A) \[{{2}^{x}}+{{2}^{y}}=c\]
B) \[{{2}^{x}}-{{2}^{y}}=c\]
C) \[\frac{1}{{{2}^{x}}}-\frac{1}{{{2}^{y}}}=c\]
D) \[x+y=c\]
Correct Answer: C
Solution :
Given \[\frac{dy}{dx}={{2}^{y-x}}\]\[=\frac{{{2}^{y}}}{{{2}^{x}}}\] or \[\frac{dy}{{{2}^{y}}}=\frac{dx}{{{2}^{x}}}\] Integrating both sides, \[\int{\frac{dy}{{{2}^{y}}}=\int{\frac{dx}{{{2}^{x}}}}}\] \[-{{2}^{-y}}\log 2=-{{2}^{-x}}\log 2+{{c}_{1}}\] \[\frac{\log 2}{{{2}^{x}}}-\frac{\log 2}{{{2}^{y}}}={{c}_{1}}\]; \[\frac{1}{{{2}^{x}}}-\frac{1}{{{2}^{y}}}=\frac{{{c}_{1}}}{\log 2}=c\].
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