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

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

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deginasiba
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
$$\displaystyle{x}^{{{2}}}{y}’+{y}^{{{2}}}={0},{y}{\left(-{1}\right)}={1}$$
$$\displaystyle{x}^{{{2}}}{y}’=-{y}^{{{2}}}\Rightarrow\frac{{\left.{d}{y}\right.}}{{y}^{{{2}}}}=-\frac{{\left.{d}{x}\right.}}{{x}^{{{2}}}}\Rightarrow\int\frac{{\left.{d}{y}\right.}}{{y}^{{{2}}}}=-\int\frac{{\left.{d}{x}\right.}}{{x}^{{{2}}}}$$
$$\displaystyle-\frac{{1}}{{y}}=\frac{{1}}{{x}}+{C}\Rightarrow-\frac{{1}}{{1}}=\frac{{1}}{{-{1}}}+{C}\Rightarrow{C}={0}$$
$$\displaystyle-\frac{{1}}{{y}}=\frac{{1}}{{x}}\Rightarrow{y}=-{x}$$
karton

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