If (dy)/(dx) = 2 ^ x + y - 2 ^ x 2 ^ y , y(0) = 1, then y ( 1 ) is...

Given that

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

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

By using Variable Separable method

$∫ \dfrac{{2}^{y}dy}{{2}^{y} - 1} = ∫ {2}^{x}dx$

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

$∫ \dfrac{dt}{t}\dfrac{1}{ℓn2} = ∫ {2}^{x}dx$

We know that $∫ \dfrac{1}{x}dx = \ln \left(x\right) + c \text{ and } ∫ {a}^{x}dx = \dfrac{{a}^{x}}{\ln \left(a\right)} + c .$

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

$⇒ \ln \left(t\right) = {2}^{x} + c$

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

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

$⇒ c = - 1 .$

$⇒ \ln \left({2}^{y} - 1\right) = {2}^{x} - 1$

$⇒ {2}^{y} - 1 = {e}^{{2}^{x} - 1}$

$⇒ {2}^{y} = 1 + {e}^{{2}^{x} - 1}$

when $x = 1 ⇒ {2}^{y} = 1 + {e}^{1} ⇒ y = \log_{2} ( 1 + e )$