What is the solution for y^'(p)= 1/9 y(p)(2y(p)-1)(p-3) ?

The solution for y^'(p)= 1/9 y(p)(2y(p)-1)(p-3) is y(p)= 1/(2-e^{\frac{p^2){18}-p/3+c_{1}}}

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y^'(p)= 1/9 y(p)(2y(p)-1)(p-3)

y^{^{'}} (p) = \frac{1}{9} y (p) (2 y (p) - 1) (p - 3)

y (p) = \frac{1}{2 - e ^{\frac{p ^{2}}{18} - \frac{p}{3} + c_{1}}}

Solution steps

y^{'} (p) = \frac{1}{9} y (p) (2 y (p) - 1) (p - 3)

Solve separable ODE:

y (p) = \frac{1}{2 - e ^{\frac{p ^{2}}{18} - \frac{p}{3} + c_{1}}}

y (p) = \frac{1}{2 - e ^{\frac{p ^{2}}{18} - \frac{p}{3} + c_{1}}}

2 y y^{^{'}} = 15 t^{2}, y (0) = 6

\frac{d y}{d x} + \frac{d y}{d x} + y = 0

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

y^{^{'}} + \frac{y}{5} = 3, y (- 1) = 0

x \frac{d y}{d x} - y = 5 x^{9} y

What is the solution for y^'(p)= 1/9 y(p)(2y(p)-1)(p-3) ?
The solution for y^'(p)= 1/9 y(p)(2y(p)-1)(p-3) is y(p)= 1/(2-e^{\frac{p^2){18}-p/3+c_{1}}}