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2023年12月10日 星期日

112年雲科大電子系碩士班-工程數學詳解

 



解答:
(a)xy+3y=2xy+3xy=2integrating factor I(x)=e(3/x)dx=x3x3y+3x2y=2x3(x3y)=2x3x3y=2x3dx=12x4+cy=12x+cx3(b)y+5y+6y=0λ2+5λ+6=0(λ+3)(λ+2)=0λ=3,2y=c1e3x+c2e2x(c)y=xmy=mxm1y=m(m1)xm2x2y+1.5xy0.5y=(m2m)xm+1.5mxm0.5xm=(m2+0.5m0.5)xm=012(2m1)(m+1)=0m=12,1y=c1x+c2xx
解答:{P(x,y)=3x2y+6xy+y22Q(x,y)=3x2+y{Py=3x2+6x+yQx=6xPyQxNOT exactPyQxQ=1 independent of xu=u integration factor u=ex(exP)dx+(exQ)dy=0potential Φ(x,y) satisfying{Φx=exPΦy=exQΦ=3x2yex+6xyex+y22exdx=3x2ex+yexdyΦ=3x2yex+y22ex+ϕ(y)=3x2yex+12y2ex+ρ(x)3x2yex+y22ex+c=0
解答:\先y+2y+y=0λ2+2λ+1=0(λ+1)2=0λ=1yh=c1ex+c2xex varation of parameter,{y1=exy2=xexr(x)=xexW=|y1y2y1y2|=e2xyp=y1y2rWdx+y2y1rWdx=exx2dx+xexxdx=16x3exy=yh+ypy=c1ex+c2xex+16x3ex


解答:(a)L{(t+2)2}=L{t2+4t+4}=2s3+4s2+4s(b)L1{3s+3}+L1{3ss2+5}=3e3t+3cos(5t)



解答:(a){u=[x1,x2,x3,x4]Tv=[y1,y2,y3,y4]Tau+bv=[ax1+by1,ax2+by2,ax3+by3,ax4+by4]TT(au+bv)=[ax1+by1,ax2+by2,ax3+by3,ax4+by4]T[2,0,2,1]T=2(ax1+by1)+2(ax3+by3)+(ax4+by4)aT(u)+bT(v)=a(2x1+2x3+x4)+b(2y1+2y3+y4)=2(ax1+by1)+2(ax3+by3)+(ax4+by4)T(au+bv)=aT(u)+bT(v)T is a linear transformationQ.E.D(b)A=[2,0,2,1](c)Ax=02x1+2x3+x4=0ker(A)={α(1002)+β(0100)+γ(0012)α,β,γR}

解答:det

解答:\mathbf{(a)}\; \det(A-\lambda I)=0 \Rightarrow (\lambda-{3\over 5})(\lambda-1)=0 \Rightarrow \lambda=1,{3\over 5} \\ \lambda_1={3\over 5} \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow v=k\begin{pmatrix} -3/2\\ 1\end{pmatrix},取v_1=\begin{pmatrix} -3/2\\ 1\end{pmatrix}\\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow v=k\begin{pmatrix} -1/2\\ 1\end{pmatrix},取v_2=\begin{pmatrix} -1/2\\ 1\end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{P=\begin{bmatrix} -3/2 & -1/2\\ 1& 1\end{bmatrix}, D=\begin{bmatrix} 3/5 & 0\\ 0 & 1\end{bmatrix}}\\ \mathbf{(b)}\; \lim_{n\to \infty}A^n = \lim_{n\to \infty}(PDP^{-1})^n = \lim_{n\to \infty}PD^nP^{-1} = \begin{bmatrix} -3/2 & -1/2\\ 1& 1\end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} \begin{bmatrix} -1 & -1/2\\ 1& 3/2\end{bmatrix} \\\qquad =\bbox[red, 2pt]{\begin{bmatrix} -1/2 & -3/4\\ 1 & 3/2\end{bmatrix}} \\\mathbf{(c)} \text{eigenvalues of }A^{-1}={1\over \lambda_1},{1\over \lambda_2} =\bbox[red, 2pt]{{5\over 3},1}

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