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

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

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

y^{^{'}} = \frac{3 t ^{2}}{4 + 2 y}

y = - 2 + t^{3} + c_{1} + 4, y = - t^{3} + c_{1} + 4 - 2

Solution steps

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

Solve separable ODE:

y = - 2 + t^{3} + c_{1} + 4, y = - t^{3} + c_{1} + 4 - 2

y = - 2 + t^{3} + c_{1} + 4, y = - t^{3} + c_{1} + 4 - 2

y^{^{'}} = y x^{2} - 1.1 y

x \frac{d y}{d x} = e^{3 y + x}

y^{^{'}} + 4 y = 2

\frac{d y}{d x} = e^{x} cos^{2} (y)

(x^{2} + 1) \frac{d y}{d x} + 3 x (y - 1) = 0, y (0) = 2

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