國立臺北科技大學110學年度碩士班招生考試
系所組別:車輛工程
科目:工程數學
解答: {y′1=−3y1+y2−6e2ty′2=y1−3y2+2e2t⇒y′=Ay+bA=[−311−3]=[−1111][−400−2][−12121212]⇒etA=[−1111][e−4t00e−2t][−12121212]homogeneous solution for y′=Ay⇒yh=etAy0⇒yh=[−1111][e−4t00e−2t][−12121212][c1c2]=[−1111][e−4t00e−2t][c3c4]⇒yh=[−e−4te−2te−4te−2t][c3c4]Apply variation of parameters, yp=[−e−4te−2te−4te−2t]∫[−e−4te−2te−4te−2t]−1[−6e2t2e2t]dt=[−e−4te−2te−4te−2t]∫[−e4t2e4t2e2t2e2t2][−6e2t2e2t]dt=[−e−4te−2te−4te−2t]∫[4e6t−2e4t]dt=[−e−4te−2te−4te−2t][23e6t−12e4t]=[−76e2t16e2t]⇒y=yh+yp⇒y=[−e−4te−2te−4te−2t][c3c4]+[−76e2t16e2t]解答: y=∞∑n=0anxn⇒y′=∞∑n=0nanxn−1⇒y″=∞∑n=0n(n−1)anxn−2y(0)=y′(0)=1⇒a0=a1=1⇒y″+4xy′+2y=∞∑n=0((4n+2)an+(n+2)(n+1)an+2)xn=0⇒an+2=−4n+2(n+2)(n+1)an,n=0,1,2,⋯⇒{a0=a1=1an+2=(2n+1−6n+2)an,n=0,1,2,…⇒y=1+x−x2−x3+56x4+1320x5+⋯
解答: F(x1,x2)=7x21+6x1x2+7x22=[x1,x2][7337][x1x2]≡xtAxA=[7337]=[−1/√21/√21/√21/√2][40010][−1/√21/√21/√21/√2]≡QDQt⇒F(x1,x2)=xtAx=xtQDQtx=(Qtx)tD(Qtx)≡ytDy⇒[y1y2]=[−1/√21/√21/√21/√2][x1x2]⇒[x1x2]=[−1/√21/√21/√21/√2][y1y2]⇒{x1=(y2−y1)/√2x2=(y2+y1)/√2F(y1,y2)=[y1,y2][40010][y1y2]=4y21+10y22=200⇒y2150+y2220=1,ellipse
解答: C1:(0,0,0)→(0,1,0)⇒{x(t)=0y(t)=tz(t)=0,0≤t≤1⇒{dx=0dy=dtdz=0⇒∫c1x2ydx+(x3+1)dy+9z2dz=∫101dt=1C2:(0,1,0)→(1,1,1)⇒{x(t)=ty(t)=1z(t)=t,0≤t≤1⇒{dx=dtdy=0dz=dt⇒∫c1x2ydx+(x3+1)dy+9z2dz=∫10t2dt+9t2dt=103⇒∫(1,1,1)(0,0,0)x2ydx+(x3+1)dy+9z2dz=1+103=133
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解題僅供參考,其他歷年試題及詳解
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