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2023年12月28日 星期四

111年雲林科大電機系-工程數學詳解

 國立雲林科技大學111學年度碩士班招生考試

系所:電機系
科目:工程數學

解答:1.y=5sin(3x)y=5sin(3x)dx=53cos(3x)+c1y=53cos(3x)+c12.I(x)=e5x4dx=ex5yI(x)+5x4yI(x)=3x2ex5yex5+5x4yex5=3x2(yex5)=3x2yex5=x3+c1y=x3ex5+c1ex53.x2y=y2+xyyyx=y2x2v=1yv=yy2y=y2vy2vyx=y2x2v+1xy=1x2v+vx=1x2xv+v=1x(xv)=1xxv=lnx+c1xy=lnx+c1y=xlnx+c1
解答:,y+9y=0λ2+9=0λ=±3iyh=c1cos(3x)+c2sin(3x)yp=Acosx+Bsinxyp=Asinx+Bcosxyp=AcosxBsinxyp+9yp=8Acosx+8Bsinx=16sinx{A=0B=2yp=2sinxy=yh+yp=c1cos(3x)+c2sin(3x)+2sinxy=3c1sin(3x)+3c2cos(3x)+2cosx{y(π)=c1=0y(0)=3c2+2=0{c1=0c2=23y=23sin(3x)+2sinx
解答:(1)F(s)=L{t2}+L{e6t}+L{sin(6t)}=2s3+1s+6+6s2+36(2)f(t)=L1{8s(s2)2}=L1{2s2s2+4(s2)2}=22e2t+4e2tt(3)L{y}+L{t0y(τ)dτ}=2L{1}2L{sint}sY(s)+Y(s)=2s2s2+1Y(s)=2s(s+1)2(s2+1)(s+1)y(t)=L1{2s(s+1)}L1{2(s2+1)(s+1)}=2L1{1s1s+1}L1{1s2+1ss2+1+1s+1}=22et(sintcost+et)y(t)=23etsint+cost


解答:y3y=te3tsin(t)L{y}3L{y}=L{te3tsin(t)}sY(s)3Y(s)=ddsL{e3tsin(t)}=dds1(s3)2+1=2s6(s26s+10)2Y(s)=2(s26s+10)2y(t)=L1{2(s26s+10)2}=L1{2((s3)2+1)2}=e3tL1{2(s2+1)2}=e3t(sinttcost)y(t)=e3t(sinttcost)
解答:(a)A=[003021201]det(AλI)=(λ+2)(λ2)(λ3)=0λ=2,2,3λ1=2(Aλ1I)v=0[203041203][x1x2x3]=0{2x1+3x3=04x2+x3=0v=[3k/2k/4k],kR,v1=[3/2/41]λ2=2(Aλ2I)v=0[203001201][x1x2x3]=0{x1=0x3=0v=[0k0],kR,v2=[010]λ3=3(Aλ3I)v=0[303011202][x1x2x3]=0{x1=x3x2=x3v=[kkk],kR,v3=[111]eigenvalues: 2,2,3 and corresponding eigenvectors are [3/2/41],[010],[111](b)X=[v1v2v3],D=[λ1000λ2000λ3]A=[3/2011/411101][200020003][3/2011/411101]1A5=[3/2011/411101][(2)50002500035][3/2011/411101]1=[3/2011/411101][3200032000243][3/2011/411101]1A5=[78016578321331100133]


解答:(a)A=[434123600]rref(A)=[100010001]linearly independent(b)[6786]+2[4641]=[219164]linearly dependent

解答:A=[142101112842]rref(A)=[102501110000]Rank(A)=2[102501110000][x1x2x3x4]=0{x12x3+5x4=0x2+x3x4=0[x1x2x3x4]=[2x35x4x3+x4x3x4]=x3[2110]+x4[5101]Null(A)=Span{[2110],[5101]},Nulity(A)=2

解答:A=[101778121776]rref(A)=[100010001000]a basis for column space of A={a1,a2,a3},where a1=[1717],a2=[0727],a3=[1816]b1=a1e1=b1b1=[1/107/101/107/10]b2=a2(a2e1)e1=[0727][1717]=[1010]e2=b2b2=[1/201/20]b3=a3(a3e1)e1(a3e2)e2=[1816][1717]0=[0101]e3=b3b3=[01/201/2]an orthonomal basis for the column space of A={[1/107/101/107/10],[1/201/20],[01/201/2]}

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解題僅供參考,其他歷年試題及詳解

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