Solve this differential equation explicitly, either by using partial fractions, or with a computer algebra system. Use the initial populations 350 and 450. Can anyone please help me with this problem and explain how you got the answers?

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1 Answer
Jun 24, 2018

For the IV: #P_o = 350#

  • # P= 100 ((e^( t/50) + 5)/(e^( t/50) - 5) + 5 )#

For the IV: #P_o = 450#

  • # P= 100 (( e^( t/50) - 3)/( e^( t/50) + 3 ) + 5 )#

Explanation:

#dot P = P/10 - P^2/10000 - 24 #

Complete square:

#= 1 - (P - 500)^2/10000#

Let #Q = P - 500 qquad dot Q = dot P#

So the DE is:

#dot Q = 1 - (Q/100)^2#

Let #R = Q/100 qquad dot R = (dotQ)/100#

#100 dot R = 1 - R^2#

This separates:

#(dR)/(1 - R^2) = 1/100dt#

Factor and use partial fractions:

#= int dR \ 1/((1 - R)(1+R)) = 1/100 int \dt#

#=1/2 int dR \ 1/ (R + 1) - 1/ (R - 1) = t/100 + C#

#=1/2( ln (R + 1) - ln (R - 1) ) = t/100 + C#

#= ln ((R + 1)/ (R - 1) ) = t/50 + C#

#R = (Ce^( t/50) + 1)/(Ce^( t/50) - 1)#

Reverse the subs:

# Q= 100 (Ce^( t/50) + 1)/(Ce^( t/50) - 1)#

#bb( P= 100 ((Ce^( t/50) + 1)/(Ce^( t/50) - 1) + 5 )#

For the IV: #P_o = 350#

#350 = 100 ((C + 1)/(C - 1) + 5 ) implies C = 1/5#

# P= 100 ((e^( t/50) + 5)/(e^( t/50) - 5) + 5 )#

For the IV: #P_o = 450#

#450 = 100 ((C + 1)/(C - 1) + 5 ) implies C = -1/3#

# P= 100 (( e^( t/50) - 3)/( e^( t/50) + 3 ) + 5 )#