What is the solution for 3x^2y^'=2y(y-3) ?

The solution for 3x^2y^'=2y(y-3) is y= 3/(e^{(-2-3c_{1)x)/x}+1}

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3x^2y^'=2y(y-3)

3 x^{2} y^{^{'}} = 2 y (y - 3)

y = \frac{3}{e ^{\frac{- 2 - 3 c _{1} x}{x}} + 1}

Solution steps

3 x^{2} y^{'} (x) = 2 y (y - 3)

Solve Bernoulli ODE:

y = \frac{3}{e ^{\frac{- 2 - 3 c _{1} x}{x}} + 1}

y = \frac{3}{e ^{\frac{- 2 - 3 c _{1} x}{x}} + 1}

\int x e^{(\frac{- x}{2})} d x

\int 8 e^{x} 9 + e^{x} d x

f (x) = \frac{x ^{7}}{7} - \frac{7}{x ^{7}}, at x = - 1

4

\frac{d}{d x} (x sin (x) - cos (x))

What is the solution for 3x^2y^'=2y(y-3) ?
The solution for 3x^2y^'=2y(y-3) is y= 3/(e^{(-2-3c_{1)x)/x}+1}