What is the solution for 2x^2yy^'+1+2xy^2=0 ?

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

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2x^2yy^'+1+2xy^2=0

2 x^{2} y y^{^{'}} + 1 + 2 x y^{2} = 0

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

Solution steps

2 x^{2} y y^{'} (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}

\int_{- 2}^{1} x ((4 - x^{2}) - (x + 2)) d x

d x = (e^{t^{2}} + 2 x t) d t

\int 1 3^{x} d x

- 2 t^{3} + 96 t

\frac{d}{d x} (9 cos (x) - 7 x^{2})

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