Question

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

First order differential equations
ANSWERED
asked 2020-12-17
Solve differential equation \(\displaystyle{y}{\left({2}{x}-{2}+{x}{y}+{1}\right)}{\left.{d}{x}\right.}+{\left({x}-{y}\right)}{\left.{d}{y}\right.}={0}\)

Answers (1)

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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