A) \[{{2}^{x}}+{{2}^{y}}=c\]
B) \[{{2}^{x}}-{{2}^{y}}=c\]
C) \[\frac{1}{{{2}^{x}}}-\frac{1}{{{2}^{y}}}=c\]
D) \[\frac{1}{{{2}^{x}}}+\frac{1}{{{2}^{y}}}=c\]
Correct Answer: C
Solution :
\[\frac{dy}{dx}={{2}^{y-x}}\Rightarrow \frac{dy}{dx}={{2}^{y}}{{.2}^{-x}}\] \[{{2}^{-y}}dy={{2}^{-x}}dx\]and integrate \[-\frac{1}{{{2}^{y}}}=\frac{-1}{{{2}^{x}}}+c\]or \[\frac{1}{{{2}^{x}}}-\frac{1}{{{2}^{y}}}=c\]
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