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= 0.11(1 – I) – 0.03I; I € (-0.1,1.1); I(0) = 0.2; I(0) = 0.9 dt dR R - 0.1R; R E (-0.1, 10); R(0) = 20; R(0) = 0 dt 1+ R dP = 0.02P(P – 100)(1200 – P) P € (-0.1, 1200); P(0) = 20; P(0) = 1000 %3D dt

= 0.11(1 – I) – 0.03I; I € (-0.1,1.1); I(0) = 0.2; I(0) = 0.9 dt dR R - 0.1R; R E (-0.1, 10); R(0) = 20; R(0) = 0 dt 1+ R dP = 0.02P(P – 100)(1200 – P) P € (-0.1, 1200); P(0) = 20; P(0) = 1000 %3D dt

Advanced Engineering Mathematics
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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For the following differential equations: a) plot the defining function over the indicated range (use any computational tools you wish) to determine the intervals on which the dependent variable is increasing and decreasing; b) find the equilibria c) determine the stability of each equilibrium; d) based on your analysis in parts a-c, sketch (by hand) plots of the solutions with the specified initial values.

dI = 0.11(1 – I) –- 0.03I; I € (-0.1,1.1); I(0) = 0.2; I(0) = 0.9 dt dR R 5. 0.1R; RE (-0.1, 10); R(0) = 20; R(0) = 0 dt 1+ R dP — 0.02P(P — 100) (1200 — Р) Р (-0.1,1200); Р(0) — 20; Р(0) — 1000 dt 6. 4.
Transcribed Image Text:dI = 0.11(1 – I) –- 0.03I; I € (-0.1,1.1); I(0) = 0.2; I(0) = 0.9 dt dR R 5. 0.1R; RE (-0.1, 10); R(0) = 20; R(0) = 0 dt 1+ R dP — 0.02P(P — 100) (1200 — Р) Р (-0.1,1200); Р(0) — 20; Р(0) — 1000 dt 6. 4.
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