Re: [理工] [工數]-拉式常係數ode
※ 引述《smallprawn (水中瑕)》之銘言：
: t-1, 1<t<2
: 3.y''+4y'+3y=r(t) y(0)=2, y'(0)=-2, r(t)={
: 0, otherwise
: 請問這題要怎麼算?!
: 麻煩哩!!
r(t)=(t-1)[H(t-1)-H(t-2)]
y''+4y'+3y=r(t)
令￡[y(t)]=Y(s)
原式取Laplace transform得
2 -s -2s
[s Y-sy(0)-y'(0)]+4[sY-y(0)]+3Y=e ￡[(t-1)+1] - e ￡[(t-1)+2]
2 -s 1 -2s 1 1
(s+4s+3)Y-2s+2-8=e ── - e (── + ──)
s^2 s^2 s
1 -s -2s 1 -2s
(s+1)(s+3)Y= 2s+6 + ──(e - e ) - ── e
s^2 s
2 1 -s -2s 1 -2s
Y= ── + ─────── (e - e ) - ─────── e
s+1 s^2(s+1)(s+3) s(s+1)(s+3)
1 A B C D
令 ─────── = ── + ─── + ─── + ───
s^2(s+1)(s+3) s s^2 s+1 s+3
去分母得
1 = As(s+1)(s+3) + B(s+1)(s+3) + Cs^2(s+3) + Ds^2(s+1)
s=-1代入得 1=C[(-1)^2]*2 , C=1/2
s=-3代入得 1=D[(-3)^2]*(-2), D=-1/18
s=0 代入得 1=B*1*3, B=1/3
等號左式之s^3之係數為0，等號右式之s^3之係數為A+C+D
所以 A+C+D=0 即A+(1/2)+(-1/18)=0, A=-4/9
1 E F G
令 ─────── = ── + ─── + ───
s(s+1)(s+3) s s+1 s+3
去分母得
1 = E(s+1)(s+3) + Fs(s+3) + Gs(s+1)
s=-1代入得 1=F(-1)*2 , F=-1/2
s=-3代入得 1=G(-3)*(-2), G=1/6
s=0 代入得 1=E*1*3, E=1/3
所以
2 -4/9 1/3 1/2 -1/18 -s -2s
Y= ── +( ── + ─── + ─── + ───) (e - e )
s+1 s s^2 s+1 s+3
1/3 -1/2 1/6 -2s
-(── + ─── + ─── )e
s s+1 s+3
2 -4/9 1/3 1/2 -1/18 -s
= ── +( ── + ─── + ─── + ───) e
s+1 s s^2 s+1 s+3
1/9 -1/3 -1/9 -2s
+(── + ─── + ───)e
s s^2 s+3
-1 -t -(t-1) -3(t-1)
y(t)=￡ [Y(s)] = 2e +[-4/9+(1/3)(t-1)+(1/2)e -(1/18)e ]H(t-1)
-3(t-2)
+[1/9-(1/3)(t-2)-(1/9)e ]H(t-2)
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01/29 19:58, , 1F
01/29 19:58, 1F
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