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

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

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

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

y = x^{2} - x^{4} - c_{1}, y = x^{2} + x^{4} - c_{1}

Solution steps

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

Solve exact ODE:

y = x^{2} - x^{4} - c_{1}, y = x^{2} + x^{4} - c_{1}

y = x^{2} - x^{4} - c_{1}, y = x^{2} + x^{4} - c_{1}

\frac{\partial}{\partial x} (\frac{64}{x ^{2}})

\frac{\partial}{\partial x} (ln (1 + x^{2}))

\int (1 - x) e^{- x} d x

\frac{d}{d x} (x \cdot arctan (x))

d er i v a t i v e f (x) = cot^{2} (sin (x))

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