臺灣綜合大學系統108學年度學士班轉學生聯合招生考試
科目名稱:工程數學
類組代碼:D04
(a)
(b)f(x)={x+π,−π<x<0−x+π,0<x<π⇒f(x)為偶函數⇒bn=0,只需計算an;a0=12π∫π−πf(x)dx=12π(∫0−πx+πdx+∫π0−x+πdx)=12π×π2=π2an=1π∫π−πf(x)cos(nx)dx=1π(∫0−π(x+π)cos(nx),dx+∫π0(−x+π)cos(nx)dx)=1π([xnsin(nx)+1n2cos(nx)+πnsin(nx)]|0−π+[−xnsin(nx)−1n2cos(nx)+πnsin(nx)]|π0)=1π⋅2n2(1−(−1)n)={4n2π,n是奇數0,n是偶數⇒f(x)=a0+∞∑n=1ancos(nx)⇒f(x)=π2+2π∞∑n=11n2(1−(−1)n)cos(nx)
解答:先求齊次解,即y″+ω2y=0⇒特徵方程式:λ2+ω2=0⇒λ=±ωi⇒yh=Acos(ωt)+Bsin(ωt)由於r(t)=sint,因此需考慮ω是否等於1;由題意知ω=0.5,1.2,2,10,皆不等於1因此yp=Csint⇒y″p=−Csint⇒−Csint+ω2Csint=sint⇒C(ω2−1)=1⇒C=1/(ω2−1)⇒y=yh+yp⇒y=Acos(ωt)+Bsin(ωt)+1ω2−1sint,A與B皆為常數
解答:y′=2xy⇒∫ydy=∫2xdx⇒12y2=x2+Cy(1)=5⇒252=1+C⇒C=232⇒y2=2x2+23
解答:f=2exyz⇒{fx=2yzexyzfy=2xzexyzfz=2xyexyz⇒{fxx=2y2z2exyzfyy=2x2z2exyzfzz=2x2y2exyz⇒∇2f=fxx+fyy+fzz=2exyz(x2y2+y2z2+z2x2)
解答:假設{→u=[abc]→v=[def]⇒→u×→v=[bf−cecd−afae−bd]⇒(→u×→v)′=[(bf−ce)′(cd−af)′(ae−bd)′]=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(1)而{→u′×→v=[a′b′c′]×[def]=[b′f−c′ec′d−a′fa′e−b′d]→u×→v′=[abc]×[d′e′f′]=[bf′−ce′cd′−af′ae′−bd′]⇒→u′×→v+→u×→v′=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(2)(1)=(2)⇒→u×→v=→u′×→v+→u×→v′,故得證
解答:A=[−1−10−1−10001]⇒det(A−λI)=0⇒−λ(λ+2)(λ−1)=0⇒特徵值{λ1=0λ2=−2λ3=1λ1=0⇒(A−λ1I)x=[−1−10−1−10001][x1x2x3]=0⇒{x1+x2=0x3=0,取v1=[−110]λ2=−2⇒(A−λ1I)x=[1−10−110003][x1x2x3]=0⇒{x1=x2x3=0,取v2=[110]λ3=1⇒(A−λ1I)x=[−2−10−1−20000][x1x2x3]=0⇒x1=x2=0,取v3=[001]⇒P=[v1v2v3]=[−110110001]⇒P−1=[−1212012120001]⇒A=P[λ1000λ2000λ3]P−1=[−110110001][0000−20001][−1212012120001]
解答:{y′1=y2+2−u(t−1)y′2=−y1+1−u(t−1)⇒[y′1y′2]=[01−10][y1y2]+[2−u(t−1)1−u(t−1)]≡y′=Ay+bA=[01−10]⇒det(A−λI)=λ2+1=0⇒λ=±iλ1=i⇒(A−λ1I)x=[−i1−1−i][x1x2]=0⇒x1=−ix2,取v1=[−i1]λ2=−i⇒(A−λ2I)x=[i1−1i][x1x2]=0⇒x1=ix2,取v2=[i1]因此y′=Ay的齊次解為⇒[y1y2]=C1eit[−i1]+C2e−it[i1]⇒{y1=−C1ieit+C2ie−ity2=C1eit+C2e−it⇒{y′1=C1eit+C2e−it+2−u(t−1)y′2=C1ieit−C2ie−it+1−u(t−1)⇒{L{y′1}=L{C1eit+C2e−it+2−u(t−1)}L{y′2}=L{C1ieit−C2ie−it+1−u(t−1)}⇒{sY1(s)−y1(0)=C1s−i+C2s+i+2s−e−s/ssY2(s)−y2(0)=C1is−i−C2is+i+1s−e−s/s⇒{Y1(s)=C1s(s−i)+C2s(s+i)+2s2+2s−e−s/s2Y2(s)=C1is(s−i)−C2is(s+i)+1s2+1s−e−s/s2⇒{Y1(s)=C1i(1s−i−1s)+C2i(1s−1s+i)+2s2+2s−e−s/s2Y2(s)=C1(1s−i−1s)−C2(1s−1s+i)+1s2+1s−e−s/s2⇒{y1=C1i(eit−1)+C2i(1−e−it)+2t+2−(t−1)u(t−1)y2=C1(eit−1)−C2(1−e−it)+t+1−(t−1)u(t−1)
解答:A=[103111010]⇒rref(A)=[100010001]⇒Rank(A)=3⇒線性獨立
解答:(2,0,a1)⋅(1,0,8)=2+8a1=0⇒a1=−14
解答:{3x+y+2z=1x−y+3z=−3y−2z=1⇒[3121−1301−2][xyz]=[1−31]≡Ax=b⇒{Ax=[112−3−1311−2]Ay=[3121−3301−2]Az=[3111−1−3011]⇒{det(Ax)=−8det(Ay)=13det(Az)=6det(A)=1⇒{x=det(Ax)/det(A)=−8y=det(Ay)/det(A)=13z=det(Az)/det(A)=6⇒{x=−8y=13z=6
解答:y′=2xy⇒∫ydy=∫2xdx⇒12y2=x2+Cy(1)=5⇒252=1+C⇒C=232⇒y2=2x2+23
解答:f=2exyz⇒{fx=2yzexyzfy=2xzexyzfz=2xyexyz⇒{fxx=2y2z2exyzfyy=2x2z2exyzfzz=2x2y2exyz⇒∇2f=fxx+fyy+fzz=2exyz(x2y2+y2z2+z2x2)
解答:假設{→u=[abc]→v=[def]⇒→u×→v=[bf−cecd−afae−bd]⇒(→u×→v)′=[(bf−ce)′(cd−af)′(ae−bd)′]=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(1)而{→u′×→v=[a′b′c′]×[def]=[b′f−c′ec′d−a′fa′e−b′d]→u×→v′=[abc]×[d′e′f′]=[bf′−ce′cd′−af′ae′−bd′]⇒→u′×→v+→u×→v′=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(2)(1)=(2)⇒→u×→v=→u′×→v+→u×→v′,故得證
解答:A=[−1−10−1−10001]⇒det(A−λI)=0⇒−λ(λ+2)(λ−1)=0⇒特徵值{λ1=0λ2=−2λ3=1λ1=0⇒(A−λ1I)x=[−1−10−1−10001][x1x2x3]=0⇒{x1+x2=0x3=0,取v1=[−110]λ2=−2⇒(A−λ1I)x=[1−10−110003][x1x2x3]=0⇒{x1=x2x3=0,取v2=[110]λ3=1⇒(A−λ1I)x=[−2−10−1−20000][x1x2x3]=0⇒x1=x2=0,取v3=[001]⇒P=[v1v2v3]=[−110110001]⇒P−1=[−1212012120001]⇒A=P[λ1000λ2000λ3]P−1=[−110110001][0000−20001][−1212012120001]
解答:{y′1=y2+2−u(t−1)y′2=−y1+1−u(t−1)⇒[y′1y′2]=[01−10][y1y2]+[2−u(t−1)1−u(t−1)]≡y′=Ay+bA=[01−10]⇒det(A−λI)=λ2+1=0⇒λ=±iλ1=i⇒(A−λ1I)x=[−i1−1−i][x1x2]=0⇒x1=−ix2,取v1=[−i1]λ2=−i⇒(A−λ2I)x=[i1−1i][x1x2]=0⇒x1=ix2,取v2=[i1]因此y′=Ay的齊次解為⇒[y1y2]=C1eit[−i1]+C2e−it[i1]⇒{y1=−C1ieit+C2ie−ity2=C1eit+C2e−it⇒{y′1=C1eit+C2e−it+2−u(t−1)y′2=C1ieit−C2ie−it+1−u(t−1)⇒{L{y′1}=L{C1eit+C2e−it+2−u(t−1)}L{y′2}=L{C1ieit−C2ie−it+1−u(t−1)}⇒{sY1(s)−y1(0)=C1s−i+C2s+i+2s−e−s/ssY2(s)−y2(0)=C1is−i−C2is+i+1s−e−s/s⇒{Y1(s)=C1s(s−i)+C2s(s+i)+2s2+2s−e−s/s2Y2(s)=C1is(s−i)−C2is(s+i)+1s2+1s−e−s/s2⇒{Y1(s)=C1i(1s−i−1s)+C2i(1s−1s+i)+2s2+2s−e−s/s2Y2(s)=C1(1s−i−1s)−C2(1s−1s+i)+1s2+1s−e−s/s2⇒{y1=C1i(eit−1)+C2i(1−e−it)+2t+2−(t−1)u(t−1)y2=C1(eit−1)−C2(1−e−it)+t+1−(t−1)u(t−1)
解答:A=[103111010]⇒rref(A)=[100010001]⇒Rank(A)=3⇒線性獨立
解答:(2,0,a1)⋅(1,0,8)=2+8a1=0⇒a1=−14
解答:{3x+y+2z=1x−y+3z=−3y−2z=1⇒[3121−1301−2][xyz]=[1−31]≡Ax=b⇒{Ax=[112−3−1311−2]Ay=[3121−3301−2]Az=[3111−1−3011]⇒{det(Ax)=−8det(Ay)=13det(Az)=6det(A)=1⇒{x=det(Ax)/det(A)=−8y=det(Ay)/det(A)=13z=det(Az)/det(A)=6⇒{x=−8y=13z=6
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