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2022年10月20日 星期四

108年台綜大轉學考-工程數學D39詳解

 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)X=[abcd][1324]X+2[2113]=[1324][abcd]+2[2113]=[a+3cb+3d2a+4c2b+4d]+[4226]=[a+3c+4b+3d+22a+4c22b+4d+6]=[1521]{{a+3c=32a+4c=4{a=12c=5{b+3d=32b+4d=7{b=33/2d=13/2X=[1233/2513/2](b)(AI3)=(123100233010342001)2r1+r2r23r1+r3r3(123100013210027301)2r2+r1r12r2+r3r3(103320013210001121)3r3+r1r13r3+r2r2(100683010573001121)r3(100683010573001121)A1=(683573121)X=A1AX=(683573121)(121)=(19164)X=(19164)det(5A)=53det(A)=125×|123233342|=125×(1)=125det(5A)=125

解答(a)x3y+2y=x3+2xy+2x3y=1+2x2,I(x)=e2x3dx=e1/x2e1x2y+2x3e1x2y=e1x2+2x2e1x2(e1x2y)=e1x2+2x2e1x2e1x2y=e1x2+2x2e1x2dx=xe1x2+Cy=x+Ce1x2y(1)=e+1e+1=1+CeC=1y=x+e1x2 (b)y=xv(x)y=v(x)+xv(x)y=2v(x)+xv(x)y=3v(x)+xv(x)x3(3v+xv)3x2(2v+xv)+(6x2)(xv+x2v)(6x2)xv=0x4vx4v=0vv=0v(x)=C1+C2ex+C3exy=C1x+C2xex+C3xex (c)y3y+2y=0λ23λ+2=0λ=1,2yh=C1ex+C2e2x:{y1=exy2=e2xW=|y1y2y1y2|=|exe2xex2e2x|=e3xyp=y1y2sin(ex)Wdx+y2y1sin(ex)Wdx=exsin(ex)exdx+e2xsin(ex)e2xdx=excos(ex)+e2x(excos(ex)sin(ex))=e2xsin(ex)y=yh+ypy=C1ex+C2e2xe2xsin(ex)
解答L1{s2s2+2s+10}=L1{(s+1)3(s+1)2+9}=L1{s+1(s+1)2+32}L1{3(s+1)2+32}=etcos(3t)etsin(3t)
解答F(f(x))=12πf(x)eiωxdx=12πex2/a2eiωxdx=12πex2/a2(cos(ωx)isin(ωx))dx=12πex2/a2cos(ωx)dx(f(x)sin(ωx)0)=22π0ex2/a2cos(ωx)dx=2π0ex2/a2cos(ωx)dx(f(x)cos(ωx)y)g(ω)=0ex2/a2cos(ωx)dx,ω0;g(ω)=0xex2/a2sin(ωx)dx{u=sin(ωx)du=ωcos(ωx)dxdv=xex2/a2dxv=a22ex2/a2g(ω)=a22ex2/a2sin(ωx)|0ωa220cos(ωx)ex2/a2dx=0ωa22g(ω)g(ω)=ωa22g(ω)(ea2ω2/4g(ω))=0g(ω)=C1ea2ω2/4F(f(x))=2πg(ω)F(f(x))=2πC1ea2ω2/4

解答{A(1,1,1)B(2,2,2)C(3,4,X){u=AB=(1,1,1)v=AC=(2,3,X1)u×v=(x4,3x,1)ABC=12(x4,3x,1)=122x214x+26=122(x(7/2))2+3/2ABC=1232=64
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