(2) log₂ (1 + e)

Explanation:

\(\frac{dy}{dx}\) = \(\frac{2^x2^y - 2^x}{2^y}\)

\(2^y \frac{dy}{dx}\) = 2^x(2^y - 1)

\(\int \frac{2^y}{2^y - 1}\)dy = ∫2^x dx

\(\frac{\ell n(2^y - 1)}{\ell n 2}\) = \(\frac{2^x}{\ell n2}\)+C

⇒ log₂(2^y - 1) = 2^xlog₂e + C

∵ y(0) = 1 ⇒ 0 = log₂e + C

C = -log₂e

⇒ log₂e(2^y - 1) = (2^x - 1)log₂e

Put x = 1, log₂(2^y - 1) = log₂e

2^y = e + 1

y = log₂(e + 1)