What is the Solution of the Differential Equation dPdt=kP−AP2?

dPdt=kP−AP2,A&gt;0This is separable:dPdt=kP(1−AkP)∫dPP(1−AkP)=∫kdtPartial Fraction decomp:1P(1−AkP)=αP+β1−AkP⇒α(1−AkP)+βP=1⇒α=1,β=Ak⇒∫1P+Ak1−AkPdP=∫kdt[ln⁡P−ln⁡(1−AkP)]PoP=[kt]0t⇒ln⁡(PPo1−AkP1−AkPo)=kt⇒PPo1−AkP1−AkPo=ekt⇒P=(1−AkP)Po1−AkPoekt⇒P(1+Ak⋅Po1−AkPoekt)=Po1−AkPoekt⇒P=Po1−AkPoekt1+Ak⋅Po1−AkPoekt=kPoekt(k−APo)+APoekt=kPoektk−APo(1−ekt)

This differential equation can be separated, sodPkP−AP2=dt1k(1P+Ak−AP)dP=dt1k(logeP−loge(k−AP))=t+C0logeP−loge(k−AP)=k(t+C0)Pk−AP=C1ekt+C0thenP=k(ek(t+C0)1+Aekt+C0)

dPdt=kP−AP2,A&gt;0This is separable:dPdt=kP(1−AkP)∫dPP(1−AkP)=∫kdtPartial Fraction decomp:1P(1−AkP)=αP+β1−AkP⇒α(1−AkP)+βP=1⇒α=1,β=Ak⇒∫1P+Ak1−AkPdP=∫kdt[ln⁡P−ln⁡(1−AkP)]PoP=[kt]0t⇒ln⁡(PPo1−AkP1−AkPo)=kt⇒PPo1−AkP1−AkPo=ekt⇒P=(1−AkP)Po1−AkPoekt⇒P(1+Ak⋅Po1−AkPoekt)=Po1−AkPoekt⇒P=Po1−AkPoekt1+Ak⋅Po1−AkPoekt=kPoekt(k−APo)+APoekt=kPoektk−APo(1−ekt)

This differential equation can be separated, sodPkP−AP2=dt1k(1P+Ak−AP)dP=dt1k(logeP−loge(k−AP))=t+C0logeP−loge(k−AP)=k(t+C0)Pk−AP=C1ekt+C0thenP=k(ek(t+C0)1+Aekt+C0)

Find out the answer to "What is the Solution of the Differential Equation..."

Kason Wong

Answered question

2022-12-21

What is the Solution of the Differential Equation

\frac{d P}{d t} = k P - A P^{2}

Answer & Explanation

CredyBetCreta109

Beginner2022-12-22Added 11 answers

\frac{d P}{d t} = k P - A P^{2}, A > 0

This is separable:

\frac{d P}{d t} = k P (1 - \frac{A}{k} P)

\int \frac{d P}{P (1 - \frac{A}{k} P)} = \int k d t

Partial Fraction decomp:

\frac{1}{P (1 - \frac{A}{k} P)} = \frac{α}{P} + \frac{β}{1 - \frac{A}{k} P}

\Rightarrow α (1 - \frac{A}{k} P) + β P = 1

\Rightarrow α = 1, β = \frac{A}{k}

\Rightarrow \int \frac{1}{P} + \frac{\frac{A}{k}}{1 - \frac{A}{k} P} d P = \int k d t

{[\ln P - \ln (1 - \frac{A}{k} P)]}_{P_{o}}^{P} = {[k t]}_{0}^{t}

\Rightarrow \ln (\frac{\frac{P}{P_{o}}}{\frac{1 - \frac{A}{k} P}{1 - \frac{A}{k} P_{o}}}) = k t

\Rightarrow \frac{\frac{P}{P_{o}}}{\frac{1 - \frac{A}{k} P}{1 - \frac{A}{k} P_{o}}} = e^{k t}

\Rightarrow P = (1 - \frac{A}{k} P) \frac{P_{o}}{1 - \frac{A}{k} P_{o}} e^{k t}

\Rightarrow P (1 + \frac{A}{k} \cdot \frac{P_{o}}{1 - \frac{A}{k} P_{o}} e^{k t}) = \frac{P_{o}}{1 - \frac{A}{k} P_{o}} e^{k t}

\Rightarrow P = \frac{\frac{P_{o}}{1 - \frac{A}{k} P_{o}} e^{k t}}{1 + \frac{A}{k} \cdot \frac{P_{o}}{1 - \frac{A}{k} P_{o}} e^{k t}}

= \frac{k P_{o} e^{k t}}{(k - A P_{o}) + A P_{o} e^{k t}}

= \frac{k P_{o} e^{k t}}{k - A P_{o} (1 - e^{k t})}

alemanica8h2

Beginner2022-12-23Added 2 answers

This differential equation can be separated, so

\frac{d P}{k P - A P^{2}} = d t

\frac{1}{k} (\frac{1}{P} + \frac{A}{k - A P}) d P = d t

\frac{1}{k} (\log_{e} P - \log_{e} (k - A P)) = t + C_{0}

\log_{e} P - \log_{e} (k - A P) = k (t + C_{0})

\frac{P}{k - A P} = C_{1} e^{k t + C_{0}}

then

P = k (\frac{e^{k (t + C_{0})}}{1 + A e^{k t + C_{0}}})

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