115年台北科技大學電機碩士班-工程數學詳解

國立臺北科技大學115學年度碩士班招生考試

系所組別:電機碩士班丙組 科目:工程數學

解答:$$\cases{P(x,y)=y\\ Q(x,y)=2xy-e^{-2y}} \Rightarrow \cases{P_y=1\\ Q_x=2y} \Rightarrow P_y \ne Q_x \Rightarrow \text{Not Exact} \\ {P_y-Q_x\over P} ={1-2y\over y} \text{ depends on $y$ only} \Rightarrow u'={2y-1\over y}u \Rightarrow {1\over u}\,du = \left( 2-{1\over y} \right)\,dy \\ \Rightarrow \int {1\over u}\,du = \int\left( 2-{1\over y} \right)\,dy \Rightarrow \ln u= 2y-\ln y \Rightarrow \text{integration factor }u={e^{2y} \over y} \\ \Rightarrow \cases{uP =e^{2y} \\uQ=2xe^{2y}-{1\over y}}  \Rightarrow (uP)_y=2e^{2y} =(uQ)_x \Rightarrow \text{ Exact} \\ \Rightarrow \Psi(x,y)= \int e^{2y}\,dx = \int \left( 2xe^{2y}-{1\over y} \right)\,dy \Rightarrow \Psi(x,y)=xe^{2y}+\rho(y)=xe^{2y}-\ln |y|+ \phi(x) \\ \Rightarrow \Psi(x,y)= \bbox[red, 2pt]{xe^{2y}-\ln |y|=C}$$

解答:$$v={y\over x} \Rightarrow y=vx \Rightarrow y'=v+v'x \Rightarrow (v+v'x)-v=\sec v \Rightarrow xv'=\sec v \\ \Rightarrow \int \cos v\,dv= \int{1\over x}\,dx \Rightarrow \sin v=\ln |x|+C \Rightarrow \bbox[red, 2pt]{\sin {y\over x}= \ln|x|+C}$$

解答:$$\det(A)= \begin{vmatrix} 2 & 0 & 1 & 3 & -2 \\ -2 & 1 & 3 & 2 & -1 \\ 1 & 0 & -1 & 2 & 3 \\ 3 & -1 & 2 & 4 & -3 \\ 1 & 1 & 3 & 2 & 0 \end{vmatrix} \xrightarrow{R_4+R_2\to R_4\\ R_5-R_2\to R_5}   \begin{vmatrix} 2 & 0 & 1 & 3 & -2 \\ -2 & 1 & 3 & 2 & -1 \\ 1 & 0 & -1 & 2 & 3 \\ 1 & 0 & 5 & 6 & -4 \\ 3 & 0 & 0 & 0 & 1 \end{vmatrix}  \\=  1 \cdot \begin{vmatrix} 2 & 1 & 3 & -2 \\ 1 & -1 & 2 & 3 \\ 1 & 5 & 6 & -4 \\ 3 & 0 & 0 & 1 \end{vmatrix} \xrightarrow{R_1+R_2\to R_2, R_3-5R_1\to R_3} \begin{vmatrix} 2 & 1 & 3 & -2 \\ 3 & 0 & 5 & 1 \\ -9 & 0 & -9 & 6 \\ 3 & 0 & 0 & 1 \end{vmatrix} =- \begin{vmatrix} 3& 5& 1\\-9& -9& 6\\3& 0& 1 \end{vmatrix} \\= 3\begin{vmatrix} 3& 5& 1\\3& 3& -2\\3& 0& 1 \end{vmatrix}= 3\times (-45)= \bbox[red, 2pt]{-135}$$
 

解答:$$\cases{T(1,0,0)= (2,0) \\ T(0,1,0)=(0,-3) \\ T(0,0,1) =(0,0)} \Rightarrow   \text{the matrix $A$ of representation of }T   = \bbox[red, 2pt]{\begin{bmatrix}2& 0& 0\\ 0&-3& 0 \end{bmatrix}} \\\Rightarrow  A \begin{bmatrix}1\\2\\3  \end{bmatrix} = \begin{bmatrix}2\\-6 \end{bmatrix} \Rightarrow T(1,2,3)= \bbox[red, 2pt]{(2,-6)}$$

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解題僅供參考，碩士班歷年試題及詳解