dy / dx = x + y , y(0) = 1
上題的“解析解”為 y = (2e^x) - x - 1
問題：
dy / dt = y - t^2 , y(0) = 1
的“解析解” 該如何解呢???????????
請大家幫幫忙
求數值解析解???
2012-01-06 7:47 am
回答 (2)
2012-01-06 8:51 pm
2012-01-06 10:49 am
[[ANS]]
齊次解
y'=y
.=> y(h)=c e^t
特解(算子法)
y'-y=-t^2
.=>(D-1)y=-t^2
.=>y(p)=1/(D-1) -t^2
=1/(1-D) t^2
=(1+D+D^2+...)t^2
=t^2+2 t+2
解
y=y(h)+y(p)
=c e^t+t^2+2t+2
y(0)=c+2=1
.=> c=-1
.=> y=-e^t+t^2+2t+2
[[DONE]]
齊次解
y'=y
.=> y(h)=c e^t
特解(算子法)
y'-y=-t^2
.=>(D-1)y=-t^2
.=>y(p)=1/(D-1) -t^2
=1/(1-D) t^2
=(1+D+D^2+...)t^2
=t^2+2 t+2
解
y=y(h)+y(p)
=c e^t+t^2+2t+2
y(0)=c+2=1
.=> c=-1
.=> y=-e^t+t^2+2t+2
[[DONE]]
收錄日期: 2021-04-13 18:27:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120105000010KK09921