Find the solution to the ordinary differential equation m'(n) + nm = m given that m(0) = 20?
回答 (2)
2016-07-06 2:32 pm
✔ 最佳答案
Use separation of variables.dm/dn + nm = m
dm/dn = m - nm
dm/dn = m(1-n)
1/m dm = (1-n) dn
Integrating:
ln m = n - ½ n² + C
Initial condition:
ln 20 = 0 - ½ (0)² + C
ln 20 = C
Solution:
ln m = n - ½ n² + ln 20
2016-07-06 2:48 pm
m'(n) + nm = m
dm/dn = m-nm
dm/dn = m(1-n)
dm/m = (1-n) dn
Integrate both sides
ln m = n - (1/2) n^2 + C
m(0) = 20
ln 20 = 0 - (1/2) (0)^2 + C
C = ln 20
ln m = n - (1/2) n^2 + ln 20
m = e^(n-n^2 /2) e^ln 20
m = 20 e^(n - n^2/2)
dm/dn = m-nm
dm/dn = m(1-n)
dm/m = (1-n) dn
Integrate both sides
ln m = n - (1/2) n^2 + C
m(0) = 20
ln 20 = 0 - (1/2) (0)^2 + C
C = ln 20
ln m = n - (1/2) n^2 + ln 20
m = e^(n-n^2 /2) e^ln 20
m = 20 e^(n - n^2/2)
收錄日期: 2021-04-21 19:16:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160706062701AAbmN6j