What is the Solution of the Differential Equation # (dP)/dt = kP - AP^2#?

3 Answers
Steve M
May 9, 2017

# P(t) = ( kmue^(kt)) / (1 + muAe^(kt)) # where #mu = P_0/(k-AP_0) #

Explanation:

We have:

# (dP)/dt = kP - AP^2 #
# :. (dP)/dt = P(k-AP) #
# :. 1/(P(k-AP))(dP)/dt = 1 #

Which is a First Order Separable DE, so we can "separate the variables" to gett:

# int \ 1/(P(k-AP)) \ dP = int \ dt \ \ \ \ \ ..... (star) #

We can decompose the LHS integrand into Partial Fractions:

# 1/(P(k-AP)) -= a/P + b/(k-AP) #
# " " -= (a(k-AP) + bP)/(P(k-AP)) #

Leading to the identity:

# 1 -= a(k-AP) + bP #

To find thee coefficients #a# and #b# we can use substitution:

Put #P = 0 \ \ => 1 = ak \ \ => a = 1/k#
Put #P = k/A => 1 = (bk)/A => b = A/k#

Hence we can re-write #(star)# as:

# int \ (1/k)/P + (A/k)/(k-AP) \ dP = int \ dt #

# :. 1/k \ int \ 1/P + A/(k-AP) \ dP = int \ dt #

Which we can now integrate to get:

# :. 1/k {lnP+A(-1/A)ln(k-AP)} = t + C #

# :. 1/k {lnP - ln(k-AP)} = t + C #

# :. 1/k ln(P/(k-AP)) = t + C #

Using the initial condition #P=P_0# when #t=0# we have:

# :. 1/k ln(P_0/(k-AP_0)) = C #

And so the solution is:

# 1/k ln(P/(k-AP)) = t + 1/k ln(P_0/(k-AP_0)) #
# :. ln(P/(k-AP)) = kt + ln(P_0/(k-AP_0)) #

We can remove the logarithm by taking exponents of both sides:

# e^(ln(P/(k-AP))) = e^(kt + ln(P_0/(k-AP_0))) #

# :. P/(k-AP) = e^(kt)e^ln(P_0/(k-AP_0))) #
# " " = e^(kt)(P_0/(k-AP_0)) #

To simplify the expression, let #mu = P_0/(k-AP_0) #, then:

# P/(k-AP) = mu e^(kt) #
# :. P = mu (k-AP) e^(kt) #
# " " = muke^(kt) - muAPe^(kt) #
# :. P + muAPe^(kt) = muke^(kt) #
# :. P(1 + muAe^(kt)) = muke^(kt) #
# :. P = ( muke^(kt)) / (1 + muAe^(kt)) #

Hence:

# P(t) = ( kmue^(kt)) / (1 + muAe^(kt)) # where #mu = P_0/(k-AP_0) #

Eddie
May 9, 2017

#(dP)/(dt) = kP - AP^2, A>0#

This is separable:

#(dP)/(dt) = kP(1 - A/kP)#

#int (dP)/(P(1 - A/kP)) = int k dt#

Partial Fraction decomp:

#1/(P(1 - A/kP) )= alpha / P + beta / (1 - A/kP)#

#implies alpha (1 - A/kP) + beta P = 1#

#implies alpha = 1, beta = A/k#

#implies int 1/ P + (A/k) / (1 - A/kP) dP = int k dt#

#[ln P - ln (1 - A/kP) ]_(P_o)^P= [ kt]_0^t #

#implies ln( (P/P_o) / ( (1 - A/kP) / (1 - A/kP_o)) )= kt #

#implies (P/P_o) / ( (1 - A/kP) / (1 - A/kP_o)) = e^(kt) #

# implies P = ( 1 - A/kP) P_o/(1 - A/kP_o) e^(kt) #

# implies P ( 1 + A/k * P_o/(1 - A/kP_o) e^(kt) )= P_o/(1 - A/kP_o) e^(kt) #

# implies P= ( P_o/(1 - A/kP_o) e^(kt) )/ ( 1 + A/k * P_o/(1 - A/kP_o) e^(kt) )#

#= ( k P_o e^(kt) )/ ( (k - A P_o) + A P_o e^(kt) )#

#= ( k P_o e^(kt) )/ ( k - A P_o (1- e^(kt) )#

That's soooo ugly, there must be a better way :(

Cesareo R.
May 9, 2017

#P=k(e^(k (t+C_0))/(1+Ae^(kt+C_0)))#

Explanation:

This is a separable differential equation so

#(dP)/(kP-AP^2) = dt# or

#1/k(1/P+A/(k-AP))dP= dt#

or

#1/k(log_e P-log_e(k-AP))= t + C_0#

or

#log_e P-log_e(k-AP)=k(t + C_0)# or

#P/(k-AP)=C_1e^(kt+C_0)#

and finally

#P=k(e^(k (t+C_0))/(1+Ae^(kt+C_0)))#

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