What is the solution for y^'=2(1+x)(1+y^2),y(0)=0 ?

The solution for y^'=2(1+x)(1+y^2),y(0)=0 is y=tan(x^2+2x)

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y^'=2(1+x)(1+y^2),y(0)=0

y^{^{'}} = 2 (1 + x) (1 + y^{2}), y (0) = 0

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

Solution steps

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

Solve separable ODE:

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

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

\int (x^{2} + 3 x - 1) \cdot e^{2 x} d x

\frac{\partial}{\partial x} (5 x^{2 y})

\frac{d}{d x} (ln (csc (x) - cot (x)))

\frac{1}{x}

\int_{- 1}^{1} \frac{1}{x ^{3}} d x

What is the solution for y^'=2(1+x)(1+y^2),y(0)=0 ?
The solution for y^'=2(1+x)(1+y^2),y(0)=0 is y=tan(x^2+2x)