A) \[{{e}^{y}}=({{e}^{x}}+1)+C{{e}^{-{{e}^{x}}}}\]
B) \[{{e}^{y}}=({{e}^{x}}-1)+C\]
C) \[{{e}^{y}}=({{e}^{x}}-1)+C{{e}^{-{{e}^{x}}}}\]
D) None of the above
Correct Answer: C
Solution :
Given equation can be rewritten as \[\frac{dy}{dx}=\frac{{{e}^{x}}}{{{e}^{y}}}({{e}^{x}}-{{e}^{y}})\] \[\Rightarrow \] \[{{e}^{y}}\frac{dy}{dx}={{e}^{2x}}-{{e}^{x}}{{e}^{y}}\] \[\Rightarrow \] \[{{e}^{y}}\frac{dy}{dx}+{{e}^{x}}{{e}^{y}}={{e}^{2x}}\] Put \[{{e}^{y}}=t\Rightarrow {{e}^{y}}\frac{dy}{dx}=\frac{dt}{dx}\] \[\therefore \] \[\frac{dt}{dx}+{{e}^{x}}t={{e}^{2x}}\] On comparing with \[\frac{dx}{dy}+Pt=Q,\]we get \[P={{e}^{x}}\] and \[Q={{e}^{2x}}\] \[\therefore \] If \[={{e}^{\int_{{}}^{{}}{pdx}}}={{e}^{\int_{{}}^{{}}{{{e}^{x}}dx}}}={{e}^{{{e}^{x}}}}\] Required solution is \[t.{{e}^{{{e}^{x}}}}=\int_{{}}^{{}}{{{e}^{2x}}{{e}^{{{e}^{x}}}}}dx+C\] \[\Rightarrow \] \[{{e}^{y}}{{e}^{{{e}^{x}}}}=({{e}^{x}}-1){{e}^{{{e}^{x}}}}+C\] \[\Rightarrow \] \[{{e}^{y}}=({{e}^{x}}-1)+C{{e}^{-{{e}^{x}}}}\]
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