
Re: [理工] [拉式] 98南交工數

: 第18題,我原本想要直接分式,但發現很難做
: 然後我讓它等於2式相乘、使用摺積定理,但逆轉換後的摺積我不會做
: 請問有人知道這題該怎麼處理嗎?
: 感謝~m(_ _)m~
(D^4+2D^2+1)y(t) = 4t*e^t , y(0)=y'(0)=y''(0)=y'''(0)=0
=>(D^2 + 1)^2 y(t) = 4*t*e^t
yh=c1*cost + c2*sint + c3*tsint + c4*tcost
1
yp=4*e^t* ----------------- t
((D+1)^2 + 1)^2
1
=4*e^t* -------- t = e^t*(1-2D)t
8D + 4
=(t-2)*e^t
=>
y(t) = c1*cost + c2*sint + c3*tsint + c4*tcost + (t-2)*e^t
由 y(0)=0
=> c1=2
y'(0)=0
=>y'(t)= -2sint + c2*cost + c3*(sint+tcost) + c4(cost-tsint) + (t-1)e^t
=> 0 = c2 +c4 -1
=> c2 + c4 = 1
y''(t)=-2cost - c2*sint +c3*(2cost - tsint) + c4(-2sint-tcost) +te^t
=> 0 = -2 + 2*c3
=> c3=1
y'''(t)=2sint -c2*cost + c3*(-3sint - tcost) +c4*(-3cost+tsint)+(t+1)e^t
0= -c2 + -3*c4 + 1
=>c1=2,c2=1,c3=1,c4=0
y(t) = c1*cost + c2*sint + c3*tsint + c4*tcost + (t-2)*e^t
= 2*cost + sint + tsint +(t-2)*e^t
--
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