What is the solution for 2x^2yy^'+y^2=2 ?

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

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2x^2yy^'+y^2=2

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

y = - e^{\frac{1}{x} - c_{1}} + 2, y = - - e^{\frac{1}{x} - c_{1}} + 2

Solution steps

2 x^{2} y y^{'} (x) + y^{2} = 2

Solve separable ODE:

y = - e^{\frac{1}{x} - c_{1}} + 2, y = - - e^{\frac{1}{x} - c_{1}} + 2

y = - e^{\frac{1}{x} - c_{1}} + 2, y = - - e^{\frac{1}{x} - c_{1}} + 2

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x y^{^{'}} + y = 3 cos (x)

\frac{\partial}{\partial x} (\frac{6}{5} - \frac{6}{5} e^{- 20 x})

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