[理工] [工數] LAPLACE求解ODE
Q: 0 , 0<=t<4
y"+4y=f(t) f(t)= {3 , t>=4
y(0)=1,y'(0)=0
ANS:
y = cos(2t) + (3/4)[1-cos2(t-4)]U(t-4)
我的答案:
y = (1/2)sin(2t) + (3/4)[1-cos2(t-4)]U(t-4)
過程:
原式取LAPLACE
3 1
Y(s)=--------e^(-4s) + -----
s(s^2+4) s^2+4
我是取LAPLACE時算錯了嗎?
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