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

The solution for y^'=1+x^2+y^2+x^2y^2 is y=tan((x^3)/3+x+c_{1})

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

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

y = tan (\frac{x ^{3}}{3} + x + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (\frac{x ^{3}}{3} + x + c_{1})

y = tan (\frac{x ^{3}}{3} + x + c_{1})

\int_{0}^{1} \frac{y}{e ^{2 y}} d y

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

\frac{d}{d x} (x^{2} - \frac{2}{x ^{2}})

\frac{\partial}{\partial x} (x \cdot Y (x))

(- 7, - 4), (- 2, 3)

What is the solution for y^'=1+x^2+y^2+x^2y^2 ?
The solution for y^'=1+x^2+y^2+x^2y^2 is y=tan((x^3)/3+x+c_{1})