What is the solution for (x^2+1)yy^'=1 ?

The solution for (x^2+1)yy^'=1 is y=sqrt(2arctan(x)+c_{1)},y=-sqrt(2arctan(x)+c_{1)}

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(x^2+1)yy^'=1

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

y = 2 arctan (x) + c_{1}, y = - 2 arctan (x) + c_{1}

Solution steps

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

Solve separable ODE:

y = 2 arctan (x) + c_{1}, y = - 2 arctan (x) + c_{1}

y = 2 arctan (x) + c_{1}, y = - 2 arctan (x) + c_{1}

\frac{d x}{d t} - 0.04 x = 5000

d y (x y - 2 x + 4 y - 8) - d x (x y + 3 x - y - 3) = 0

(4 y - 5) d x + (7 x - 2) d y = 0

y^{^{'}} + y^{2} - y = 0

(6 y^{2} + 4 x y^{3}) d x + (6 x y + 4 x^{2} y^{2}) d y = 0

What is the solution for (x^2+1)yy^'=1 ?
The solution for (x^2+1)yy^'=1 is y=sqrt(2arctan(x)+c_{1)},y=-sqrt(2arctan(x)+c_{1)}