The general solution of the differential equation \(\displaystyle\frac{dy}{dx} = 2^{x + y}\) is (C is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2^{x} + 2^{-y} = C\)
Since \(2^{x + y} = 2^{x} \cdot 2^{y}\), the variables separate:
\( 2^{-y}\,dy = 2^{x}\,dx \)
Using \(\int a^{x}\,dx = \frac{a^{x}}{\log a}\):
\( \displaystyle -\frac{2^{-y}}{\log 2} = \frac{2^{x}}{\log 2} + C_1 \)
Multiplying by \(\log 2\) and taking both terms to one side:
\( 2^{x} + 2^{-y} = C \)
where \(C = -C_1\log 2\).
Why the other options are incorrect
Option (1) \(2^{x} - 2^{-y} = C\) — Incorrect. The minus sign from \(\int 2^{-y}\,dy = -\frac{2^{-y}}{\log 2}\) is lost. Differentiating gives \(\frac{dy}{dx} = -2^{x + y}\).
Option (3) \(2^{x} + 2^{y} = C\) — Incorrect. Differentiating gives \(\frac{dy}{dx} = -2^{x - y}\).
Option (4) \(2^{x + y} = C\) — Incorrect. This means x + y is constant, so \(\frac{dy}{dx} = -1\).