If frac{dy}{dx} + frac{2^{x-y}(2^y - 1)}{2^x | vedclass

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(C) Given the differential equation: $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$.Rearranging the terms,we get: $\frac{dy}{dx} = -\frac{2^x

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$2 - \log_2 3$

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(C) Given the differential equation: $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$.Rearranging the terms,we get: $\frac{dy}{dx} = -\frac{2^x \cdot 2^{-y}(2^y - 1)}{2^x - 1} = -\frac{2^x(2^y - 1)}{2^y(2^x - 1)}$.Separating the variables: $\frac{2^y}{2^y - 1} dy = -\frac{2^x}{2^x - 1} dx$.Integrating both sides: $\int \frac{2^y}{2^y - 1} dy = -\int \frac{2^x}{2^x - 1} dx$.Using the substitution $u = 2^y - 1$,$du = 2^y \ln 2 \, dy$,we get $\frac{1}{\ln 2} \ln|2^y - 1| = -\frac{1}{\ln 2} \ln|2^x - 1| + C$.Multiplying by $\ln 2$: $\ln|2^y - 1| + \ln|2^x - 1| = C_1$,where $C_1 = C \ln 2$.This simplifies to $(2^y - 1)(2^x - 1) = K$,where $K = e^{C_1}$.Given $y(1) = 1$,we have $(2^1 - 1)(2^1 - 1) = K \implies (1)(1) = K \implies K = 1$.So,$(2^y - 1)(2^x - 1) = 1$.For $x = 2$,$(2^y - 1)(2^2 - 1) = 1 \implies (2^y - 1)(3) = 1$.$2^y - 1 = \frac{1}{3} \implies 2^y = 1 + \frac{1}{3} = \frac{4}{3}$.Taking $\log_2$ on both sides: $y = \log_2(\frac{4}{3}) = \log_2 4 - \log_2 3 = 2 - \log_2 3$.

If $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$,$x, y > 0$,and $y(1) = 1$,then $y(2)$ is equal to

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