What is the solution for y^'=(4x^3+1)/(2y-6) ?

The solution for y^'=(4x^3+1)/(2y-6) is y=3+sqrt(x^4+x+c_{1)+9},y=3-sqrt(x^4+x+c_{1)+9}

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y^'=(4x^3+1)/(2y-6)

y^{^{'}} = \frac{4 x ^{3} + 1}{2 y - 6}

y = 3 + x^{4} + x + c_{1} + 9, y = 3 - x^{4} + x + c_{1} + 9

Solution steps

y^{'} (x) = \frac{4 x ^{3} + 1}{2 y - 6}

Solve separable ODE:

y = 3 + x^{4} + x + c_{1} + 9, y = 3 - x^{4} + x + c_{1} + 9

y = 3 + x^{4} + x + c_{1} + 9, y = 3 - x^{4} + x + c_{1} + 9

\frac{d}{d t} (\frac{t sin ( t )}{1 + t})

\int 0.1 x^{2} d x

\int \frac{6}{x ^{2} - x - 2} d x

y^{^{'}} (1 - x y^{1985}) = y^{1986}

\int \frac{5 x ^{3} - 3 x}{x} d x

What is the solution for y^'=(4x^3+1)/(2y-6) ?
The solution for y^'=(4x^3+1)/(2y-6) is y=3+sqrt(x^4+x+c_{1)+9},y=3-sqrt(x^4+x+c_{1)+9}