What is the solution for (2xy^{-1})dx+(2-4x^2y^{-2})dy=0 ?

The solution for (2xy^{-1})dx+(2-4x^2y^{-2})dy=0 is

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(2xy^{-1})dx+(2-4x^2y^{-2})dy=0

(2 x y^{- 1}) d x + (2 - 4 x^{2} y^{- 2}) d y = 0

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

Solution steps

(2 x y^{- 1}) d x + (2 - 4 x^{2} y^{- 2}) d y = 0

Solve exact ODE:

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

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

y = 2 ln (x), at x = 4

\frac{\partial}{\partial x} (x y e^{- x y})

\frac{d}{d x} (ln (3 x + 2 y))

y^{^{''}} + 7 y^{^{'}} + 23 y = 0

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

What is the solution for (2xy^{-1})dx+(2-4x^2y^{-2})dy=0 ?
The solution for (2xy^{-1})dx+(2-4x^2y^{-2})dy=0 is