What is the solution for xy+(x^2+y^2)((dy)/(dx))=0 ?

The solution for xy+(x^2+y^2)((dy)/(dx))=0 is

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xy+(x^2+y^2)((dy)/(dx))=0

x y + (x^{2} + y^{2}) (\frac{d y}{d x}) = 0

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

x y + (x^{2} + y^{2}) (\frac{d y}{d x}) = 0

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

s \to - 3 lim (3 \frac{- 4 + 5 s + 3 s ^{2}}{- s ^{3} + s + 40})

f (x) = sin (x)

(\frac{1}{x + 2})^{^{'}}

\int_{- 1}^{4} 3 y - (y^{2} - 4) d y

x \to 0 lim (\frac{( 1 + x ) ^{2} + 1}{x})

What is the solution for xy+(x^2+y^2)((dy)/(dx))=0 ?
The solution for xy+(x^2+y^2)((dy)/(dx))=0 is