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

x \to - 1 lim (\frac{2 x}{∣ x ∣})

\int \frac{e ^{x^{\frac{1}{2}}}}{x ^{3}} d x

\int (- 2 T + x) \cdot e^{- i \cdot z \cdot x} d x

(t^{2}) \cdot (sin (2 \cdot t))

x \to 25 lim (\frac{\frac{1}{x} - \frac{1}{5}}{x - 25})

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}