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

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

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

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

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

Solution steps

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

\frac{d}{d x} (3 ln (x + 1))

\frac{d}{d x} (35 x ln (13 x))

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

x \to 0 - lim (\frac{2}{x} - \frac{2}{∣ x ∣})

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