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:∮C→F⋅d→r=∫2π0(sinθ,cosθ,cosθ+sinθ)⋅(−sinθ,cosθ,0)dt=∫2π0−sin2θ+cos2θdθ=∫2π0cos(2θ)dθ=0解答:(a)∮Cz2−3z+4iz+2i=2πi×[z2−3z+4i|z=−2i]=2πi×(−4+10i)=−20π−8πi(b)2πi×d2dz2(e−zsin(z))|z=0=2πi×−2e−zcosz|z=0=−4πi
解答:(a)L{y″}+5L{y′}+6L{y}=L{δ(t−π/2}+L{u(t−π)sin(t)}⇒s2Y(s)+5sY(s)+6Y(s)=e−πs/2+e−πs⋅−1s2+1⇒Y(s)=1s2+5s+6⋅(e−πs/2−e−πs⋅1s2+1)=(1s+2−1s+3)e−πs/2−(1s+2−1s+3)⋅1s2+1⋅e−πs⇒y(t)=u(t−π/2)(e−2(t−π/2)−e−3(t−π/2))−u(t−π)(15e−2(t−π−110e−3(t−π)−110cos(t−π)+110sin(t−π))(b)L{y(t)}+L{∫t0y(τ)sinh(t−τ)dτ}=3L{t}+L{et}⇒Y(s)+Y(s)⋅1s2−1=3s2+1s−1⇒Y(s)=3s2+1s−1−3s4−1s2(s−1)⇒y(t)=L−1{3s2}+L−1{1s−1}−L−1{3s4}−L−1{1s2(s−1)}=3t+et−12t3−et+t+u(t)⇒y(t)=−12t3+4t+u(t)
解答:先求齊次解,y″+2y′+3y=0⇒λ2+2λ+0⇒λ=−1±√2i⇒yh=c1e−xcos(√2x)+c2e−xsin(√2x)令yp=asinx+bcosx+cx+d⇒y′p=acosx−bsinx+c⇒y″p=−asinx−bcosx⇒y″p+2y′p+3yp=(2a+2b)cosx+(2a−2b)sinx+3cx+(3d+2c)=sinx−x⇒{2a+2b=02a−2b=13c=−13d+2c=0⇒{a=1/4b=−1/4c=−1/3d=2/9⇒yp=14sinx−14cosx−13x+29⇒y=yh+yp=c1e−xcos(√2x)+c2e−xsin(√2x)+14sinx−14cosx−13x+29⇒y′=−e−x(c1cos(√2x)+c2sin(√2x))+e−x(−√2c2sin(√2x)+√2c2cos(√2x))+14cosx+14sinx−13{y(0)=c1−14+29=0y′(0)=−c1+√2c2+14−13=0⇒{c1=1/36c2=1/9√2⇒y=136e−xcos(√2x)+19√2e−xsin(√2x)+14sinx−14cosx−13x+29
解答:先求齊次解,即x3y‴−3x2y″+6xy′−6y=0,令y=xm⇒{y′=mxm−1y″=m(m−1)xm−2y‴=m(m−1)(m−2)xm−3⇒x3y‴−3x2y″+6xy′−6y=(m−1)(m−2)(m−3)xm=0⇒m=1,2,3⇒yh=c1x+c2x2+c3x3接著利用 variations of parameter 求yp,令{y1=xy2=x2y3=x3⇒W=|y1y2y3y′1y′2y′3y″1y″2y″3|=|xx2x312x3x2026x|=2x3⇒{W1=|0x2x302x3x2126x|=x4W2=|x0x3103x2016x|=−2x3W3=|xx2012x0021|=x2⇒yp=x∫x4lnx2x3dx+x2∫(−2x3)lnx2x3+x3∫x2lnx2x3dx=x2(12x2lnx−14x2)−x2(xlnx−x)+x32(12ln2(x))⇒yp=14x3ln2(x)−34x3ln(x)+78x3
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