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2023年12月24日 星期日

112年成大航太碩士班-工程數學詳解

國立成功大學112學年度碩士班招生考試

系所: 航空太空工程學系、民航研究所、能源工程國際碩士學位學程

解答:CFdr=2π0(sinθ,cosθ,cosθ+sinθ)(sinθ,cosθ,0)dt=2π0sin2θ+cos2θdθ=2π0cos(2θ)dθ=0
解答:(a)Cz23z+4iz+2i=2πi×[z23z+4i|z=2i]=2πi×(4+10i)=20π8πi(b)2πi×d2dz2(ezsin(z))|z=0=2πi×2ezcosz|z=0=4πi
解答:(a)L{y}+5L{y}+6L{y}=L{δ(tπ/2}+L{u(tπ)sin(t)}s2Y(s)+5sY(s)+6Y(s)=eπs/2+eπs1s2+1Y(s)=1s2+5s+6(eπs/2eπs1s2+1)=(1s+21s+3)eπs/2(1s+21s+3)1s2+1eπsy(t)=u(tπ/2)(e2(tπ/2)e3(tπ/2))u(tπ)(15e2(tπ110e3(tπ)110cos(tπ)+110sin(tπ))(b)L{y(t)}+L{t0y(τ)sinh(tτ)dτ}=3L{t}+L{et}Y(s)+Y(s)1s21=3s2+1s1Y(s)=3s2+1s13s41s2(s1)y(t)=L1{3s2}+L1{1s1}L1{3s4}L1{1s2(s1)}=3t+et12t3et+t+u(t)y(t)=12t3+4t+u(t)
解答:,y+2y+3y=0λ2+2λ+0λ=1±2iyh=c1excos(2x)+c2exsin(2x)yp=asinx+bcosx+cx+dyp=acosxbsinx+cyp=asinxbcosxyp+2yp+3yp=(2a+2b)cosx+(2a2b)sinx+3cx+(3d+2c)=sinxx{2a+2b=02a2b=13c=13d+2c=0{a=1/4b=1/4c=1/3d=2/9yp=14sinx14cosx13x+29y=yh+yp=c1excos(2x)+c2exsin(2x)+14sinx14cosx13x+29y=ex(c1cos(2x)+c2sin(2x))+ex(2c2sin(2x)+2c2cos(2x))+14cosx+14sinx13{y(0)=c114+29=0y(0)=c1+2c2+1413=0{c1=1/36c2=1/92y=136excos(2x)+192exsin(2x)+14sinx14cosx13x+29
解答:,x3y3x2y+6xy6y=0,y=xm{y=mxm1y=m(m1)xm2y=m(m1)(m2)xm3x3y3x2y+6xy6y=(m1)(m2)(m3)xm=0m=1,2,3yh=c1x+c2x2+c3x3 variations of parameter yp,{y1=xy2=x2y3=x3W=|y1y2y3y1y2y3y1y2y3|=|xx2x312x3x2026x|=2x3{W1=|0x2x302x3x2126x|=x4W2=|x0x3103x2016x|=2x3W3=|xx2012x0021|=x2yp=xx4lnx2x3dx+x2(2x3)lnx2x3+x3x2lnx2x3dx=x2(12x2lnx14x2)x2(xlnxx)+x32(12ln2(x))yp=14x3ln2(x)34x3ln(x)+78x3
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解題僅供參考,其他歷年試題及詳解 

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