K01 koz = k03 K04 = = = P(1) ≈ P₁ = Iteration 2: K11 = K12 K13 = K14 = = P(2) ≈ P₂ =

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
Suppose that the
population of
Jacobsbaai, a small
coastal town on the
West Coast of South
Africa, was 1000 at
the end of the year
2000. Suppose further
that the population
increased
(continuously or
steadily) by
approximately 5% per
year. Then the
population can be
described by the DE
dP
dt
= 0.05P.
Applying two
iterations of the 4th
Order Runge-Kutta
Method with P(0) = 100
57
Transcribed Image Text:Suppose that the population of Jacobsbaai, a small coastal town on the West Coast of South Africa, was 1000 at the end of the year 2000. Suppose further that the population increased (continuously or steadily) by approximately 5% per year. Then the population can be described by the DE dP dt = 0.05P. Applying two iterations of the 4th Order Runge-Kutta Method with P(0) = 100 57
K01
koz
=
=
k03 =
K04
=
P(1) ≈ P₁ =
Iteration 2:
K11
K12 =
K13
=
K₁4 =
P(2) ≈ P₂ =
(55
Transcribed Image Text:K01 koz = = k03 = K04 = P(1) ≈ P₁ = Iteration 2: K11 K12 = K13 = K₁4 = P(2) ≈ P₂ = (55
Expert Solution
Step 1

Given P=0.05P,  P(0)=100,   h=1

4th order Runge-Kutta method


k01=hP(t0,y0)=(1)P(0,100)=(1)(5)=5

k02=hP(t0+h/2,P0+k01/2)=(1)P(0.5,102.5)=(1)(5.125)=5.125

k03=hP(t0+h/2,P0+k02/2)=(1)P(0.5,102.5625)=(1)(5.1281)=5.1281

k04=hP(t0+h,P0+k03)=(1)P(1,105.1281)=(1)(5.2564)=5.2564

P1=P0+16(k01+2k02+2k03+k04)

P1=100+16[5+2(5.125)+2(5.1281)+(5.2564)]

P1=105.1271

 ∴P(1)=105.1271

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