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=xm⇒y′=mxm−1⇒y″=m(m−1)xm−2⇒x2y″−2xy′+4y=(m2−3m+4)xm=0⇒m2−3m+4=0⇒m=3±√7i2⇒yh=c1x3/2cos(√72lnx)+c2x3/2sin(√72lnx)..待解(b)y=x顯然為其一解,因此令{y1=xy=v+y1=v+x代回原式⇒x(v′+1)−(v+x)+2(v+x)2=2x2⇒xv′−v+2v2+4xv=0⇒v′+(4−1x)v=−2xv2取u=1v⇒u′=−v′v2=−u2v′⇒v′=−u′u2⇒−u′u2+(4−1x)1u=−2x⋅1u2⇒u′+(1x−4)u=2x⇒ integrating factor I(x)=e∫((1/x)−4)dx=xe−4x⇒xe−4xu′+(1−4x)e−4xu=2e−4x⇒(xe−4xu)′=2e−4x⇒xe−4xu=∫2e−4xdx=−12e−4x+c1⇒u=−12x+c1xe4x=c2e4x−12x⇒v=1u=2xc2e4x−1⇒y=v+x⇒y=2xc2e4x−1+x

解答: (a)
解答:u(x,t)=X(x)T(t)⇒XT″=X″T⇒T″T=X″X=λλ≥0⇒X=0⇒u=0λ<0⇒λ=−ρ2⇒X″+ρ2=0⇒X=c1cosρx+c2sinρx⇒{X(0)=c1=0X(L=π)=c2sinρπ=0⇒ρ=1,2,⋯⇒Xn=sinnxT″+ρ2T=0⇒T=c1cosρt+c2sinρt⇒Tn=c1cosnt+c2sinnt⇒u(x,t)=∞∑n=1(Ancosnt+Bnsinnt)sin(nx)⇒ut(x,t)=∞∑n=1(−nAnsinnt+nBncosnt)sin(nx)⇒ut(x,0)=∞∑n=1nBnsin(nx)⇒Bn=2nπ∫π0g(x)sin(nx)dx=2nπ(∫π/200.01xsin(nx)dx+∫ππ/20.01(π−x)sin(nx)dx)=0.02nπ([sin(nx)−nxcos(nx)n2]|π/20+[−sin(nx)+n(π−x)cos(nx)n2]|ππ/2)=0.02nπ⋅2sin(nπ/2)n2=0.04n3πsin(nπ/2)⇒Bn=0.04n3πsin(nπ/2)u(x,0)=0⇒u(x,t)=∞∑n=1Ansin(nx)=0⇒An=0⇒u(x,t)=∞∑n=10.04n3πsin(nπ/2)sin(nt)sin(nx)
解答:(a)|→i→j→k∂∂x∂∂y∂∂zy2cosx+z32ysinx−43xz2+2|=2ycosx→k+3z2→j−2ycosx→k−3z2→j=0⇒→F is a conservative field.QED(b)Let ∇Φ=F, then Φx=y2cosx+z3⇒Φ=∫(y2cosx+z3)dx⇒Φ=y2sinx+xz3+ϕ(y,z)⇒Φy=2ysinx+ϕy=2ysinx−4⇒ϕy=−4⇒ϕ(y,z)=−4y+ρ(z)⇒Φ=y2sinx+xz3−4y+ρ(z)⇒Φz=3xz2+ρ′(z)=3xz2+2⇒ρ′(z)=2⇒ρ(z)=2z+c1⇒Φ=y2sinx+xz3−4y+2z+c1(c)Φ(π/2,−1,2)−Φ(1,1,−1)=(1+4π+4+4)−(sin1−1−4−2)=4π+2−sin1

(b)f(t)=π3(u(t)−u(t−2π))+u(t−4π)2πsin(4t)(c)L{f(t)}=∫∞0f(t)e−stdt=∫2π0π3e−stdt+∫∞4π2πsin(4t)e−stdt=π3[−1se−st]|2π0+2π[−e−sts2+16(ssin(4t)+4cos(4t)]|∞0=π3s(1−e−2πs)+2π(4s2+16)
解答:(a)A=[10−2−121−31−24120202]R2−2R1→R2,R3+2R1→R3→[10−2−1011304−300202]R3−4R2→R3,R4−2R2→R4→[10−2−1011300−7−1200−2−4]R3/(−7)→R3→[10−2−1011300112700−2−4]R1+2R3→R1,R2−R3→R2,R4+2R3→R4→[10017701097001127000−47]−(7/4)R4→R4→[100177010970011270001]R1−(17/7)R4→R1,RO2−(9/7)R4→R2,R3−(12/7)R4→R3→[1000010000100001]⇒rref(A)=[1000010000100001]⇒rank(A)=4(b)A=[10−2−121−31−24120202]R2−2R1→R2,R3+2R1→R3→[10−2−1011304−300202]R3−4R2→R3,R4−2R2→R4→[10−2−1011300−7−1200−2−4]R4−(2/7)R3→R4→[10−2−1011300−7−12000−47]⇒det(A)=|10−2−1011300−7−12000−47|=1⋅1⋅(−7)⋅(−47)=4(c)[A∣I]=[10−2−1100021−310100−2412001002020001]R2−2R1→R2,R3+2R1→R3→[10−2−110000113−210004−30201002020001]R3−4R2→R3,R4−2R2→R4→[10−2−110000113−210000−7−1210−41000−2−44−201]−(1/7)R3→R3→[10−2−110000113−2100001127−10747−17000−2−44−201]R1+2R3→R1,R2−R3→R2,R4+2R3→R4→[100177−13787−27001097−4737170001127−10747−170000−4787−67−271]−(7/4)R4→R4→[100177−13787−27001097−4737170001127−10747−1700001−23212−74]R1−(17/7)R4→R1,R2−(9/7)R4→R2,R3−(12/7)R4→R3→[10003−52−3217401002−32−129400102−2−130001−23212−74]⇒A−1=[3−52−321742−32−12942−2−13−23212−74]
解答:(a)|→i→j→k∂∂x∂∂y∂∂zy2cosx+z32ysinx−43xz2+2|=2ycosx→k+3z2→j−2ycosx→k−3z2→j=0⇒→F is a conservative field.QED(b)Let ∇Φ=F, then Φx=y2cosx+z3⇒Φ=∫(y2cosx+z3)dx⇒Φ=y2sinx+xz3+ϕ(y,z)⇒Φy=2ysinx+ϕy=2ysinx−4⇒ϕy=−4⇒ϕ(y,z)=−4y+ρ(z)⇒Φ=y2sinx+xz3−4y+ρ(z)⇒Φz=3xz2+ρ′(z)=3xz2+2⇒ρ′(z)=2⇒ρ(z)=2z+c1⇒Φ=y2sinx+xz3−4y+2z+c1(c)Φ(π/2,−1,2)−Φ(1,1,−1)=(1+4π+4+4)−(sin1−1−4−2)=4π+2−sin1

解答:→r(u,v)=[u,v,6u−9v]⇒{→ru=[1,0,6]→rv=[0,1,−9]⇒→ru×→rv=[−6,9,1]→F(u,v)=[5u,v3,0]⇒∫→F⋅(→ru×→rv)dudv=∫1−1∫20−30u+9v3dudv=∫1−1−60+18v3dv=−120
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