On substituting

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={y}^{{t}}\)

On rewriting in the form of an exact differential equation

\(\displaystyle-{x}+{y}+{1}+{\left({x}+{y}+{3}\right)}{y}^{{t}}={0}\)

To verify the following condition

\(\displaystyle{\frac{{\partial{M}{\left({x},{y}\right)}}}{{\partial{y}}}}={\frac{{\partial{N}{\left({x},{y}\right)}}}{{\partial{x}}}}\)

Thus the above condition is true

\(\displaystyle{3}{y}+{x}{y}+{\frac{{{y}^{{2}}}}{{{2}}}}-{\frac{{{x}^{{2}}}}{{{2}}}}+{x}+{c}_{{1}}={c}_{{2}}\)

\(\displaystyle{3}{y}+{x}{y}+{\frac{{{y}^{{2}}}}{{{2}}}}-{\frac{{{x}^{{2}}}}{{{2}}}}+{x}={c}_{{1}}\)

PSKy= -3-x+\sqrt{2x^2+4x+c_1+9}PSK

\(\displaystyle{y}=-{x}-\sqrt{{{2}{x}^{{2}}+{4}{x}+{c}_{{1}}+{9}}}-{3}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={y}^{{t}}\)

On rewriting in the form of an exact differential equation

\(\displaystyle-{x}+{y}+{1}+{\left({x}+{y}+{3}\right)}{y}^{{t}}={0}\)

To verify the following condition

\(\displaystyle{\frac{{\partial{M}{\left({x},{y}\right)}}}{{\partial{y}}}}={\frac{{\partial{N}{\left({x},{y}\right)}}}{{\partial{x}}}}\)

Thus the above condition is true

\(\displaystyle{3}{y}+{x}{y}+{\frac{{{y}^{{2}}}}{{{2}}}}-{\frac{{{x}^{{2}}}}{{{2}}}}+{x}+{c}_{{1}}={c}_{{2}}\)

\(\displaystyle{3}{y}+{x}{y}+{\frac{{{y}^{{2}}}}{{{2}}}}-{\frac{{{x}^{{2}}}}{{{2}}}}+{x}={c}_{{1}}\)

PSKy= -3-x+\sqrt{2x^2+4x+c_1+9}PSK

\(\displaystyle{y}=-{x}-\sqrt{{{2}{x}^{{2}}+{4}{x}+{c}_{{1}}+{9}}}-{3}\)