What is the solution for (dy)/(dt)=(3t^2+4t+2)/(2(y-1)),y(0)=-1 ?

The solution for (dy)/(dt)=(3t^2+4t+2)/(2(y-1)),y(0)=-1 is y=1-sqrt((t^2+2)(t+2))

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(dy)/(dt)=(3t^2+4t+2)/(2(y-1)),y(0)=-1

\frac{d y}{d t} = \frac{3 t ^{2} + 4 t + 2}{2 ( y - 1 )}, y (0) = - 1

y = 1 - (t^{2} + 2) (t + 2)

Solution steps

\frac{d y}{d t} = \frac{3 t ^{2} + 4 t + 2}{2 ( y - 1 )}

Solve separable ODE:

y = 1 - (t^{2} + 2) (t + 2)

y = 1 - (t^{2} + 2) (t + 2)

y^{^{'}} = \frac{e ^{- x} - e ^{x}}{3 + 4 y}

y + 2 y^{^{'}} - 3 y = 0

4 (1 + y^{2}) d x - x y d y = 0, y (2) = 15

2 x y d x = (2 - x^{2}) d y, y (4) = 3

(1 + y) d y - x d x = 0

What is the solution for (dy)/(dt)=(3t^2+4t+2)/(2(y-1)),y(0)=-1 ?
The solution for (dy)/(dt)=(3t^2+4t+2)/(2(y-1)),y(0)=-1 is y=1-sqrt((t^2+2)(t+2))