What is the solution for y^'=-(xy)/(x^2+y^2) ?

The solution for y^'=-(xy)/(x^2+y^2) is

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y^'=-(xy)/(x^2+y^2)

y^{^{'}} = - \frac{x y}{x ^{2} + y ^{2}}

y = - x^{2} + x^{4} + c_{1}, y = - - x^{2} + x^{4} + c_{1}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x ^{2}} x, y = - - \frac{x ^{2} + x ^{4} + c _{1}}{x ^{2}} x

Solution steps

y^{'} (x) = - \frac{x y}{x ^{2} + y ^{2}}

Solve using substitution:

y = - x^{2} + x^{4} + c_{1}, y = - - x^{2} + x^{4} + c_{1}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x ^{2}} x, y = - - \frac{x ^{2} + x ^{4} + c _{1}}{x ^{2}} x

y = - x^{2} + x^{4} + c_{1}, y = - - x^{2} + x^{4} + c_{1}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x ^{2}} x, y = - - \frac{x ^{2} + x ^{4} + c _{1}}{x ^{2}} x

\int \frac{x ^{2}}{1 + x ^{2}} d x

x y^{^{'}} + 4 y = 8 x^{4}, y (1) = 2

x^{2} + 3 x + 9, at x = 4

\frac{d}{d x} ((x + \frac{1}{x})^{2} - \frac{5}{3 x})

\frac{\partial}{\partial x} (x e^{3 y} sin (5 z))

What is the solution for y^'=-(xy)/(x^2+y^2) ?
The solution for y^'=-(xy)/(x^2+y^2) is