What is the solution for (dP)/(dt)=P(5-P)-4 ?

The solution for (dP)/(dt)=P(5-P)-4 is P=-(4e^{3t-3c_{1}})/(1-e^{3(t-c_{1))}}+1/(1-e^{3(t-c_{1))}}

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(dP)/(dt)=P(5-P)-4

\frac{d P}{d t} = P (5 - P) - 4

P = - \frac{4 e ^{3 t - 3 c_{1}}}{1 - e ^{3 (t - c_{1})}} + \frac{1}{1 - e ^{3 (t - c_{1})}}

Solution steps

\frac{d P}{d t} = P (5 - P) - 4

Substitute

\frac{d P}{d t}

with

P^{'} (t)

P^{^{'}} (t) = P (5 - P) - 4

Solve for

t, P^{'} (t) = \frac{d P}{d t} = \frac{1}{\frac{d t}{d P}} = \frac{1}{t ^{^{'}}}

(\frac{1}{t ^{^{^{'}}}}) = P (5 - P) - 4

Isolate

t^{^{'}} : t^{^{'}} = \frac{1}{5 P - P ^{2} - 4}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{1}{5 P - P ^{2} - 4} d P

\int \frac{1}{5 P - P ^{2} - 4} d P = \frac{1}{3} (ln (- \frac{2}{3} + \frac{2}{3} P) - ln (- \frac{8}{3} + \frac{2}{3} P)) + c_{1}

t = \frac{1}{3} (ln (- \frac{2}{3} + \frac{2}{3} P) - ln (- \frac{8}{3} + \frac{2}{3} P)) + c_{1}

Simplify

t = \frac{1}{3} ln (\frac{- 1 + P}{- 4 + P}) + c_{1}

Isolate

P : P = - \frac{4 e ^{3 t - 3 c_{1}}}{1 - e ^{3 (t - c_{1})}} + \frac{1}{1 - e ^{3 (t - c_{1})}}

P = - \frac{4 e ^{3 t - 3 c_{1}}}{1 - e ^{3 (t - c_{1})}} + \frac{1}{1 - e ^{3 (t - c_{1})}}

x \to 2 lim (x^{3} \cdot 2 x^{2})

\frac{\partial}{\partial x} (x y + x^{2} y + x y^{2})

\frac{\partial}{\partial x} (e^{- 2 x^{2}})

\frac{\partial}{\partial s} (4 s^{3} t^{2} + 2 s t)

8 y^{^{''}} - 32 y = 0

What is the solution for (dP)/(dt)=P(5-P)-4 ?
The solution for (dP)/(dt)=P(5-P)-4 is P=-(4e^{3t-3c_{1}})/(1-e^{3(t-c_{1))}}+1/(1-e^{3(t-c_{1))}}