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

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Differential Equations

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

x, y > 0

y (1) = 1

, then

y (2)

is equal to :

Answer: (C)

Separate variables:

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

. Integrating (since

\frac{d}{d t} (2^{t} - 1) = 2^{t} ln 2

) gives

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

, i.e.

(2^{y} - 1) (2^{x} - 1) = K

. With

y (1) = 1

(2 - 1) (2 - 1) = 1 \Rightarrow K = 1

. At

x = 2

(2^{y} - 1) (3) = 1 \Rightarrow 2^{y} = \frac{4}{3} \Rightarrow y = 2 - lo g_{2} 3

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0$ , $y (0) = 1$ . Then the value of $y (1)$ is :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation $sec x \frac{dy}{dx} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2})$ , $y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to:
JEE Main 2026View solution →
Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to
JEE Main 2026View solution →
If $y = y (x)$ satisfies the differential equation
$16 (x + 9 x) (4 + 9 + x) cos y d y = (1 + 2 sin y) d x$ , $x > 0$ and $y (256) = \frac{π}{2}$ , $y (49) = α$ , then $2 sin α$ is equal to :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation
$x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0$ , $x > 0$ , $y (\frac{π}{2}) = 0$ .

Then $π^{4} y (\frac{π}{3})$ is equal to :
JEE Main 2026View solution →
If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0, x > 2$ , passes through the point (3,15), then the local maximum value of f is
JEE Main 2026View solution →