國立臺北科技大學108學年度碩士班招生考試
系所組別:2131 電機工程系碩士班丙組
第一節 工程數學 試題(選考)
解答:{P(x,y)=4xy2−2Q(x,y)=y3⇒{Px=4y2Qy=3y2⇒Px≠Qy⇒NonExact⇒u′=−Qy−PxQu=1yu⇒(1yu)′=0⇒1yu=c1⇒積分因子u=y⇒{uP=4xy3−2yuQ=y4⇒(uP)x=4y3=(uQ)y⇒Exact⇒Φ(x,y)=∫(4xy3−2y)dy=∫y4dx⇒xy4−y2+ρ(x)=xy4+ϕ(y)⇒Φ(x,y)=xy4−y2+c1=0解答:y″+4y′+4y=0⇒λ2+4λ+4=0⇒(λ+2)2=0⇒λ=−2⇒yh=e−2x(c1+c2x)Let {y1=e−2xy2=xe−2x, then{y′1=−2e−2xy′2=e−2x−2xe−2x⇒W=|y1y2y′1y′2|=|e−2xxe−2x−2e−2xe−2x−2xe−2x|=e−4xBy variations of parameters, yp=−e−2x∫xe−2x⋅e−2xlnxe−4xdx+xe−2x∫e−2x⋅e−2xlnxe−4xdx⇒yp=−e−2x∫xlnxdx+xe−2x∫lnxdx=−e−2x(12x2lnx−14x2)+xe−2x(xlnx−x)⇒yp=12x2e−2xlnx−34x2e−2x⇒y=yh+yp⇒y=e−2x(c1+c2x)+12x2e−2xlnx−34x2e−2x
解答:[dx1(t)dxdx2(t)dx]=[σω−ωσ][x1(t)x2(t)]⇒{x′1(t)=σx1(t)+ωx2(t)x′2(t)=−ωx1(t)+σx2(t)⇒{L{x′1(t)}=L{σx1(t)+ωx2(t)}L{x′2(t)}=L{−ωx1(t)+σx2(t)}⇒{sX1(s)−X1(0)=σX1(s)+ωX2(s)sX2(s)−X2(0)=−ωX1(s)+σX2(s)⇒{(s−σ)X1(s)=ωX2(s)+A(s−σ)X2(s)=−ωX1(s)+B⇒X1(s)=Bω−s−σωX2(s)=Bω−s−σω(s−σωX1(s)−Aω)⇒(1+(s−σ)2ω2)X1(s)=Bω+A(s−σ)ω2⇒X1(s)=1ω2+(s−σ)2(Bω+A(s−σ))⇒x1(t)=L−1{X1(s)}=Beσtsin(ωt)+Aeσtcos(ωt)⇒x′1(t)=(Aσ+Bω)eσtcos(ωt)+(Bσ−Aω)eσtsin(ωt)⇒x2(t)=1ω(x′1(t)−σx1(t))=Beσtcos(ωt)−Aeσtsin(ωt)⇒{x1(t)=eσt(Acos(ωt)+Bsin(ωt))x2(t)=eσt(Bcos(ωt)−Asin(ωt))
解答:(a) Let A=[000000000], then −AT=A. That is, A is a 3x3 skew-symmetrix matrix.Suppose A∈S, then A=−AT⇒cA=−(cA)T⇒cA∈S.Suppose A,B∈S, then A=−AT and B=−BT⇒A+B=−AT−BT=−(A+B)T⇒A+B∈S.Therefore, V is a subspace of V.QED.(b) A∈S⇒A=[0a1,2a1,3−a1,20a2,3−a1,3−a2,30]=a1,2[010−100000]+a1,3[001000−100]+a2,3[0000010−10], then {[010−100000],[001000−100],[0000010−10]} is a basis
解答:(a) A=[2−4404−10−14]⇒det(A−λI)=−λ3+10λ2−31λ+30=−(λ−2)(λ−3)(λ−5)=0⇒eigenvalues: 2,3,5(b) λ1=2⇒(A−λ1I)v=0⇒[0−4402−10−12][x1x2x3]=0⇒{x2=0x3=0⇒v=x1(100),choose v1=(100)λ2=3⇒(A−λ2I)v=0⇒[−1−4401−10−11][x1x2x3]=0⇒{x1=0x2=x3⇒v=x3(011),choose v2=(011)λ3=5⇒(A−λ3I)v=0⇒[−3−440−1−10−1−1][x1x2x3]=0⇒{3x1=8x3x2+x3=0⇒v=x3(8/3−11),choose v3=(8/3−11)⇒P=[v1v2v3]⇒P=[108301−1011]
解答:{4x1−2x2+3x3=03x1−4x2+2x3=06x1+2x2+5x3=0⇒[4−233−42625][x1x2x3]=0Let A=[4−233−42625], then rref(A)=[104501110000]⇒{x1+45x3=0x2+110x3=0⇒solutions of Ax=0:{k(−45,−110,1)∣k∈R}⇒{ a basis: {(8,1,−10)}nullity(A)=1
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解題僅供參考,碩士班歷年試題及詳解
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