What is the solution for tan^2(y)sec^2(y)y^'=x^2 ?

The solution for tan^2(y)sec^2(y)y^'=x^2 is y=arctan(\sqrt[3]{x^3+c_{1}})+pin

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tan^2(y)sec^2(y)y^'=x^2

tan^{2} (y) sec^{2} (y) y^{^{'}} = x^{2}

Solution steps

tan^{2} (y) sec^{2} (y) y^{'} (x) = x^{2}

Solve separable ODE:

\frac{d y}{d x} y = 2 x + 1

2 x y d x - x^{2} d y + d y = 0

2 x y y^{^{'}} = 4 y^{2} + 3 y^{2}

2 x y + 6 x + (x^{2} - 9) \frac{d y}{d x} = 0

\frac{d y}{d x} = \frac{4 ( y + 1 )}{x}

What is the solution for tan^2(y)sec^2(y)y^'=x^2 ?
The solution for tan^2(y)sec^2(y)y^'=x^2 is y=arctan(\sqrt[3]{x^3+c_{1}})+pin