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}

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))