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

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2^{\mathrm{x}}{2}^{\mathrm{y}}-2^{\mathrm{x}}}{2^{\mathrm{y}}}\\ 2^{\mathrm{y}}\frac{\mathrm{dy}}{\mathrm{dx}}=2^{\mathrm{x}}\left(2^{\mathrm{y}}-1\right)\\ \int \frac{2^{\mathrm{y}}}{2^{\mathrm{y}}-1}\mathrm{dy}=\int 2^{\mathrm{x}}\mathrm{dx}\\ \frac{\mathrm{\ell n}\left(2^{\mathrm{y}}-1\right)}{\mathrm{\ell ln}2}=\frac{2^{\mathrm{x}}}{\mathrm{\ell ln}2}+\mathrm{C}\\ \Rightarrow {\log }_2\left(2^{\mathrm{y}}-1\right)=2^{\mathrm{x}}{\log }_2\mathrm{e}+\mathrm{C}\\ \because \mathrm{y}\left(0\right)=1\Rightarrow 0={\log }_2\mathrm{e}+\mathrm{C}\end{array}$
$C = -{\log }_{2}e$
$\Rightarrow$${\log }_2(2y-1)=(2x-1) {\log }_{2}e$
$put x = 1, {\log }_2(2y-1)={\log }_{2}e$
$2^y = e + 1\Rightarrow y={\log }_2(e + 1) .$