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(2) In this problem we will solve the logistic equation with an initial growth rate of = 0.40, a carrying capacity of K = 300, and an initial value of No = 10. (a) The logistic equation is an autonomous differential equation. Begin by taking the reciprocal of both sides of the equation and integrate both sides. To evaluate 1 1 1 the integral -dN = you will need to N N 1 300 0.40 (.40)N ( 1 300 - perform partial fraction decomposition. After you integrate you should be left with N 1 t = 0.40 In |N| – In |1 - 300 +C In order to continue, you must use the logarithmic difference rule ln |x|– In |y| = In |x/y| to rewrite the righthand side as: 1 t = 0.40 N + C N In 300 (b) Use the initial value N(0) = 10 to solve for C in the above equation. You should 300 get C = – In 29 (c) Finish by solving for N. Your answer should look like: 300 e(0.40)t 29 N(t) = 300 e 29 (0.40)t 1+ 300

(2) In this problem we will solve the logistic equation with an initial growth rate of = 0.40, a carrying capacity of K = 300, and an initial value of No = 10. (a) The logistic equation is an autonomous differential equation. Begin by taking the reciprocal of both sides of the equation and integrate both sides. To evaluate 1 1 1 the integral -dN = you will need to N N 1 300 0.40 (.40)N ( 1 300 - perform partial fraction decomposition. After you integrate you should be left with N 1 t = 0.40 In |N| – In |1 - 300 +C In order to continue, you must use the logarithmic difference rule ln |x|– In |y| = In |x/y| to rewrite the righthand side as: 1 t = 0.40 N + C N In 300 (b) Use the initial value N(0) = 10 to solve for C in the above equation. You should 300 get C = – In 29 (c) Finish by solving for N. Your answer should look like: 300 e(0.40)t 29 N(t) = 300 e 29 (0.40)t 1+ 300

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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(2) In this problem we will solve the logistic equation with an initial growth rate of r = 0.40, a carrying capacity of K = 300, and an initial value of No (a) The logistic equation is an autonomous differential equation. Begin by taking the reciprocal of both sides of the equation and integrate both sides. To evaluate 10. 1 1 1 the integral / you will need to N 0.40 N (.40)N (1 300 N ( 1 300 perform partial fraction decomposition. After you integrate you should be left with 1 N In |N| – In |1 t = 0.40 300 In order to continue, you must use the logarithmic difference rule In |x|– In |y| = In |x/y| to rewrite the righthand side as: 1 t = In 0.40 N + C N 300 (b) Use the initial value N(0) = 10 to solve for C in the above equation. You should 300 get C = – In 29 (c) Finish by solving for N. Your answer should look like: 300 e(0.40)t 29 N(t) = 300 e(0.40)t 29 1+ 300 (d) Using a graphing utility graph the function N(t) found in part (c), and compare it to the graph sketched in 1.(c).
Transcribed Image Text:(2) In this problem we will solve the logistic equation with an initial growth rate of r = 0.40, a carrying capacity of K = 300, and an initial value of No (a) The logistic equation is an autonomous differential equation. Begin by taking the reciprocal of both sides of the equation and integrate both sides. To evaluate 10. 1 1 1 the integral / you will need to N 0.40 N (.40)N (1 300 N ( 1 300 perform partial fraction decomposition. After you integrate you should be left with 1 N In |N| – In |1 t = 0.40 300 In order to continue, you must use the logarithmic difference rule In |x|– In |y| = In |x/y| to rewrite the righthand side as: 1 t = In 0.40 N + C N 300 (b) Use the initial value N(0) = 10 to solve for C in the above equation. You should 300 get C = – In 29 (c) Finish by solving for N. Your answer should look like: 300 e(0.40)t 29 N(t) = 300 e(0.40)t 29 1+ 300 (d) Using a graphing utility graph the function N(t) found in part (c), and compare it to the graph sketched in 1.(c).
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