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

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

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(dy)/(dx)=(3x^2+4x+2)/(2y+1),y(0)=-1

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

y = \frac{- 1 - 1 - 4 ( - 2 x - 2 x ^{2} - x ^{3} )}{2}

Solution steps

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

Solve separable ODE:

y = \frac{- 1 - 1 - 4 ( - 2 x - 2 x ^{2} - x ^{3} )}{2}

y = \frac{- 1 - 1 - 4 ( - 2 x - 2 x ^{2} - x ^{3} )}{2}

\frac{d v}{d t} = 30 - \frac{9}{100} v, v = 0, t = 0

(2 y + 2) d y - (4 x^{3} + 6) d x = 0

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

\frac{y ^{^{'}}}{x} = y

(1 + x^{2}) y^{^{'}} + x e^{- y} = 0

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