What is the solution for y^'=(-(3y^2+1)x)/(x^2+1) ?

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

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y^'=(-(3y^2+1)x)/(x^2+1)

y^{^{'}} = \frac{- ( 3 y ^{2} + 1 ) x}{x ^{2} + 1}

y = \frac{tan ( \frac{3 ( - l n ( x ^{2} + 1 ) + c _{1} )}{2} )}{3}

Solution steps

y^{'} (x) = \frac{- ( 3 y ^{2} + 1 ) x}{x ^{2} + 1}

Solve separable ODE:

y = \frac{tan ( \frac{3 ( - l n ( x ^{2} + 1 ) + c _{1} )}{2} )}{3}

y = \frac{tan ( \frac{3 ( - l n ( x ^{2} + 1 ) + c _{1} )}{2} )}{3}

\frac{d}{d x} (7 x^{2} + 8 ln (x y)) = 14

y^{^{'}} = \frac{4 x}{1 + 2 y}, y (2) = 0

y^{^{'}} = - \frac{9}{4} x^{6} y^{4}

y^{^{'}} + 4 y = 5, y (0) = 1

\frac{d y}{d x} = 140 y x^{13}

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