What is the solution for 2x^2y(dy)/(dx)+1+2xy^2=0 ?

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

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2x^2y(dy)/(dx)+1+2xy^2=0

2 x^{2} y \frac{d y}{d x} + 1 + 2 x y^{2} = 0

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

Solution steps

2 x^{2} y \frac{d y}{d x} + 1 + 2 x y^{2} = 0

Solve Bernoulli ODE:

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

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

\frac{\partial}{\partial x} (x y + \frac{27}{x})

\frac{d}{d x} (e^{5 x + πi})

\int (6 + x^{5})^{2} 5 x^{4} d x

\frac{\partial}{\partial x} (x^{3} + y^{3} - 3 x^{2} + 3 y^{2})

\frac{d}{d a} (1 + cos^{4} (a) - sin^{4} (a))

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