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

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}