1/a b/a dP = dt P а — БР, 1 – In \PI 1 - In la – bP| = t + c а а P In = at + ac а bP P Gjea а — БР follows from the last equation that ac¡ed 1 + bcje“ acī P(t) = bc, + e¯at* P(0) = Po, Po ± alb, we find c= Po/(a – bPo), and so, after substituting and simplifying, the
1/a b/a dP = dt P а — БР, 1 – In \PI 1 - In la – bP| = t + c а а P In = at + ac а bP P Gjea а — БР follows from the last equation that ac¡ed 1 + bcje“ acī P(t) = bc, + e¯at* P(0) = Po, Po ± alb, we find c= Po/(a – bPo), and so, after substituting and simplifying, the
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Please show all the steps including partial fractions,
1st image: Solve the first equation to get the second equation shown and Please explain all the steps.[separation of variables]
2nd image: The necessary steps are shown below but I don't get why the sign changed in the second step and where did the b go from the numerator in the first step go.
![(1/a
b/a
= dt
dP
а — БР,
P
1
- In|P]
1
- In la – bP] = t + c
-
a
а
P
= at + ac
In
a
bP
jea
a
bP
It follows from the last equation that
acje
1 + bcje
ac
P(t)
bc, + e¯at*
If P(0) = Po, Po + alb, we find c¡ = Po/(a – bPo), and so, after substituting and simplifying, the
solution becomes
aPo
P(t)
(5)
bP + (a
bPe-at•
|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb6da5ff5-4f9e-46f1-b6af-425e0be2530f%2F6aa2d9f9-d2da-41da-b50b-9eb694d6339b%2F4qi5u6_processed.png&w=3840&q=75)
Transcribed Image Text:(1/a
b/a
= dt
dP
а — БР,
P
1
- In|P]
1
- In la – bP] = t + c
-
a
а
P
= at + ac
In
a
bP
jea
a
bP
It follows from the last equation that
acje
1 + bcje
ac
P(t)
bc, + e¯at*
If P(0) = Po, Po + alb, we find c¡ = Po/(a – bPo), and so, after substituting and simplifying, the
solution becomes
aPo
P(t)
(5)
bP + (a
bPe-at•
|
Transcribed Image Text:Since, we have logistic equation. So, we can write dN
dt
- Ν(α - bN) :
On solving, we get N(t)
аNo
=
bNo + (a – bNo)e-at
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