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2023年12月16日 星期六

112年台科大機械工程碩士班甲組-工程數學詳解

 


解答:(a)[123100211010111001]2R1+R2R2,R1+R3R3[123100035210012101]2R3+R1R1,3R3+R2R2[101102001113012101]R2R3[101102012101001113]R3+R1R1,2R3+R2R2[100011010125001113]R2R2[100011010125001113]A1=[011125113](b)(a)A=[123211111]R22R1R2[123035111]R31R1R3[123035012]R313R2R3[123035001/3]A=LU=[10021011/31][123035001/3]

解答:



解答:u=yu3u=8e3x+4integration factor I(x)=e3dx=e3xue3x3e3xu=8+4e3x(e3xu)=8+4e3xe3xu=8x43e3x+c1u=y=8xe3x43+c1e3xy=83xe3x89e3x43x+13c1e3x+c2y=83xe3x43x+c3e3x+c2
解答:{u(x,y,z)=2x2+3y2+z2u=(ux,uy,uz)=(4x,6y,2z)v=i2k=(1,0,2)vv=15(1,0,2)Dvu(2,1,3)=u(2,1,3)vv=(8,6,6)15(1,0,2)=45

解答:(a)r(θ,ϕ)=(2cosϕsinθ,2sinϕsinθ,2cosθ)n=rθ×rϕ=(4sin2θcosϕ,4sin2θsinϕ,4sinθcosθ)S(Fn)dA=πθ=02πϕ=0(14sinθcosϕ,0,2cosθ)(4sin2θcosϕ,4sin2θsinϕ,4sinθcosθ)dθ=πθ=02πϕ=056cos2ϕsin2θ8cos2θsinθdϕdθ=64π(b)F=7xizkdivF=Fx+Fy+Fz=7+01=6S(Fn)dA=RdivFdV=R6dV=643π23=64π

解答:

u(x,t)=F(x)G(t){utt=FGuxx=FGFG=c2FGGc2G=FFGc2GtFFxkGc2G=FF=k:{u(0,t)=F(0)G(t)=0u(L,t)=F(L)G(t)=0G(t)=0u(x,t)=0F(0)=F(L)=0FkF=0F=c1ekx+c2ekx{F(0)=c1+c2=0(1)F(L)=c1ekL+c2ekL=0(2)(1)c2=c1(2)c1(ekLekL)=0c1=0ekLekLc1=0c2=0F=0;ekLekLekLekL,k>0;k=0F=0k<0,F(x)=c1(ei|k|xei|k|x)=2ic1sin(|k|x)F(L)=2ic1sin(|k|L)=0|k|L=nπ,nN|k|=nπ/LF(x)=2ic1sin(nπx/L)=Asin(nπLx),Ak=n2π2L2G+n2π2L2c2G=0G(t)=c3cos(nπLct)+c4sin(nπLct)u(x,t)=Asin(nπLx)(c3cos(nπLct)+c4sin(nπLct))=n=1sin(nπLx)(Ancos(nπLct)+Bnsin(nπLct))
1:u(x,0)=f(x)f(x)=n=1Ansin(nπLx)L0f(x)sin(kπLx)dx=n=1L0Ansin(kπLx)sin(nπLx)dx=L0Aksin2(kπLx)dx=L2AkAk=2LL0f(x)sin(kπLx)dxAn=2LL0f(x)sin(nπLx)dx=2L(L/202kLxsin(nπLx)dx+LL/22kL(Lx)sin(nπLx)dx)An=8knπ(1nπsin(nπ2)cos(nπ2)+cos(nπ))

2:ut|t=0=0n=1(cnπLAnsin(cnπLt)+cnπLBncos(cnπLt))sin(nπLx)|t=0=n=1cnπLBnsin(nπLx)=0Bn=0u(x,t)=n=1sin(nπLx)(Ancos(nπLct))u(x,t)=n=18knπ(1nπsin(nπ2)cos(nπ2)+cos(nπ))sin(nπLx)cos(nπLct)

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