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}}}

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}}}