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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

a x^{^{'}} + b x = 0, x (0) = e^{r t}

(4 x^{5} - 16 x^{3} - 2 x) (1 - y^{2}) - 2 y (x^{2} - 4) y^{^{'}} = 0

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = 1

y^{^{'}} = \frac{1 + x}{x y}, y (1) = - 4

\frac{d y}{d x} + 4 x y = 4 e^{6} x

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