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

Bihar CECEBihar CECE 2014

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}}}} $