If d y   d x = 2 x + y - 2 x 2 y , y 0 = 1 , then y ( 1 ) is equal to :

Given that

$\frac{dy}{dx}=\frac{2^{x+y}-2^x}{2^y}=2^x\left(\frac{2^y-1}{2^y}\right)$

$\Rightarrow \left(\frac{2^y}{2^y-1}\right)dy = 2^{x}dx$

By using Variable Separable method

$\int \frac{2^{y}dy}{2^y-1}=\int 2^{x}dx$

Let $2^y-1=t\Rightarrow 2^y\ln \left(2\right)dy=dt$

$\int \frac{dt}{t}\frac{1}{\ell n2}=\int 2^{x}dx$

We know that $\int \frac{1}{x}dx = \ln \left(x\right) + c \& \int a^{x}dx = \frac{a^x}{\ln \left(a\right)}+c.$

$\Rightarrow \frac{\ln \left(t\right)}{\ln \left(2\right)}=\frac{2^x}{\ln \left(2\right)}+\frac{c}{\ln \left(2\right)}$

$\Rightarrow \ln \left(t\right)=2^{\mathrm{x}}+c$

$\Rightarrow \ln \left(2^y-1\right)=2^{\mathrm{x}}+c$

when $x=0, y=1 \Rightarrow \ln \left(2-1\right) = 2^0 + c$

$\Rightarrow c = -1.$

$\Rightarrow \ln \left(2^y-1\right)=2^x-1$

$\Rightarrow 2^y-1=e^{2^x-1}$

$\Rightarrow 2^y=1+e^{2^x-1}$

when $x=1\Rightarrow 2^y=1+e^1\Rightarrow y={\log }_2(1+e)$