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

The solution for 2xydx+(x^2+2y)dy=0 is y=(-x^2+sqrt(x^4+c_{1)})/2 ,y=(-x^2-sqrt(x^4+c_{1)})/2

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2xydx+(x^2+2y)dy=0

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

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

Solution steps

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

Solve exact ODE:

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

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

y = tan^{2} (x)

\int \frac{1}{sin ( t )} d t

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\int_{0}^{\infty} y^{2} e^{- y} d y

\int \frac{x ^{4}}{x ^{3} + x ^{2} - x - 1} d x

What is the solution for 2xydx+(x^2+2y)dy=0 ?
The solution for 2xydx+(x^2+2y)dy=0 is y=(-x^2+sqrt(x^4+c_{1)})/2 ,y=(-x^2-sqrt(x^4+c_{1)})/2