Solution:

\(\displaystyle{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}+{x}{e}^{{{y}}}{\left.{d}{y}\right.}={0},{y}{\left({1}\right)}={0}\)

Compare above with \(\displaystyle{M}{\left.{d}{x}\right.}+{N}{\left.{d}{y}\right.}={0}\)

\(\displaystyle{M}={2}{x}+{e}^{{{y}}},{N}={x}{e}^{{{y}}}\)

\(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={e}^{{{y}}},{\frac{{\partial{N}}}{{\partial{x}}}}={e}^{{{y}}}\)

Since \(\displaystyle{\frac{{\partial{M}}}{{\partial{y}}}}={\frac{{\partial{N}}}{{\partial{y}}}}={\frac{{\partial{N}}}{{\partial{x}}}}\), therefore differential equation is exact

\(\displaystyle{\frac{{\partial{f}}}{{\partial{x}}}}={2}{x}+{e}^{{{y}}}\) (i)

\(\displaystyle{\frac{{\partial{f}}}{{\partial{y}}}}={x}{e}^{{{y}}}\) (ii)

integrate (i) wrt x, we get

\(\displaystyle{f{{\left({x},{y}\right)}}}=\int{\left({2}{x}+{e}^{{{y}}}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{2}}}+{x}{e}^{{{y}}}+{h}{\left({y}\right)}\)

Differentiate above wrt y, we get

\(\displaystyle{\frac{{\partial{f}}}{{\partial{y}}}}={x}{e}^{{{y}}}+{h}'{\left({y}\right)}\)

compare above and (ii), we get

\(\displaystyle{x}{e}^{{{y}}}+{h}'{\left({y}\right)}={x}{e}^{{{y}}}\)

\(\displaystyle{h}'{\left({y}\right)}={0}\)

integrate both sides wrt y, we get \(\displaystyle{h}{\left({y}\right)}={C}\)

therefore, we get

\(\displaystyle{f{{\left({x},{y}\right)}}}={x}^{{{2}}}+{x}{e}^{{{y}}}+{C}\)

or we can write as \(\displaystyle{x}^{{{2}}}+{x}{e}^{{{y}}}+{C}={0}\)