※ 引述《t5d (t5d)》之銘言：
: 1.y'' + 4y = f(t) ,y(0)=0 ,y'(0)=1 ,f(t)=t ,0<t<1 and f(t+1)=f(t)
1 -st
∫ t e dt
0 -1 -s 1 -s 1 1
L[f(t)] = ------------------ = [---- e - ---- e + ---- ] ---------
-s s 2 2 -s
1 - e s s 1 - e
-1 -s 1 -s 1
= [---- e + ---- (1- e ) ] -----------
s 2 -s
s 1 - e
-s
-1 e 1
= ------ ---------- + -----------
s -s 2
1 - e s
-1 1 1
= ----- [ -1 + ----------] + ----------
s -s 2
1 - e s
1 1 1 1
= ---- - ---- ------------ + -----------
s s -s 2
1 - e s
2 1 1 1 1
(s + 4) Y(s) = ---- - --- ---------- + ------------ + 1
s s -s 2
1 - e s
1 1 1 1 1
Y(s) = ----------- - ----------- --------- + ------- + --------
2 2 -s 2 2 2
s (s + 4 ) s(s + 4) 1 - e s (s+4) s + 4
1 1 1 s 1 s 1
= --- --- - ---- -------- - [--- - --------- ] ------
4 s 4 2 4s 2 -s
s + 4 4(s + 4 ) 1 - e
1 1 1 1 1
+ --- ---- - ----- ---------- + --------------
4 2 4 2 2
s s + 4 s + 4
1 1 1 3
y(t) = --- - ---- cos2t + ----t + ----- sin2t + h(t)
4 4 4 8
1 1
where h(t) = - ----t + ----- cos2t , h(t) = h(t+1)
4 4
希望沒錯 = =
有錯請指正 感謝
(s+2)^2 -4
: 2.f(t) = L^-1 {--------------} (DONE)
: [(s+2)^2 +4]^2
: 這題我會算 我只是想問這題要怎麼拆解來算比較快@@
: 3.偏微用d代表 d^2 u d^2 u
: ----- = c^2----- ,0<x<∞ ,t>∞ ,u(0,t)=0 ,u(x,0)=0 u_t(x,0)=f(x)
: dt^2 dx^2
: 4.In Rc circuit
: 1 t
: Ri + ---∫i(t)dt = E(t) R=20Ω ,c=0.25f ,E=4(t^2 +t) ,find i(t) (DONE)
: c 0
: t
: 5.y(t) - ∫(1+x) y(t-x)dx = 1 - sinht (DONE)
: 0
: ↑
: 這個地方是迴旋積分的形式嗎?
: 謝謝
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