國立臺北科技大學113學年度碩士班招生考試
系所組別: 自動化科技研究所
第一節 工程數學
解答:y″−y′−2y=0⇒λ2−λ−2=0⇒(λ−2)(λ+1)=0⇒λ=−1,2⇒yh=c1e−x+c2e2xyp=Ax2+Bx+C⇒y′p=2Ax+B⇒y″p=2A⇒y″p−y′p−2yp=−2Ax2−2(A+B)x+2A−B−2C=2x2+5⇒{−2A=2A+B=02A−B−2C=5⇒{A=−1B=1C=−4⇒yp=−x2+x−4y=yh+yp⇒y=c1e−x+c2e2x−x2+x−4解答:A=[131211−22−1]⇒det(A)=3≠0B=[130210−220]⇒det(B)=0⇒(a) is invertible
解答:(a)The plane is the null space of the matrix A=[13−1000000].The special solutions: v1=[3−10],v2=[101].Then {v1,v2}={[3−10],[101]} is a basis for the plane(b)x+3y−z=0⇒→n=(1,3,−1)⇒L:x1=y−13=z−2−1P∈L⇒P(t,3t+1,−t+2)⇒P∈x+3y−z=0⇒t+3(3t+1)−(−t+2)=0⇒t=−111⇒P=(−111,811,2311)
解答:y=xm⇒y′=mxm−1⇒y″=m(m−1)xm−2⇒x2y″−2xy′+2y=m(m−1)xm−2mxm+2xm=(m2−3m+2)xm=0⇒m2−3m+2=(m−2)(m−1)=0⇒m=2,1⇒yh=c1x+c2x2Let {y1=xy2=x2r(x)=lnx+1x2⇒W=|xx212x|=x2By variation of parameters,yp=−x∫(lnx+1)x2dx+x2∫lnx+1x3dx=x⋅lnx+2x−x2⋅2lnx+34x2=12lnx+54⇒y=yh+yp⇒y=c1x+c2x2+12lnx+54
解答:A=[3432]=[−14311][−1006][−37473737]⇒eAt=[3432]=[−14311][e−t00e6t][−37473737]=[4e(6t)+3e−t74e(6t)−4e−t73e(6t)−3e−t73e(6t)+4e−t7]{y′1=3y1+4y2y′2=3y1+2y2y1(0)=6y2(0)=1⇒y′=Ay,y(0)=[61]⇒y=eAty(0)⇒y=[4e(6t)+3e−t74e(6t)−4e−t73e(6t)−3e−t73e(6t)+4e−t7][61]=[4e(6t)+2e−t3e(6t)−2e−t]⇒{y1(t)=4e6t+2e−ty2(t)=3e6t−2e−t
==================== END ======================
解題僅供參考,其他歷年試題及詳解
沒有留言:
張貼留言