國立中正大學111學年度碩士班招生考試
科目:工程數學
系所組別: 機械工程學系乙組
解答: (a)d3ydx3−y=0⇒λ3−1=0⇒(λ−1)(λ2+λ+1)=0⇒λ=1,−1±√3i2⇒yh=c1ex+e−x/2(c2cos√32x+c3sin√32x)(b)yp=Ae−x⇒y′p=−Ae−x⇒y″p=Ae−x⇒y‴p=−Ae−x⇒y‴p−yp=−2Ae−x=3e−x⇒A=−32⇒yp=−32e−x⇒y=yh+yp⇒y=c1ex+e−x/2(c2cos√32x+c3sin√32x)−32e−x⇒y′=c1ex+e−x/2[(−12c2+√32c3)cos√32x+(−√32c2−12c2)sin√32x]+32e−x⇒y″=c1ex+e−x/2[(−12c2−√32c3)cos√32x+(√32c2−12c3)sin√32x]−32e−xinitial conditions:{y(0)=0y′(0)=1y″(0)=0⇒{c1+c2−3/2=0c1−c2/2+√3c3/2+3/2=1c1−c2/2−√3c3/2−3/2=0⇒{c1=5/6c2=2/3c3=−2√3/3⇒y=56ex+e−x/2(23cos√32x−2√33sin√32x)−32e−x解答: (a)L{y″}−L{y}=L{e−t}⇒s2Y(s)−3s−Y(s)=1s+1⇒Y(s)=1(s+1)(s2−1)+3ss2−1=3s2+3s+1(s+1)(s2−1)=74(s−1)+54(s+1)−12(s+1)2⇒y(t)=L−1{Y(s)}=74L−1{1s−1}+54L−1{1s+1}−12L−1{1(s+1)2}⇒y(t)=74et+54e−t−12te−t(b)L{y′}+2L{y}=L{2(u(t)−u(t−2))}⇒sY(s)+2Y(s)=2s(1−e−2s)⇒Y(s)=2s2+2s−2e−2ss2+2s=1s−1s+2−(1s−1s+2)e−2s⇒y(t)=L−1{Y(s)}⇒y(t)=u(t)−e−2t−u(t−2)(1−e−2(t−2))
解答: (a)f(s)=|2−s−1−10−13−s−1−1−1−13−s−10−1−12−s|R1−R4→R1,R2−R4→R2,R3−R4→R3→|2−s00s−2−14−s0s−3−104−ss−30−1−12−s|⇒f(s)=(2−s)|4−s0s−304−ss−3−1−12−s|−(s−2)|−14−s0−104−s0−1−1|=(s−4)2(s−2)2+2(s−2)(s−3)(s−4)−2(s−2)(s−4)=s4−10s3+32s2−32s,QED(b)f(s)=(s−4)2(s−2)2+2(s−2)(s−3)(s−4)−2(s−2)(s−4)=(s−2)(s−4)[(s−2)(s−4)+2(s−3)−2]=(s−2)(s−4)(s2−4s)=s(s−2)(s−4)2⇒If f(s)=0, then s=0,2,4⇒eigenvalues of M:0,2,4(c)det(M)=0⇒M is NOT invertible(d)(M−4I)v=0⇒[−2−1−10−1−1−1−1−1−1−1−10−1−1−2][x1x2x3x4]=0⇒{x1−x4=0x2+x3+2x4=0⇒v=x3[0−110]+x4[1−201]⇒eigenvectors: [0−110],[1−201](e)M=[1−10110−1−210101101][0000020000400004][14141414−120012−14−1434−1414−14−1414]⇒Yes, M can be diagonalized.
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解題僅供參考,其他歷年試題及詳解
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