If $\frac{d y}{d x}=\frac{2^{x+y}-2^{x}}{2^{y}}, y(0)=1$, then $y(1)$ is equal to :
Correct Option: , 2
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{\mathrm{x}} 2^{\mathrm{y}}-2^{\mathrm{x}}}{2^{\mathrm{y}}}$
$2^{y} \frac{d y}{d x}=2^{x}\left(2^{y}-1\right)$
$\int \frac{2^{y}}{2^{y}-1} d y=\int 2^{x} d x$
$\frac{\ln \left(2^{y}-1\right)}{\ln 2}=\frac{2^{x}}{\ln 2}+C$
$\Rightarrow \log _{2}\left(2^{y}-1\right)=2^{x} \log _{2} e+C$
$\because y(0)=1 \Rightarrow 0=\log _{2} e+C$
$\mathrm{C}=-\log _{2} \mathrm{e}$
$\Rightarrow \log _{2}\left(2^{y}-1\right)=\left(2^{x}-1\right) \log _{2} e$
put $x=1, \log _{2}\left(2^{y}-1\right)=\log _{2} e$
$2^{y}=e+1$
$\mathrm{y}=\log _{2}(\mathrm{e}+1)$ Ans.
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