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

The solution for (dy)/(dx)=(3x^2-2x)/(2y+1),y(0)=2 is y=(-1+sqrt(4x^3-4x^2+25))/2

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

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

y = \frac{- 1 + 4 x ^{3} - 4 x ^{2} + 25}{2}

Solution steps

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

Solve separable ODE:

y = \frac{- 1 + 4 x ^{3} - 4 x ^{2} + 25}{2}

y = \frac{- 1 + 4 x ^{3} - 4 x ^{2} + 25}{2}

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

y^{^{'}} = t \cdot e^{y}, y (2) = 0

\frac{d y}{d x} = y cos (x), y (0) = 1

x^{2} \cdot y^{^{'}} + y = 0

y cos (x) d x + sin (x) d y = 0

What is the solution for (dy)/(dx)=(3x^2-2x)/(2y+1),y(0)=2 ?
The solution for (dy)/(dx)=(3x^2-2x)/(2y+1),y(0)=2 is y=(-1+sqrt(4x^3-4x^2+25))/2