Question

The solution of the differential equation $ \frac{dy}{dx}={{e}^{x-y}}({{e}^{x}}-{{e}^{y}}) $ is

• 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

Answer 1

Answer: (C)

$ \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}}}} $