# Solve differential equation y(2x-2+xy+1)dx+(x-y)dy=0

Question
Solve differential equation $$\displaystyle{y}{\left({2}{x}-{2}+{x}{y}+{1}\right)}{\left.{d}{x}\right.}+{\left({x}-{y}\right)}{\left.{d}{y}\right.}={0}$$

2020-12-18
$$\displaystyle{y}{\left({2}{x}-{2}+{x}{y}+{1}\right)}{\left.{d}{x}\right.}+{\left({x}-{y}\right)}{\left.{d}{y}\right.}={0}$$
$$\displaystyle{y}{\left({2}{x}-{2}+{x}{y}+{1}\right)}+{\left({x}-{y}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0}$$
$$\displaystyle{y}{\left({2}{x}-{2}+{x}{y}+{1}\right)}+{\left({x}-{y}\right)}{y}'={0}$$
$$\displaystyle{y}'=-{\frac{{{y}{\left({2}{x}^{{2}}-{x}{y}+{1}\right)}}}{{{x}-{y}}}}$$
$$\displaystyle{y}={x}{v}\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}+{x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle{y}'=-{\frac{{{y}{\left({2}{x}^{{2}}-{x}{y}+{1}\right)}}}{{{x}-{y}}}}\Rightarrow{\left({x}-{y}\right)}{y}'+{y}{\left({2}{x}^{{2}}-{x}{y}+{1}\right)}={0}$$
$$\displaystyle{\left({x}-{y}\right)}{b}{i}{g}{g{{\left({v}+{x}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}{b}{i}{g}{g}\right)}}}+{x}{v}{\left({2}{x}^{{2}}-{x}{y}+{1}\right)}={0}$$
$$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{-{v}^{{2}}-{x}^{{2}}{v}^{{2}}+{2}{v}+{2}{x}^{{2}}{v}}}{{{x}{\left({v}-{1}\right)}}}}$$
$$\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{\left({x}^{{2}}+{1}\right)}{\left({v}-{2}\right)}{v}}}{{{x}{\left({v}-{1}\right)}}}}$$
$$\displaystyle{\frac{{{1}}}{{{v}}}}{b}{i}{g}{g{{\left({\frac{{{v}-{1}}}{{{v}-{2}}}}{b}{i}{g}{g}\right)}}}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}-{\frac{{{x}^{{2}}+{1}}}{{{x}}}}$$
Integrating both sides with respect to x
$$\displaystyle{\frac{{{1}}}{{{v}}}}{b}{i}{g}{g{{\left({\frac{{{v}-{1}}}{{{v}-{2}}}}{b}{i}{g}{g}\right)}}}{\left\lbrace{d}{v}\right\rbrace}=-\int{\frac{{{\left({x}^{{2}}+{1}\right)}}}{{{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}{\log{{\left(-{v}+{2}\right)}}}+{\frac{{{1}}}{{{2}}}}{\log{{v}}}=-{\frac{{{x}^{{2}}}}{{{2}}}}={\log{{x}}}+{c}_{{1}}$$
$$\displaystyle{v}{\left({x}\right)}=-{\frac{{{e}^{{-{x}^{{2}}\sqrt{{{e}^{{{x}^{{2}}}}{\left(-{e}^{{{2}{c}_{{1}}}}+{e}^{{{x}^{{2}}}}{x}^{{2}}\right)}}}}}}}{{{x}}}}+{1}\ \text{or}\ {v}{\left({x}\right)}={\frac{{{e}^{{-{x}^{{2}}\sqrt{{{e}^{{{x}^{{2}}}}{\left(-{e}^{{{2}{c}_{{1}}}}+{e}^{{{x}^{{2}}}}{x}^{{2}}\right)}}}}}}}{{{x}}}}+{1}$$

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