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:$$y''-y'-2y=0 \Rightarrow \lambda^2-\lambda-2=0 \Rightarrow (\lambda-2)(\lambda +1)=0 \Rightarrow \lambda=-1,2\\ \Rightarrow y_h=c_1e^{-x} +c_2e^{2x} \\ y_p= Ax^2+Bx+C \Rightarrow y_p'= 2Ax+B \Rightarrow y_p''=2A \\ \Rightarrow y_p''-y_p'-2y_p=-2Ax^2-2(A+B)x +2A-B-2C=2x^2+5 \\ \Rightarrow \cases{-2A=2\\ A+B=0\\ 2A-B-2C=5} \Rightarrow \cases{A=-1\\ B=1\\ C=-4} \Rightarrow y_p=-x^2+x-4\\ y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{-x}+ c_2e^{2x}-x^2+x-4}$$
解答:$$A=\begin{bmatrix}1 & 3 & 1 \\2 & 1 & 1 \\ -2 & 2 & -1 \end{bmatrix} \Rightarrow \det(A)=3 \ne 0 \\ B=\begin{bmatrix}1 & 3 & 0 \\2 & 1 & 0 \\-2 & 2 & 0 \end{bmatrix} \Rightarrow \det(B)=0\\ \Rightarrow \bbox[red, 2pt]{(a)} \text{ is invertible}$$
解答:$$\textbf{(a)}\;\text{The plane is the null space of the matrix }A=\begin{bmatrix}1 & 3 &-1\\ 0 & 0& 0 \\ 0 & 0& 0 \end{bmatrix}.\\ \text{The special solutions: }v_1=\begin{bmatrix}3 \\ -1 \\ 0 \end{bmatrix}, v_2= \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}. \\\text{Then }\{v_1, v_2\} = \bbox[red, 2pt]{\left\{ \begin{bmatrix}3 \\ -1 \\ 0 \end{bmatrix}, \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix} \right\}} \text{ is a basis for the plane} \\\textbf{(b)}\; x+3y-z=0 \Rightarrow \vec n=(1,3,-1) \Rightarrow L:{x\over 1}={y-1\over 3}={z-2\over -1}\\\quad P\in L \Rightarrow P(t,3t+1,-t+2) \Rightarrow P\in x+3y-z=0 \Rightarrow t+3(3t+1)-(-t+2)=0 \\ \Rightarrow t=-{1\over 11} \Rightarrow P= \bbox[red, 2pt]{(-{1\over 11},{8\over 11},{23\over 11})}$$
解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow x^2y''-2xy'+2y= m(m-1)x^m-2mx^m+2x^m= (m^2-3m+2)x^m=0\\ \Rightarrow m^2-3m+2 =(m-2)(m-1)=0 \Rightarrow m=2,1 \Rightarrow y_h=c_1 x+c_2 x^2 \\ \text{Let }\cases{y_1=x\\ y_2=x^2 \\ r(x)={\ln x+1 \over x^2}} \Rightarrow W=\begin{vmatrix}x & x^2 \\1 & 2x \end{vmatrix} =x^2\\ \text{By variation of parameters}, y_p =-x \int {(\ln x+1)\over x^2}\,dx +x^2 \int{ \ln x+1 \over x^3} \,dx \\ =x \cdot {\ln x+2 \over x} -x^2\cdot {2\ln x+3\over 4x^2} ={1\over 2}\ln x+{5\over 4} \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1 x+c_2 x^2 +{1\over 2}\ln x+{5\over 4}}$$
解答:$$A=\begin{bmatrix} 3 & 4 \\3 & 2 \end{bmatrix} = \begin{bmatrix}-1 & \frac{4}{3} \\1 & 1 \end{bmatrix} \begin{bmatrix}-1 & 0 \\0 & 6 \end{bmatrix} \begin{bmatrix}\frac{-3}{7} & \frac{4}{7} \\\frac{3}{7} & \frac{3}{7} \end{bmatrix} \\ \Rightarrow e^{At} = \begin{bmatrix} 3 & 4 \\3 & 2 \end{bmatrix} = \begin{bmatrix}-1 & \frac{4}{3} \\1 & 1 \end{bmatrix} \begin{bmatrix}e^{-t} & 0 \\0 & e^{6t} \end{bmatrix} \begin{bmatrix}\frac{-3}{7} & \frac{4}{7} \\\frac{3}{7} & \frac{3}{7} \end{bmatrix} =\begin{bmatrix} \frac{4e^\left(6t\right)+3e^{-t}}{7} & \frac{4e^\left(6t\right)-4e^{-t}}{7 } \\\frac{3e^\left(6t\right)-3e^{-t}}{7} & \frac{3e^\left(6t\right)+4e^{-t}}{7} \end{bmatrix} \\ \cases{y_1'=3y_1+4y_2\\ y_2'=3y_1+2y_2\\ y_1(0)=6\\ y_2(0)=1} \Rightarrow \mathbf y'=A\mathbf y, \mathbf y(0) =\begin{bmatrix} 6 \\1 \end{bmatrix} \Rightarrow \mathbf y=e^{At} \mathbf y(0) \\ \Rightarrow \mathbf y= \begin{bmatrix} \frac{4e^\left(6t\right)+3e^{-t}}{7} & \frac{4e^\left(6t\right) -4e^{-t}}{7 } \\ \frac{3e^\left(6t \right)-3e^{-t}}{7} & \frac{3e^\left(6t\right)+4e^{-t}}{7} \end{bmatrix} \begin{bmatrix}6 \\1 \end{bmatrix} =\begin{bmatrix} {4e^\left(6t\right)+2e^{-t}} \\ 3e^\left(6t \right)-2e^{-t} \end{bmatrix} \Rightarrow \bbox[red, 2pt] {\cases{y_1(t)= 4e^{6t} +2e^{-t}\\ y_2(t)= 3e^{6t} -2e^{-t}}}$$
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