Identify the following linear differential equation. x^{2}y'+y^{2}=0

quiquenobi2v6 2022-01-03 Answered
Identify the following linear differential equation.
\(\displaystyle{x}^{{{2}}}{y}'+{y}^{{{2}}}={0}\)

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deginasiba
Answered 2022-01-04 Author has 1904 answers
Given \(\displaystyle{x}^{{{2}}}{y}'+{y}^{{{2}}}={0}\)
\(\displaystyle{x}^{{{2}}}{y}'+{y}^{{{2}}}={0}\)
\(\displaystyle{x}^{{{2}}}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{y}^{{{2}}}\)
\(\displaystyle\Rightarrow{\frac{{{\left.{d}{x}\right.}}}{{{x}^{{{2}}}}}}=-{\frac{{{\left.{d}{y}\right.}}}{{{y}^{{{2}}}}}}\)
Integrate on both sides
\(\displaystyle\Rightarrow\int{\frac{{{\left.{d}{x}\right.}}}{{{x}^{{{2}}}}}}=-\int{\frac{{{\left.{d}{y}\right.}}}{{{y}^{{{2}}}}}}+{C}\)
\(\displaystyle\Rightarrow\int{x}^{{-{2}}}{\left.{d}{x}\right.}=-\int{y}^{{-{2}}}{\left.{d}{y}\right.}+{C}\)
\(\displaystyle\Rightarrow{\frac{{{x}^{{-{2}+{1}}}}}{{-{2}+{1}}}}=-{\left({\frac{{{y}^{{-{2}+{1}}}}}{{-{2}+{1}}}}\right)}+{C}\)
\(\displaystyle\Rightarrow{\frac{{{x}^{{-{1}}}}}{{-{1}}}}=-{\left({\frac{{{y}^{{-{1}}}}}{{-{1}}}}\right)}+{C}\)
\(\displaystyle\Rightarrow-{\frac{{{1}}}{{{x}}}}={\frac{{{1}}}{{{y}}}}+{C}\)
\(\displaystyle\Rightarrow{\frac{{{1}}}{{{x}}}}+{\frac{{{1}}}{{{y}}}}+{C}={0}\)
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Fasaniu
Answered 2022-01-05 Author has 642 answers
\(\displaystyle{x}^{{{2}}}{y}’+{y}^{{{2}}}={0},{y}{\left(-{1}\right)}={1}\)
\(\displaystyle{x}^{{{2}}}{y}’=-{y}^{{{2}}}\Rightarrow\frac{{\left.{d}{y}\right.}}{{y}^{{{2}}}}=-\frac{{\left.{d}{x}\right.}}{{x}^{{{2}}}}\Rightarrow\int\frac{{\left.{d}{y}\right.}}{{y}^{{{2}}}}=-\int\frac{{\left.{d}{x}\right.}}{{x}^{{{2}}}}\)
\(\displaystyle-\frac{{1}}{{y}}=\frac{{1}}{{x}}+{C}\Rightarrow-\frac{{1}}{{1}}=\frac{{1}}{{-{1}}}+{C}\Rightarrow{C}={0}\)
\(\displaystyle-\frac{{1}}{{y}}=\frac{{1}}{{x}}\Rightarrow{y}=-{x}\)
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karton
Answered 2022-01-09 Author has 8454 answers

\(\begin{array}{}-dy/y^{2}=dx/x^{2}Z \\1/y-C=-1/x \\y=1/(C-1/x) \\=-1с=0 \\y=-x \end{array}\)

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