What is the solution for x^2y^'=1-x^2+y^2-x^2y^2 ?

The solution for x^2y^'=1-x^2+y^2-x^2y^2 is y=tan(-x-1/x+c_{1})

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x^2y^'=1-x^2+y^2-x^2y^2

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

y = tan (- x - \frac{1}{x} + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (- x - \frac{1}{x} + c_{1})

y = tan (- x - \frac{1}{x} + c_{1})

\int x^{2} cos (\frac{1}{3} x) d x

\frac{d y}{d x} = - sin^{2} (x + y)

\int_{- 1}^{1} x + 2 d x

x \to 2 lim (\frac{2 - x}{2 x ^{2} - 4})

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