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2022年9月1日 星期四

109年台聯大轉學考-工程數學詳解

臺灣聯合大學系統109學年度學士班轉學生聯合招生考試

科目名稱:工程數學
類組別:A5

解答(a)y+xy=xy1ydydx+xy2=xydydx=x(1y2)y1y2dy=xdx12ln|1y2|=12x2+C1ln|1y2|=x2+C21y2=C3ex2y=±1C3ex2,y(0)=3C3=8y=1+8ex2(b)y=xmy=mxm1y=m(m1)xm2x2y3xy+3y=0m(m1)xm3mxm+3x3=0(m24m+3)xm=0(m3)(m1)=0m=1,3yh=C1x+C2x3yh=C1+3C2x{y(1)=0y(0)=1yh{0=C1+C2C1=1{C1=1C2=1yh=xx3yp=Alnxyp=Axyp=Ax2x2yp3xyp+3yp=A3A+3Alnx=3lnx4A=1yp=lnxy=yh+ypy=xx3+lnx
解答yy=tL{y}L{y}=L{t}s2Y(s)sy(0)y(0)Y(s)=1ss2Y(s)s1Y(s)=1s(s21)Y(s)=s+1+1sY(s)=s2+s+1s(s21)=1s+121s+1+321s1y(t)=L1{Y}=L{1s}+12L{1s+1}+32L{1s1}=1+12et+32ety(t)=1+12et+32et
解答{m=1k=1F(t)=1t2/π2ky+F(t)=myy+y=1t2/π2y=C1cost+C2sintt2π2+1+2π2y=C1sint+C2cost2tπ2{y(0)=0y(0)=0{C1+1+2π2=0C2=0{C1=12/π2C2=0y+y=1t2/π2y(t)=(12/π2)sintt2/π2+1+2/π2
解答A=[2λ2321λ6120λ]det(AλI)=|2λ2321λ6120λ|=(λ5)(λ+3)2=0λ1=5,λ2=3λ1=5(Aλ1I)x=[723246125][x1x2x3]=0{x1+x3=0x2+2x3=0,v1=[121]λ2=3(Aλ2I)x=[123246123][x1x2x3]=0x1+2x2=3x3,v2=[210],v3=[301]53[121],[210][301]
解答(a){x(t)=acos3ty(t)=asin3t{x(t)=3acos2tsinty(t)=3asin2tcost{(x(t))2=9a2cos4tsin2t(y(t))2=9a2sin4tcos2tπ/20(x(t))2+(y(t))2dt=π/209a2cos2tsin2t(cos2t+sin2t)dt=π/209a2cos2tsin2tdt=π/203acostsintdt=π/2032asin(2t)dt=[34acos(2t)]|π/20=34a+34a=32a(b)f(x,y)=16x2y2=399f=(fx,fy)=(32x,2y)f(1/8,1)=(4,2)(4,2)42+22=(25,15):Psurface?!(c)f=x2+y2+z2f=(fx,fy,fz)=(2x,2y,2z)f(2,2,1)=(4,4,2)f(2,2,1)a=(4,4,2)(1,1,3)=4+46=2
解答{Ax=e1Ay=enAz=2e1+3en=2Ax+3Ayz=2x+3y
解答SundA=VudV=1/21/21/21/21/21/2(x,y,z)(x,y,z)dxdydz=1/21/21/21/21/21/23dxdydz=3
解答d'Alembert's solution,utt=c2uxxu(x,t)=F(x+ct)+G(xct)=F(x+t)+G(xt)=12(f(x+t)+f(xt))+12x+txtg(s)ds=12(sin(π(x+t))+3sin(3π(x+t))+sin(π(xt))+3sin(3π(xt))+0u(x,t)=12(sin(π(x+t))+3sin(3π(x+t))+sin(π(xt))+3sin(3π(xt))
解答u(x,t)=X(x)T(t){ut=XTuxx=XTut=uxxuXT=XTXTTT=XXX=kkX(k+1)X=0{u(0,t)=X(0)T(t)=0u(1,t)=X(1)T(t)=0,T(t)=0u(x,t)=0;X(1)=X(0)=0Case I: k=1X=0X=ax+b{X(1)=a+b=0X(0)=b=0X=0u(x,t)=0Case II: k>1X=C1ek+1x+C2ek+1x{X(1)=C1ek+1+C2ek+1=0X(0)=C1+C2=0C1=C2=0X=0u(x,t)=0Case III: k<1k+1=μ2X=Acosμx+Bsinμx{Acosμ+Bsinμ=0A=0{A=B=0A=0sinμ=0sinu=0u=nπXn(x)=sinnπx,nNTT=kTkT=0T+(u2+1)T=0T(t)=Ae(u2+1)tTn(t)=Ane(n2π2+1)tu(x,t)=Ane(n2π2+1)tsin(nπx);u(x,0)=sin(πx)Ansin(nπx)=sin(πx){A1=1Am=0,m2u(x,t)=e(π2+1)tsin(πx)


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