 1}{u} \cdot \frac{d u}{( u + 1 ) ln 2} = \frac{d u}{u ln 2}

. Integrating:

\frac{ln ∣ 2 ^{y} - 1∣}{ln 2} = \frac{2 ^{x}}{ln 2} + C

, which gives

ln (2^{y} - 1) = 2^{x} + C^{'}

. Using

y (0) = 1

ln (2 - 1) = 1 + C^{'}

, so

C^{'} = - 1

. Thus

2^{y} - 1 = e^{2^{x} - 1}

, giving

2^{y} = 1 + e^{2^{x} - 1}

. At

x = 1

2^{y (1)} = 1 + e

, so

y (1) = lo g_{2} (1 + e)

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option