solvethe-system of linear d.e

2013-03-20 6:42 pm
Q.1

(D-3)x -6y =0
3x+(D+3)y=18e^ (-3t)

Q.2

(D-3)x = 2cos3t
3x+(D+3)y=2sin3t

回答 (1)

2013-03-20 9:48 pm
✔ 最佳答案
Q.1(D-3)x-6y=0 => dx/dt=3x+6y3x+(D+3)y=18e^(-3t) => dy/dt=-3x-3y+18e^(-3t)合併成矩陣微分方程式:{x'}=[.3 .6]*{x}+{0........}
{y'}.[-3 -3].{y}.{18e^(-3t)}Y'=A*Y+B => Y'-AY=B積分因子: ln(F)=-∫Adt=-At => F=e^(-At)Y(t)=e^(At)[∫e^(-At)*Bdt+c]e^(-At)*B=[e^-3t..e^-6t]*{0........}
.[e^3t ..e^3t.].{18e^(-3t)}={e^-9t}*18
.{1....}Y=e^(At)[∫{18e^(-9t),18}^T*dt+c]......^T=倒置矩陣=e^(At)[{-2e^(-9t),18t}^T+c]=[e^3t ..e^6t]*{-2e^(-9t)}+e^(At)*c
.[e^-3t.e^-3t].{18t......}={-2e^(-6t)+18te^(6t)..}+[e^3t .e^6t.]*{c1}
.{-2e^(-12t)+18te^(-3t)}.[e^-3t e^-3t].{c2}{x}={-2e^(-6t)+18te^(6t)..}+{c1*e^3t+c2*e^6t.}
{y}.{-2e^(-12t)+18te^(-3t)}.{(c1+c2)*e^(-3t).}x(t)={-2*e^(-6t)+18t*e^(6t)}+{c1*e^(3t)+c2*e^(6t)}........ansy(t)={-2*e^(-12t)+18t*e^(-3t)}+{(c1+c2)*e^(-3t)}..........ans
Q.2(D-3)x=2cos3t => x'-3x=2*cos3t3x+(D+3)y=2sin3t => y'+3x+3y=2sin3tLet Y={x y}^T, B={2cos3t 2sin3t}^T矩陣微分方程式: Y'+AY=BA=[-3 0]
..[.3 3] => F=e^(At)Y=e^(-At)[∫e^(At)Bdt+c]=2e^(-At)[∫{e^(-3t)cos3t+sin3t,e^3tcos3t+e^3tsin3t}^Tdt+c]=e^(At)[{e^(-3t)(sin3t-cos3t)-2cos3t}+c]/3
.......[{2e^(3t)sin3t...............}+c]/3={-e^(-3t)(sin3t-cos3t)-2cos3t............}/3+[e^(-3t) 0...]{c1}
.{e^(-3t)(sin3t-cos3t)-2cos3t+2e^(3t)sin3t}/3.[e^3t ...e^3t]{c2}x(t)=-[e^(-3t)(sin3t-cos3t)-2cos3t]/3+c1*e^(-3t)y(t)=[e^(-3t)(sin3t-cos3t)-2cos3t+2e^(3t)sin3t]/3+(c1+c2)e^3t計算繁雜.請自己再驗算一次.看看是否有錯!!


2013-03-20 17:55:07 補充:
第1題修正: 使用比較簡單的方法 => 個別運算

Let c=cos3t, s=sin3t

x'=3x+6y................a

y'=-3x-3y+18e^(-3t).....b

a+b: x'+y'=3y+18e^(-3t).....c

b': y"+3x'+3y'=-54e(-3t)....d

3c-d: y"+9y=-108e^(-3t)

D^2+9=0 => D=+-3j

齊次解: yh=k1*c+k2*s

2013-03-20 17:56:04 補充:
特別解: yp=g*e(-3t)

yp'=-3g*e(-3t)

yp"=9g*e^(-3t)

yp"+9yp=(9g+9g)e^(-3t)

=18g*e^(-3t)

=-108e^(-3t)

g=-108/18=-6

yp=-6e^(-3t)

通解: y=k1*c+k2*s-6e^(-3t)

2013-03-20 17:56:33 補充:
Eq.a: x'-3x=6y=6k1*c+6k2*s-36e^(-3t)

積分因子: F=e^(-3t)

xe^(-3t)=∫6(k1*c+k2*s)e^(-3t)dt-∫36e^(-6t)dt+k3

=6e^(-3t)[3k1(s-c)-3k2(s+c)]/18+6e^(-6t)+k3

x=[k1(s-c)-k2(s+c)]+6e^(-3t)+k3*e^(3t)........ans

2013-03-20 17:57:39 補充:
請繼續參考意見欄

2013-03-20 17:57:57 補充:
第2題修正: 使用比較簡單的方法 => 個別運算

Let c=cos3t, s=sin3t

x'-3x=2c

積分因子: F=e^(-3t)

x*e^(-3t)=2∫e^(-3t)*c*dt+k

=2e^(-3t)(3s-3c)/18+k

=e^(-3t)(s-c)/3+k

x=(s-c)/3+ke^(3t)........ans

2013-03-20 17:58:45 補充:
代入第2式:

y'+3y=2s-3x

=2s+(c-s)-3ke^(3t)

=c+s-3k^(3t)

積分因子: F=e^(3t)

ye^(3t)=∫e^(3t)(c+s)dt-∫3ke^(6t)dt+k1

=e^(3t)(3s+3c+3s-3c)-ke^(6t)/2+k1

y=6s-ke^(3t)+k1*e^(-3t).........ans


收錄日期: 2021-04-11 19:43:22
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