If frac{dy}{dx} = frac{2^{x+y} - 2^{x}}{2^{y} | vedclass

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{2^{x} \cdot 2^{y} - 2^{x}}{2^{y}}$.Separate the variables: $\frac{2^{y}}{2^{y}-1} dy = 2^{

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$\log_{2}(1+e)$

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(B) Given the differential equation: $\frac{dy}{dx} = \frac{2^{x} \cdot 2^{y} - 2^{x}}{2^{y}}$.Separate the variables: $\frac{2^{y}}{2^{y}-1} dy = 2^{x} dx$.Integrate both sides: $\int \frac{2^{y}}{2^{y}-1} dy = \int 2^{x} dx$.Let $u = 2^{y}-1$,then $du = 2^{y} \ln(2) dy$,so $\int \frac{du}{u \ln(2)} = \frac{2^{x}}{\ln(2)} + C$.This simplifies to: $\frac{1}{\ln(2)} \ln(2^{y}-1) = \frac{2^{x}}{\ln(2)} + C$.Multiplying by $\ln(2)$: $\ln(2^{y}-1) = 2^{x} + C'$.Using $y(0) = 1$: $\ln(2^{1}-1) = 2^{0} + C' \Rightarrow \ln(1) = 1 + C' \Rightarrow 0 = 1 + C' \Rightarrow C' = -1$.So,$\ln(2^{y}-1) = 2^{x} - 1$.For $x=1$: $\ln(2^{y}-1) = 2^{1} - 1 = 1$.$2^{y}-1 = e^{1} \Rightarrow 2^{y} = e+1$.Taking $\log_{2}$ on both sides: $y = \log_{2}(e+1)$.

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

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