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1、第二章 微分方程式定性法、向量平面向量、矩陣、固有值矩陣相乘 C=AB Cjk=ajm bmk ajm j=列，m=行 a11 a12A22=cjk=列 a21 a2 行A22B21=C21A43B32=C42A43B23=沒有乘積a11 a12 a13 b11 b12a21 a22 a23 b21 b22a31 a32 a33 33 b31 b32 32 a11b11+a12b21+a13b31 a11b12+a12b22+a13b32=a21b11+a22b21+a13b31 a21b12+a22b22+a23b32 a31b11+a32b21+a33b31 a31b12+a32b22+a
2、33b32 32AB=BA=A Xm1 mn n1j=ajnxm 1 a11 a12 a13 x1 2 =a21 a22 a23 x2 3 a31 a32 a33 x31=a11x1+a12x2+a13x3 2=a21x1+a22x2+a23x3 3=a31x1+a32x2+a33x3 轉矩陣(Transpose)a11 a12 a13 a11 a21 a31A=a21 a22 a23 AT=a12 a22 a32 a31 a32 a33 a13 a23 a33 1=2 21T=1 2 12 逆矩陣、反矩陣(Invense)如果AB=BA=I B=A-1 1 a11 a12 a11 a12A-
3、1=detA=detA a21 a22 a21 a22 a11 a12 a13A=a21 a22 a23 a31 a32 a33 a22 a23 a23 a21 a12 a13 a32 a33 a33 a31 a22 a23 A-1=1 a13 a12 a11 a13 a13 a11 detA a33 a32 a31 a33 a23 a21 a21 a22 a12 a11 a11 a12 a31 a32 a32 a31 a21 a22固有值、固有向量Ax=x (A-I)x=0 a11 a12 x1 0 =a21 a22-x x2 0 a11 a12det(A-I)=a21 a22-=2-(a1
4、1+a22)+a11a22-a12a21=0 特徵方程式由特徵方程式 1 2 x(1)x(2)Example1(p152)應用例介紹A-1=(1/detA)adjAadjA=B=bijbij=(-1)i+j AjiAji 為去除i行及j列之小行列式 a11 a12 a13A=a21 a22 a23 a31 a32 a33A22=(-1)2+2 a11 a13 a31 a33A32=(-1)1+3 a12 a13 a22 a23A12=(-1)2+1 a12 a13 a32 a33A32=(-1)3+2 a11 a12 a31 a32範例 2 2 0A=-2 1 1 之逆矩陣A-1 3 0 1(
5、1)A-1=(1/detA)adjA 2 2 0detA=-2 1 1 =2+6+4=12 3 0 1A-1=(1/12)1 1 -2 0 2 0 =(1/12)1-2 2 0 1 0 1 1 1 5 2 -2 -2 1 2 0 -2 0 -3 6 6 3 1 3 1 -2 1 -2 1 -2 2 2 2 =1/12 -1/6 1/6 3 0 3 0 -2 1 5/12 1/6 -1/6 -1/4 1/2 1/2(2)2-2 0 -2 1-1 =-3+42-9+12 3 0 1-A3+4A2-9A+12I=0除A -A2+4A-9I+12A-1=0A-1=1/12(A2-4A+9I)0 6 2
6、 2 2 0 1 0 0=1/12 -3-3 2 -4 -2 1 1 +9 0 1 0 9 6 1 3 0 1 0 0 1 1/12 -1/6 1/6=5/12 1/6 -1/6 ok -1/3 1/2 1/2(3)(A|I)(I|A-1)row operation methoddetA=a11 a12 a13 a14 a22 a23 a24 a21 a23 a24 a21 a22 a23 a24 =a11 a32 a33 a34 -a12 a31 a33 a34 a31 a32 a33 a34 a42 a43 a44 a41 a43 a44 a41 a42 a43 a44 a21 a22 a
7、24 a21 a22 a23 +a13 a31 a32 a34 -a14 a31 a32 a33 a41 a42 a44 a41 a42 a43 detA=a11 a12 a1n =aj1cj1+aj2cj2+ajncjn a21 a22 a2n .j=1.2.3.4.n .an1 an2 ann cjk=(-1)j+k Mjk 2 0-4-6 =2 5 1 0 -0 4 1 04 5 1 0 2 6-1 0 6-10 2 6 1 8 9 1 -3 9 1 -3 8 9 1 -4 4 5 0 -6 4 5 1 0 2-1 0 2 6 -3 8 1 -3 8 9 =1134基本觀念與理論一皆微分
8、方程式系統y1=f(t,y1,y2,yn)y2=f(t,y1,y2,yn)yn=f(t,y1,y2,yn)有解 y1=h1(t),y2=h2(t),yn=hn(t)y=h(t)向量解初始值問題 I.Cy1(t0)=k1,y2(t0)=k2,yn(t0)=kn y(t0)=k 初始值向量線性系統y1=a11(t)y1+a1n(t)yn+g1(t)y2=a21(t)y1+a2n(t)yn+g2(t).yn=an1(t)y1+ann(t)yn+gn(t)y=Ay+g 非齊次式 a11 a1n y1 y1 .y2 y2 A=.y=.g=.an1 ann yn yn 如果g=0 齊次式 y=Ay(定理)
9、如果y(1),y(2)均為齊次式之解，則 y=y(1)+y(2)應為其解。y =c1y(1)+c2y(2)=c1y(1)+c2y(2)=c1Ay(1)+c2Ay(2)=A(c1y(1)+c2y(2)=Ay龍思金y=c1y(1)+c2y(2)+cny(n)為通解y=y(1),y(n)基本矩陣dety=y1(1)y1(2)y1(n)y2(1)y2(2)y2(n).=W(y(1)y(2)y(n).yn(1)yn(2)yn(n)linearly independent y(1),y(2)y(n)為基底Example 3-1(p152)二階ODE方程式(sec 2-10)W=y1 y2 =z1(1)z1
10、(2)y1 y2 z2(1)z2(2)常係數之齊次系統方程組y=Ay homogeneous linear ODE systemA=ajk 為常數矩陣，不為t之函數y=ky y=cekty=xet 值得一式y=xet =Ay=AxetAx=x (A-I)x=0 固有值 x 固有向量由n個 值可求得n個 xy(1)=x(1)e1t ,y(2)=x(2)e2t ,y(n)=x(n)ent x1(1)e1t x1(n)ent x1(1)x1(n)x2(1)e1t x2(n)ent .W(y(1),y(2),y(n)=.=et+nt .xn(1)e1t xn(n)ent xn(1)xn(n)Gener
11、al solutiony=c1x(1)e1t+c2x(2)e2t+cnx(n)ent x(1),x(2),x(n)為 如果固有值有雙重根時即 det(A-I)=0 (-)2=0則 y(1)=xet y(2)=xtet+uet y(2)=xet+xtet+uet =Ay(2)=A(xtet+uet)又 Ax=xxet+Axtet+uet=Axtet+Auet(x+u)et=Auetdet(A-I)=0 u 另一組向量Example 6(p168)如果固有值有三重根即 det(A-I)=0 (-)3=0y(1)=xet y(2)=xtet+uety(3)=1/2xt2et+utet+ety=c1y
12、(1)+c2y(2)+c3y(3)(A-I)=u(A-I)u=uMixing problem 混合問題Tank 1：20 l 水，溶有 150 g cl(氣)Tank 1：10 l 水，溶有 50 g cl(氣)Check TANK1 Qin=3+3=6 l/min Qout=2+4=6 l/min V1=20 l TANK2 Qin=4 l/min Qout=3+1=4 l/min V2=10 l因為容量不變，而純水一直流進來稀釋原有 cl 含量，最終將達到兩槽為純水之情況。令 Yj(t)為第j槽在時間t時之cl含量(克)TANK1 y1=dy1/dt=Qin Qout =3*0+3Y2/1
13、0-2Y1/20+4Y1/20 =-(6/20)y1+(3/10)y2TANK2 y2=dy2 /dt=Qin Qout =4y1/20-3y2/10+y2/10 =4/20y1 4/20y2y=Ay y1=-6/20 3/10 y1 y2 4/20 -4/10 y2det(A-I)=-6/20 3/10 4/20 -4/10 =-1/10,-3/5x(1)=3/2 x(2)=-1 1 1y=y1 =c1 3/2 e(-1/10)t +c2 -1 e(-3/5)t y2 1 1(Example)二階齊次式 y-6y+9y=02-6+9=0 =3，3 y=c1e3t+c2te3t 如轉換為一階方
14、程組 令 y1=y，y2=y=y1 y1=y2y=y2=6y-9y=6y2+9y1 y2=-9y1+6y2 y =0 1 y -9 6 A=0 1 -9 6 det(A-I)=0 -1 -9 6-=2-6+9=0=3，3=3 (A-I)(x)=0-3 1 x1 =0-9 3 x2 0-3x1+x2=0 令 x1=1 x2=3x =1 3 y(1)=1 e3t 3重根(A-I)(u)=(x)-3 1 u1 =1-9 3 u2 3 -3u1+u2=1 令 u1=0 u2=1u =0 1 y(2)=1 te3t +0 e3t 3 1 y =y1 =c1 1 e3t+c2 t e3t y2 3 3t+
15、1 y1=y=c1 e3t+c2te3t y=3 c1 e3t+3c2te3t+c2e3t O.K.c2(3t+1)e3t y2=y=3 c1 e3t+c2(3t+1)e3t 對於-3u1+u2=1 如選 u2=0 u1=-(1/3)u =-1/3 0 y =y1 =c1 1 e3t+c2 t-(1/3)e3t y2 3 3t y1=y=c1e3t+c2(t-1/3)e3t y=3 c1e3t+3c2(t-1/3)e3t+c2e3t 3c2t e3t-c2e3t+c2e3ty2=y=3 c1e3t+3c2t e3t O.K.另外 由y1式比較 c1=c1-1/3 c2 (對e3t項)c2=c2
16、 (對t e3t項)c1=c1+(1/3)c2=c1+(1/3)c2y2=3 c1e3t+3c2t e3t=3(c1+(1/3)c2)e3t+3c2t e3t =3 c1e3t+c2(3t+1)e3t=舊y2 O.K.對於 -3u1+u2=1 如果 u1=-1 u2=-2 -1(u)=-2y=y1 =c1e3t+c2(t-1)e3ty=3c1e3t+3c2(t-1)e3t+c2e3ty2=y=3c1e3t+c2(3t-2)e3t係數比較 由y1式 c1=c1 c2 (對e3t項)c2=c2 (對te3t項)c1=c1+c2由於 y2=3c1e3t+c2(3t-2)e3t =3(c1+c2)e3
17、t+c2(3t-2)e3t =3c1e3t+3c2te3t+c2e3t =3c1e3t+c2(3t+1)e3t =舊y2 O.K範例y=Ay+gA=1 0 g=et 0 -1 tdet(A-I)=0 1-0 =0 6 -1-=1,-1=1 x=1 =-1 x=0 3 1yh=c1 1 et+c2 0 e-t 3 1(1)未定係數法(過程省略)yp=1 tet+0 et+0 t+0 3 -3/2 1 -1 y=yh+yp =c1 1 et+c2 0 e-t+1 tet+0 et+0 t+0 3 1 3 -3/2 1 -1(2)參數變異法(過程省略)yp=1 tet+0 et+0 t+0 3 -3
18、/2 1 -1 y=yh+yp =c1 1 et+c2 0 e-t+1 tet+0 et+0 t+0 3 1 3 -3/2 1 -1 c2=c2 +5/2(3)對角線化解法(過程省略)y=yh+yp =c1 1 et+c2 0 e-t+1 tet+0 et+0 t+0 3 1 3 -3/2 1 -1 c2=c2 +5/2範例y=Ay+g其中 A=1 0 1 g=e-t 1 1 0 0 -2 0 -1 0y=？變換係數Exampley+(ex-1)y+e2xy=0 非線性齊次 求通解令 t=ex y=(dy/dx)=(dy/dt)(dt/dx)=Yex y=dy/dx=d(Yex)/dx=Ye2
19、x+Yex上式 Ye2x+Ye2x+(ex-1)Yex+e2x Y=0 Ye2x+Yex+Ye2x-Yex+e2xY=0 (Y+Y+Y)e2x=0 Y+Y+Y=0 2+1=0 =(-1+3i)/2,(-1-3i)/2 Y=e(-1/2)tAcos(3/2)t+Bsin(3/2)t t=ex Y=e(-1/2)exAcos(3/2)ex+Bsin(3/2)ex 未定係數法y+py+gy=r(x)之特解 yp 基本規則之增列Exampley+4y+4y=7x-3cos(2x)+5xe-2x2+4+4=0 (+2)2=0 =-2,-2yh=c1e-2x+c2xe-2xyp 之預測(1)7x Kx+I
20、 (2)3cos(2x)Mcos(2x)+Nsin(2x)(3)5xe-2x x2(Px+Q)e-2xyp=Kx+I+Mcos(2x)+Nsin(2x)+x2(Px+Q)e-2xyp=K2Msin2x+2Ncos2x 2e-2x(Px3+Qx2)+(3Px2+2Qx)e-2xyp=-4Mcos2x4Nsin2x+4e-2x(Px3+Qx2)2e-2x(3Px2+2Qx)+(6Px+2Q)e-2x 2e-2x(3Px2+2Qx)代入 y p+4yp+4yp=0比較 cos2x -4M+8N+4M=-3 N=-3/8 sin2x -4N 8M+4N=0 M=0 x 4K=7 K=7/4 const
21、ant 4K+4I=0 I=-7/4 e-2x 2Q=0 Q=0 xe-2x -4Q+6P 4Q+8Q=5 P=5/6 x2e-2x -6P 6P+4Q+12P 8Q+4Q=0 OK x3e-2x 4P 8P+4p=0 OKyp=7/4(x-1)3/8sin2x +5/6x3e-2xy=(c1+c2x)e-2x+7/4(x-1)3/8sin2x+5/6x3e-2x變數變換法Exampley-3y+2y=cos(e-x)求 yh=2-3+2=0 =1,2yh=c1ex+c2e2xW=ex e2x ex 2e2xu=-(y2r/W)dx=-(e2xcos(e-x)/e3x)dx =-e-xcose-x dx=cos(e-x)d(e-x)=sin(e-x)=(y1r/W)dx=(excos(e-x)/e3x)dx =e-2xcos(ex)dx=-(e-x)cos(e-x)d(e-x)=-(cos(e-x)+e-xsin(e-x)yp=uy1+y2=exsin(e-x)+e2x(-cos(e-x)-e-xsin(e-x)=-e2xcos(e-x)y=c1ex+c2e2x e2xcos(e-x)彈簧系統之平衡