Laplace Transform for Simultaneous Equation?
Laplace Transform for Simultaneous Equation?
How to solve the following equations by using Laplace Transform?
x' + x - y = 1 + sint
y' - x' + y = t - sint
x(0)=0,y(0)=1
回答 (1)
x' + x - y = 1 + sint
y' - x' + y = t - sint
x(0)=0,y(0)=1
Laplace Transform
s X(s) - x(0) + X(s) - Y(s) = 1/s + 1/(s^2+1)
s Y(s) - y(0) - [ s X(s) - x(0) ] + Y(s) = 1/s^2 - 1/(s^2+1)
(s+1) X(s) - Y(s) = 1/s + 1/(s^2+1) .....(1)
- s X(s) + (s+1) Y(s) = 1+ 1/s^2 - 1/(s^2+1) .....(2)
(1)*(s+1) + (2)
(s^2+s+1) X(s) = 1/s + 1 + 1/(s^2+1) + s/(s^2+1) + 1+ 1/s^2 - 1/(s^2+1)
= 1/s^2 + 1/s + 2 + s/(s^2+1)
X(s) = 1/[s^2(s^2+s+1)] + 1/[s(s^2+s+1)] + 2/(s^2+s+1) + s/[(s^2+1)(s^2+s+1)]
= (1+s+2s^2)/[s^2(s^2+s+1)] + 1/(s^2+1) - 1/(s^2+s+1)
note:
set (1+s+2s^2)/[s^2(s^2+s+1)] = a/s + b/s^2 + (cs+d)/(s^2+s+1)
1+s+2s^2= as(s^2+s+1) + b(s^2+s+1) + s^2(cs+d)
a+c=0
2=a+b+d
1=a+b
1=b
a=0, c=0, d=1
X(s) = 1/[s^2(s^2+s+1)] + 1/[s(s^2+s+1)] + 2/(s^2+s+1) + s/[(s^2+1)(s^2+s+1)]
= 1/s^2 + 1/(s^2+s+1) + 1/(s^2+1) - 1/(s^2+s+1)
= 1/s^2 + 1/(s^2+1)
x(t) = t + sin t
sub X(s) = 1/s^2 + 1/(s^2+1) into (1)
(s+1)/s^2 + (s+1)/(s^2+1) - Y(s) = 1/s + 1/(s^2+1)
Y(s) = 1/s^2 + s/(s^2+1)
y(t) = t + cos t
Ans: x(t) = t + sin t, y(t) = t + cos t
收錄日期: 2021-04-18 17:56:47
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https://hk.answers.yahoo.com/question/index?qid=20171116004954AADXX2e
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