臺灣綜合大學系統108學年度學士班轉學生聯合招生考試
科目名稱:工程數學
類組代碼:D39

解答:(a)令X=[abcd]⇒[1324]X+2[21−13]=[1324][abcd]+2[21−13]=[a+3cb+3d2a+4c2b+4d]+[42−26]=[a+3c+4b+3d+22a+4c−22b+4d+6]=[152−1]⇒{{a+3c=−32a+4c=4⇒{a=12c=−5{b+3d=32b+4d=−7⇒{b=−33/2d=13/2⇒X=[12−33/2−513/2](b)(A∣I3)=(1−231002−330103−42001)−2r1+r2→r2−3r1+r3→r3→(1−2310001−3−21002−7−301)2r2+r1→r1−2r2+r3→r3→(10−3−32001−3−21000−11−21)−3r3+r1→r1−3r3+r2→r2→(100−68−3010−57−300−11−21)−r3→(100−68−3010−57−3001−12−1)⇒A−1=(−68−3−57−3−12−1)X=A−1AX=(−68−3−57−3−12−1)(−121)=(19164)⇒X=(19164)det(5A)=53det(A)=125×|1−232−333−4−2|=125×(−1)=−125⇒det(5A)=−125
解答:(a)x3y′+2y=x3+2x⇒y′+2x3y=1+2x2,取積分因子I(x)=e∫2x3dx=e−1/x2⇒e−1x2y′+2x3e−1x2y=e−1x2+2x2e−1x2⇒(e−1x2y)′=e−1x2+2x2e−1x2⇒e−1x2y=∫e−1x2+2x2e−1x2dx=xe−1x2+C⇒y=x+Ce1x2將y(1)=e+1代入上式⇒e+1=1+Ce⇒C=1⇒y=x+e1x2 (b)令y=xv(x)⇒y′=v(x)+xv′(x)⇒y″=2v′(x)+xv″(x)⇒y‴=3v″(x)+xv‴(x)⇒x3(3v″+xv‴)−3x2(2v′+xv″)+(6−x2)(xv+x2v′)−(6−x2)xv=0⇒x4v‴−x4v′=0⇒v‴−v′=0⇒v(x)=C1+C2ex+C3e−x⇒y=C1x+C2xex+C3xe−x (c)先求齊次解,即y″−3y′+2y=0⇒λ2−3λ+2=0⇒λ=1,2⇒yh=C1ex+C2e2x再用參數變換法:令{y1=exy2=e2x⇒W=|y1y2y′1y′2|=|exe2xex2e2x|=e3x⇒yp=−y1∫y2sin(e−x)Wdx+y2∫y1sin(e−x)Wdx=−ex∫sin(e−x)exdx+e2x∫sin(e−x)e2xdx=−excos(e−x)+e2x(e−xcos(e−x)−sin(e−x))=−e2xsin(e−x)⇒y=yh+yp⇒y=C1ex+C2e2x−e2xsin(e−x)
解答:L−1{s−2s2+2s+10}=L−1{(s+1)−3(s+1)2+9}=L−1{s+1(s+1)2+32}−L−1{3(s+1)2+32}=e−tcos(3t)−e−tsin(3t)
解答:F(f(x))=1√2π∫∞−∞f(x)e−iωxdx=1√2π∫∞−∞e−x2/a2e−iωxdx=1√2π∫∞−∞e−x2/a2(cos(ωx)−isin(ωx))dx=1√2π∫∞−∞e−x2/a2cos(ωx)dx(∵f(x)sin(ωx)為奇函數,其積分值為0)=2√2π∫∞0e−x2/a2cos(ωx)dx=√2π∫∞0e−x2/a2cos(ωx)dx(∵f(x)cos(ωx)對稱y軸)取g(ω)=∫∞0e−x2/a2cos(ωx)dx,其中ω≥0;⇒g′(ω)=∫∞0−xe−x2/a2sin(ωx)dx由{u=sin(ωx)⇒du=ωcos(ωx)dxdv=−xe−x2/a2dx⇒v=a22e−x2/a2⇒g′(ω)=a22e−x2/a2sin(ωx)|∞0−ωa22∫∞0cos(ωx)e−x2/a2dx=0−ωa22g(ω)⇒g′(ω)=−ωa22g(ω)⇒(ea2ω2/4g(ω))′=0⇒g(ω)=C1e−a2ω2/4⇒F(f(x))=√2πg(ω)⇒F(f(x))=√2πC1e−a2ω2/4解答:{A(1,1,1)B(2,2,2)C(3,4,X)⇒{→u=→AB=(1,1,1)→v=→AC=(2,3,X−1)⇒→u×→v=(x−4,3−x,1)⇒△ABC面積=12‖(x−4,3−x,1)‖=12√2x2−14x+26=12√2(x−(7/2))2+3/2⇒△ABC面積最小值=12√32=√64
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