What is the solution for y^'=((4t^3+1))/(2y-6) ?

The solution for y^'=((4t^3+1))/(2y-6) is y=3+sqrt(t^4+t+c_{1)+9},y=3-sqrt(t^4+t+c_{1)+9}

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y^'=((4t^3+1))/(2y-6)

y^{^{'}} = \frac{( 4 t ^{3} + 1 )}{2 y - 6}

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

Solution steps

y^{'} (t) = \frac{( 4 t ^{3} + 1 )}{2 y - 6}

Solve separable ODE:

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

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

\int tan^{3} (x) sec^{2} (x) d x

y = 15 - x^{2}, y = x^{2} - 3

\int_{1}^{5} \frac{1}{1 + 2 x} d x

\frac{\partial}{\partial x} (\frac{x ^{2} y}{x ^{4} + y ^{2}})

∣ x - 2 ∣, 0 = x - 3 y + 6

What is the solution for y^'=((4t^3+1))/(2y-6) ?
The solution for y^'=((4t^3+1))/(2y-6) is y=3+sqrt(t^4+t+c_{1)+9},y=3-sqrt(t^4+t+c_{1)+9}