※ 引述《ICRO (聖文大色胚)》之銘言：
: y"+9y=tcost
: 這題我想好久
: 但想不出來
: 還有
: x^4y"+2x^3y'-y=8e^3/x
: 麻煩大家了
y"+9y=tcos(t)
yh = c1 cos(3t) + c2 sin(3t)
用參數變異法(應該是這名子吧@@)
設yp = u1(t) cos(3t) + u2(t) sin(3t) = u1 y1 + u2 y2
yp' = u1'y1 + u2'y2 + u1y1' + u2y2' => 令 u1'y1 + u2'y2 = 0 ----(1)
yp" = u1'y1' + u2'y2' + u1y1" + u2y2"
將yp , yp' , yp" 代回 y"+9y=tcos(t) 得 u1'y1'+u2'y2' = tcos(t) ----(2)
y1=cos(3t) , y2=sin(3t) , y1'=-3sin(3t) , y2'=3cos(3t)
(1) , (2) 得
u1' = -[t cos(t) sin(3t)] / 3
u2' = {t [cos(t)^2]} / 3
化簡
u1' = [-tsin(4t)]/6 + [-tsin2t]/6
u2' = t/6 + [tcos(2t)]/6
u1 = ∫{[-tsin(4t)]/6 + [-tsin2t]/6}dt
= [tcos(4t)]/24 + [sin(4t)]/96 + [tcos(2t)]/12 +[sin(2t)]/24
u2 = ∫{t/6 + [tcos(2t)]/6}dt
= (t^2)/12 + [tsin(2t)]/12 - [cos(2t)]/24
yp = u1y1+u2y2
= {[tcos(4t)]/24 + [sin(4t)]/96 + [tcos(2t)]/12 +[sin(2t)]/24} cos(3t) +
{(t^2)/12 + [tsin(2t)]/12 - [cos(2t)]/24} sin(3t)
y = yh + yp
={c1 cos(3t)+c2 sin(3t)} +
{ {[tcos(4t)]/24 + [sin(4t)]/96 +[tcos(2t)/12] + [sin(2t)]/24} cos(3t) +
{(t^2)/12 + [tsin(2t)]/12 - [cos(2t)]/24} sin(3t) }
#
只是這樣cos sin再合成化簡就不知道裡面會不會出現齊次解
不過也可能不用再化簡了XD 我是懶的繼續做下去 ~.~
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