What is the solution for (dP)/(dt)=P(9-P)-81/4 ?

The solution for (dP)/(dt)=P(9-P)-81/4 is P= 1/(t+c_{1)}+(9t)/(2(t+c_{1))}+(9c_{1})/(2(t+c_{1))}

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(dP)/(dt)=P(9-P)-81/4

\frac{d P}{d t} = P (9 - P) - \frac{81}{4}

P = \frac{1}{t + c _{1}} + \frac{9 t}{2 ( t + c _{1} )} + \frac{9 c _{1}}{2 ( t + c _{1} )}

Solution steps

\frac{d P}{d t} = P (9 - P) - \frac{81}{4}

Solve separable ODE:

P = \frac{1}{t + c _{1}} + \frac{9 t}{2 ( t + c _{1} )} + \frac{9 c _{1}}{2 ( t + c _{1} )}

P = \frac{1}{t + c _{1}} + \frac{9 t}{2 ( t + c _{1} )} + \frac{9 c _{1}}{2 ( t + c _{1} )}

\int \frac{3 - 4 x ^{3} + 6 x ^{6}}{x ^{6}} d x

\frac{d y}{d x} = 1 - e^{x - y + 8}

- y^{^{'''}} + 2 y^{^{''}} + 3 y^{^{'}} + 3 = 0

h (x) = \frac{1}{x ^{10}}

\int \frac{1}{x ^{- 1} + 1} d x

What is the solution for (dP)/(dt)=P(9-P)-81/4 ?
The solution for (dP)/(dt)=P(9-P)-81/4 is P= 1/(t+c_{1)}+(9t)/(2(t+c_{1))}+(9c_{1})/(2(t+c_{1))}