Loading [MathJax]/jax/output/CommonHTML/jax.js

網頁

2024年4月23日 星期二

113年中山機電碩士班-工程數學詳解

 國立中山大學113學年度碩士班招生考試

科目名稱:工程數學【機電系碩士班乙組、丙組】


解答:(a)y=xmy=mxm1y=m(m1)xm2x2y2xy+4y=(m23m+4)xm=0m23m+4=0m=3±7i2yh=c1x3/2cos(72lnx)+c2x3/2sin(72lnx)..(b)y=x,{y1=xy=v+y1=v+xx(v+1)(v+x)+2(v+x)2=2x2xvv+2v2+4xv=0v+(41x)v=2xv2u=1vu=vv2=u2vv=uu2uu2+(41x)1u=2x1u2u+(1x4)u=2x integrating factor I(x)=e((1/x)4)dx=xe4xxe4xu+(14x)e4xu=2e4x(xe4xu)=2e4xxe4xu=2e4xdx=12e4x+c1u=12x+c1xe4x=c2e4x12xv=1u=2xc2e4x1y=v+xy=2xc2e4x1+x
解答: (a)

(b)f(t)=π3(u(t)u(t2π))+u(t4π)2πsin(4t)(c)L{f(t)}=0f(t)estdt=2π0π3estdt+4π2πsin(4t)estdt=π3[1sest]|2π0+2π[ests2+16(ssin(4t)+4cos(4t)]|0=π3s(1e2πs)+2π(4s2+16)

解答:(a)A=[1021213124120202]R22R1R2,R3+2R1R3[1021011304300202]R34R2R3,R42R2R4[10210113007120024]R3/(7)R3[102101130011270024]R1+2R3R1,R2R3R2,R4+2R3R4[1001770109700112700047](7/4)R4R4[100177010970011270001]R1(17/7)R4R1,RO2(9/7)R4R2,R3(12/7)R4R3[1000010000100001]rref(A)=[1000010000100001]rank(A)=4(b)A=[1021213124120202]R22R1R2,R3+2R1R3[1021011304300202]R34R2R3,R42R2R4[10210113007120024]R4(2/7)R3R4[102101130071200047]det(A)=|102101130071200047|=11(7)(47)=4(c)[AI]=[10211000213101002412001002020001]R22R1R2,R3+2R1R3[10211000011321000430201002020001]R34R2R3,R42R2R4[1021100001132100007121041000244201](1/7)R3R3[10211000011321000011271074717000244201]R1+2R3R1,R2R3R2,R4+2R3R4[1001771378727001097473717000112710747170000478767271](7/4)R4R4[100177137872700109747371700011271074717000012321274]R1(17/7)R4R1,R2(9/7)R4R2,R3(12/7)R4R3[100035232174010023212940010221300012321274]A1=[35232174232129422132321274]


解答:u(x,t)=X(x)T(t)XT=XTTT=XX=λλ0X=0u=0λ<0λ=ρ2X+ρ2=0X=c1cosρx+c2sinρx{X(0)=c1=0X(L=π)=c2sinρπ=0ρ=1,2,Xn=sinnxT+ρ2T=0T=c1cosρt+c2sinρtTn=c1cosnt+c2sinntu(x,t)=n=1(Ancosnt+Bnsinnt)sin(nx)ut(x,t)=n=1(nAnsinnt+nBncosnt)sin(nx)ut(x,0)=n=1nBnsin(nx)Bn=2nππ0g(x)sin(nx)dx=2nπ(π/200.01xsin(nx)dx+ππ/20.01(πx)sin(nx)dx)=0.02nπ([sin(nx)nxcos(nx)n2]|π/20+[sin(nx)+n(πx)cos(nx)n2]|ππ/2)=0.02nπ2sin(nπ/2)n2=0.04n3πsin(nπ/2)Bn=0.04n3πsin(nπ/2)u(x,0)=0u(x,t)=n=1Ansin(nx)=0An=0u(x,t)=n=10.04n3πsin(nπ/2)sin(nt)sin(nx)


解答:(a)|ijkxyzy2cosx+z32ysinx43xz2+2|=2ycosxk+3z2j2ycosxk3z2j=0F is a conservative field.QED(b)Let Φ=F, then Φx=y2cosx+z3Φ=(y2cosx+z3)dxΦ=y2sinx+xz3+ϕ(y,z)Φy=2ysinx+ϕy=2ysinx4ϕy=4ϕ(y,z)=4y+ρ(z)Φ=y2sinx+xz34y+ρ(z)Φz=3xz2+ρ(z)=3xz2+2ρ(z)=2ρ(z)=2z+c1Φ=y2sinx+xz34y+2z+c1(c)Φ(π/2,1,2)Φ(1,1,1)=(1+4π+4+4)(sin1142)=4π+2sin1


解答:r(u,v)=[u,v,6u9v]{ru=[1,0,6]rv=[0,1,9]ru×rv=[6,9,1]F(u,v)=[5u,v3,0]F(ru×rv)dudv=112030u+9v3dudv=1160+18v3dv=120

==================== END ======================
解題僅供參考,其他歷年試題及詳解

沒有留言:

張貼留言