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

x \to 2 lim (\frac{x ^{3} - 4}{x - 2})

8 x^{2} y^{^{'}} = y^{^{'}} + 2 x e^{- y}

\frac{s + 3}{s ^{2} + s + 3}

\int x^{3} (x^{4} - 1)^{2} d x

\int 8 cos (4 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}