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)+c1⇒y=−53cos(3x)+c12.積分因子I(x)=e∫5x4dx=ex5⇒y′I(x)+5x4yI(x)=3x2e−x5⇒y′ex5+5x4yex5=3x2⇒(yex5)′=3x2⇒yex5=x3+c1⇒y=x3e−x5+c1e−x53.x2y′=y2+xy⇒y′−yx=y2x2為白努利方程式因此取v=1y⇒v′=−y′y2⇒y′=−y2v′代入原式⇒−y2v′−yx=y2x2⇒v′+1xy=−1x2⇒v′+vx=−1x2⇒xv′+v=−1x⇒(xv)′=−1x⇒xv=−lnx+c1⇒xy=−lnx+c1⇒y=x−lnx+c1解答:先求齊次解,y″+9y=0⇒λ2+9=0⇒λ=±3i⇒yh=c1cos(3x)+c2sin(3x)yp=Acosx+Bsinx⇒y′p=−Asinx+Bcosx⇒y″p=−Acosx−Bsinx⇒y″p+9yp=8Acosx+8Bsinx=16sinx⇒{A=0B=2⇒yp=2sinx⇒y=yh+yp=c1cos(3x)+c2sin(3x)+2sinx⇒y′=−3c1sin(3x)+3c2cos(3x)+2cosx⇒{y(π)=−c1=0y′(0)=3c2+2=0⇒{c1=0c2=−23⇒y=−23sin(3x)+2sinx
解答:(1)F(s)=L{t2}+L{e−6t}+L{sin(6t)}=2s3+1s+6+6s2+36(2)f(t)=L−1{8s(s−2)2}=L−1{2s−2s−2+4(s−2)2}=2−2e2t+4e2tt(3)L{y′}+L{∫t0y(τ)dτ}=2L{1}−2L{sint}⇒sY(s)+Y(s)=2s−2s2+1⇒Y(s)=2s(s+1)−2(s2+1)(s+1)⇒y(t)=L−1{2s(s+1)}−L−1{2(s2+1)(s+1)}=2L−1{1s−1s+1}−L−1{1s2+1−ss2+1+1s+1}=2−2e−t−(sint−cost+e−t)⇒y(t)=2−3e−t−sint+cost
解答:y′−3y=te3tsin(t)⇒L{y′}−3L{y}=L{te3tsin(t)}⇒sY(s)−3Y(s)=−ddsL{e3tsin(t)}=−dds1(s−3)2+1=2s−6(s2−6s+10)2⇒Y(s)=2(s2−6s+10)2⇒y(t)=L−1{2(s2−6s+10)2}=L−1{2((s−3)2+1)2}=e3tL−1{2(s2+1)2}=e3t(sint−tcost)⇒y(t)=e3t(sint−tcost)
解答:(a)A=[003021201]⇒det(A−λI)=−(λ+2)(λ−2)(λ−3)=0⇒λ=−2,2,3λ1=−2⇒(A−λ1I)v=0⇒[203041203][x1x2x3]=0⇒{2x1+3x3=04x2+x3=0⇒v=[−3k/2−k/4k],k∈R,取v1=[−3/2−/41]λ2=2⇒(A−λ2I)v=0⇒[−20300120−1][x1x2x3]=0⇒{x1=0x3=0⇒v=[0k0],k∈R,取v2=[010]λ3=3⇒(A−λ3I)v=0⇒[−3030−1120−2][x1x2x3]=0⇒{x1=x3x2=x3⇒v=[kkk],k∈R,取v3=[111]⇒eigenvalues: −2,2,3 and corresponding eigenvectors are [−3/2−/41],[010],[111](b)X=[v1v2v3],D=[λ1000λ2000λ3]⇒A=[−3/201−1/411101][−200020003][−3/201−1/411101]−1⇒A5=[−3/201−1/411101][(−2)50002500035][−3/201−1/411101]−1=[−3/201−1/411101][−3200032000243][−3/201−1/411101]−1⇒A5=[78016578321331100133]

解答:(a)A=[−4341−23600]⇒rref(A)=[100010001]⇒linearly independent(b)−[6−786]+2[46−41]=[219−16−4]⇒linearly dependent

解答:A=[1421011−1−2−8−4−2]⇒rref(A)=[10−25011−10000]⇒Rank(A)=2又[10−25011−10000][x1x2x3x4]=0⇒{x1−2x3+5x4=0x2+x3−x4=0⇒[x1x2x3x4]=[2x3−5x4−x3+x4x3x4]=x3[2−110]+x4[−5101]⇒Null(A)=Span{[2−110],[−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=a1⇒e1=b1‖b1‖=[1/107/101/107/10]b2=a2−(a2⋅e1)e1=[0727]−[1717]=[−1010]⇒e2=b2‖b2‖=[−1/√201/√20]b3=a3−(a3⋅e1)e1−(a3⋅e2)e2=[1816]−[1717]−0=[010−1]⇒e3=b3‖b3‖=[01/√20−1/√2]⇒an orthonomal basis for the column space of A={[1/107/101/107/10],[−1/√201/√20],[01/√20−1/√2]}
==================== END ===========================
解題僅供參考,其他歷年試題及詳解
沒有留言:
張貼留言