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2025年1月11日 星期六

108年北科大電機碩士班丙組-工程數學詳解

 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)=4xy22Q(x,y)=y3{Px=4y2Qy=3y2PxQyNonExactu=QyPxQu=1yu(1yu)=01yu=c1u=y{uP=4xy32yuQ=y4(uP)x=4y3=(uQ)yExactΦ(x,y)=(4xy32y)dy=y4dxxy4y2+ρ(x)=xy4+ϕ(y)Φ(x,y)=xy4y2+c1=0
解答:y+4y+4y=0λ2+4λ+4=0(λ+2)2=0λ=2yh=e2x(c1+c2x)Let {y1=e2xy2=xe2x, then{y1=2e2xy2=e2x2xe2xW=|y1y2y1y2|=|e2xxe2x2e2xe2x2xe2x|=e4xBy variations of parameters, yp=e2xxe2xe2xlnxe4xdx+xe2xe2xe2xlnxe4xdxyp=e2xxlnxdx+xe2xlnxdx=e2x(12x2lnx14x2)+xe2x(xlnxx)yp=12x2e2xlnx34x2e2xy=yh+ypy=e2x(c1+c2x)+12x2e2xlnx34x2e2x
解答:[dx1(t)dxdx2(t)dx]=[σωωσ][x1(t)x2(t)]{x1(t)=σx1(t)+ωx2(t)x2(t)=ωx1(t)+σx2(t){L{x1(t)}=L{σx1(t)+ωx2(t)}L{x2(t)}=L{ωx1(t)+σx2(t)}{sX1(s)X1(0)=σX1(s)+ωX2(s)sX2(s)X2(0)=ωX1(s)+σX2(s){(sσ)X1(s)=ωX2(s)+A(sσ)X2(s)=ωX1(s)+BX1(s)=BωsσωX2(s)=Bωsσω(sσωX1(s)Aω)(1+(sσ)2ω2)X1(s)=Bω+A(sσ)ω2X1(s)=1ω2+(sσ)2(Bω+A(sσ))x1(t)=L1{X1(s)}=Beσtsin(ωt)+Aeσtcos(ωt)x1(t)=(Aσ+Bω)eσtcos(ωt)+(BσAω)eσtsin(ωt)x2(t)=1ω(x1(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 AS, then A=ATcA=(cA)TcAS.Suppose A,BS, then A=AT and B=BTA+B=ATBT=(A+B)TA+BS.Therefore, V is a subspace of V.QED.(b) ASA=[0a1,2a1,3a1,20a2,3a1,3a2,30]=a1,2[010100000]+a1,3[001000100]+a2,3[000001010], then {[010100000],[001000100],[000001010]} is a basis 
解答:(a) A=[244041014]det(AλI)=λ3+10λ231λ+30=(λ2)(λ3)(λ5)=0eigenvalues: 2,3,5(b) λ1=2(Aλ1I)v=0[044021012][x1x2x3]=0{x2=0x3=0v=x1(100),choose v1=(100)λ2=3(Aλ2I)v=0[144011011][x1x2x3]=0{x1=0x2=x3v=x3(011),choose v2=(011)λ3=5(Aλ3I)v=0[344011011][x1x2x3]=0{3x1=8x3x2+x3=0v=x3(8/311),choose v3=(8/311)P=[v1v2v3]P=[1083011011]
解答:{4x12x2+3x3=03x14x2+2x3=06x1+2x2+5x3=0[423342625][x1x2x3]=0Let A=[423342625], then rref(A)=[104501110000]{x1+45x3=0x2+110x3=0solutions of Ax=0:{k(45,110,1)kR}{ 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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