Solve for G.S / P.S forbthe following differential equations using the method of solutions for linear equations of first order, first degree (y-x)(y+x)(y^{2}+x^{2})(xdy-ydx)=x^{6}dx

Question
Forms of linear equations
asked 2021-01-22
Solve for G.S / P.S forbthe following differential equations using the method of solutions for linear equations of first order, first degree
\(\displaystyle{\left({y}-{x}\right)}{\left({y}+{x}\right)}{\left({y}^{{{2}}}+{x}^{{{2}}}\right)}{\left({x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}\right)}={x}^{{{6}}}{\left.{d}{x}\right.}\)

Answers (1)

2021-01-23
\(\displaystyle{\left({y}-{x}\right)}{\left({y}+{x}\right)}{\left({y}^{{{2}}}+{x}^{{{2}}}\right)}{\left({x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}\right)}={x}^{{{6}}}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{\left({y}^{{{2}}}-{x}^{{{2}}}\right)}{\left({y}^{{{2}}}-{x}^{{{2}}}\right)}{\left({x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}\right)}={x}^{{{6}}}{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{\left({y}^{{{4}}}-{x}^{{{4}}}\right)}{\left({x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}\right)}={x}^{{{4}}}{\left.{d}{x}\right.}{\left[{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{{2}}}-{b}^{{{2}}}\right]}\)
\(\displaystyle\Rightarrow{\frac{{{\left({y}^{{{4}}}-{x}^{{{4}}}\right\rbrace}{\left\lbrace{x}^{{{4}}}\right\rbrace}{\frac{{{x}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}}}{{{x}^{{{2}}}}}}={\left.{d}{x}\right.}}}{}}\)
\(\displaystyle{\left[{\left({\frac{{{y}}}{{{x}}}}\right)}^{{{4}}}-{1}\right]}{d}{\frac{{{y}}}{{{x}}}}={\left.{d}{x}\right.}\)
\(\displaystyle\int{\left[{\left({\frac{{{y}}}{{{x}}}}\right)}^{{{4}}}-{1}\right]}{d}{\frac{{{y}}}{{{x}}}}=\int{\left.{d}{x}\right.}+{e}{\left({e}={c}{o}{n}{s}{\tan{{t}}}\right)}\)
\(\displaystyle\Rightarrow{\frac{{{\left\lbrace{\frac{{{y}}}{{{x}}}}\right\rbrace}}}{^}}{\left\lbrace{4}+{1}\right\rbrace}{\left\lbrace{4}+{1}\right\rbrace}-{\frac{{{y}}}{{{x}}}}={x}+{e}\)
\(\displaystyle\Rightarrow{\frac{{{y}^{{{5}}}}}{{{5}{x}^{{{5}}}}}}-{\frac{{{y}}}{{{x}}}}={x}+{e}\)
\(\displaystyle\Rightarrow{y}^{{{5}}}-{5}{x}{6}{\left\lbrace{4}\right\rbrace}{y}={5}{x}^{{{5}}}{\left({x}+{e}\right)}\)
\(\displaystyle\Rightarrow{y}{\left({y}^{{{4}}}-{5}{x}^{{{4}}}\right)}={5}{x}^{{{5}}}{\left({x}+{e}\right)}\)
Therefore the sol is
\(\displaystyle{y}{\left({y}^{{{4}}}-{5}{x}^{{{4}}}\right)}={5}{x}^{{{5}}}{\left({x}+{e}\right)}\)
0

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