A) \[{{e}^{y}}=({{e}^{x}}+1)+C{{e}^{-x}}\]
B) \[{{e}^{y}}=({{e}^{x}}-1)+C\]
C) \[{{e}^{y}}=({{e}^{x}}-1)+C{{e}^{-x}}\]
D) None of the above
Correct Answer: C
Solution :
\[\frac{dy}{dx}=\frac{{{e}^{x}}}{{{e}^{y}}}({{e}^{x}}-{{e}^{y}})\] \[\Rightarrow \] \[{{e}^{y}}\frac{dy}{dx}+{{e}^{x}}\cdot {{e}^{y}}={{e}^{x}}\cdot {{e}^{x}}\] Let \[{{e}^{y}}=t\] \[\Rightarrow \] \[{{e}^{y}}\frac{dy}{dx}\cdot \frac{y}{x}=\frac{dt}{dx}\] Then, given equation reduces to \[\frac{dt}{dx}+{{e}^{x}}t={{e}^{2x}}\] Here, \[P={{e}^{x}}\] and \[Q={{e}^{2x}}\] \[\therefore \] \[IF={{e}^{\int{Pdx}}}={{e}^{\int{{{e}^{x}}dx}}}={{e}^{{{e}^{x}}}}\] Required solution is \[t\cdot {{e}^{{{e}^{x}}}}=\int{{{e}^{2x}}\cdot }\,\,{{e}^{{{e}^{x}}}}dx+C\] \[\Rightarrow \] \[{{e}^{y}}=({{e}^{x}}-1)+C\cdot {{e}^{-{{e}^{x}}}}\]
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