108年北科大電機碩士班丙組-工程數學詳解

 國立臺北科技大學108學年度碩士班招生考試

系所組別:2131 電機工程系碩士班丙組
第一節 工程數學 試題(選考)

解答: $$\cases{P(x,y)= 4xy^2-2\\ Q(x,y)= y^3} \Rightarrow \cases{P_x=4y^2\\ Q_y= 3y^2} \Rightarrow P_x \ne Q_y \Rightarrow \text{NonExact} \\ \Rightarrow u'=-{Q_y-P_x\over Q}u ={1\over y}u \Rightarrow ({1\over y} u)'=0 \Rightarrow {1\over y}u=c_1 \Rightarrow 積分因子u=y \\ \Rightarrow \cases{uP=4xy^3-2y\\ uQ=y^4} \Rightarrow (uP)_x=4y^3=(uQ)_y  \Rightarrow \text{Exact} \\ \Rightarrow \Phi(x,y)=\int (4xy^3-2y)\,dy = \int y^4\,dx \Rightarrow xy^4-y^2+ \rho(x) =xy^4 +\phi(y) \\ \Rightarrow \Phi(x,y)= \bbox[red, 2pt] {xy^4-y^2 +c_1=0}$$

解答: $$y''+4y'+4y=0 \Rightarrow \lambda^2+ 4\lambda+4=0 \Rightarrow (\lambda+2)^2=0 \Rightarrow \lambda=-2 \Rightarrow y_h= e^{-2x}(c_1+ c_2x) \\ \text{Let }\cases{y_1=e^{-2x } \\y_2= xe^{-2x}}, \text{ then} \cases{y_1'=-2e^{-2x} \\ y_2'=e^{-2x}-2xe^{-2x}} \Rightarrow W= \begin{vmatrix}y_1& y_2\\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} e^{-2x}& xe^{-2x}\\ -2e^{-2x} & e^{-2x}-2xe^{-2x} \end{vmatrix} = e^{-4x} \\ \text{By variations of parameters, } y_p =-e^{-2x} \int{ xe^{-2x} \cdot e^{-2x}\ln x \over e^{-4x}} \,dx +xe^{-2x} \int {e^{-2x}\cdot e^{-2x}\ln x\over e^{-4x}}\,dx \\ \Rightarrow y_p=-e^{-2x} \int x\ln x\,dx +xe^{-2x} \int \ln x\,dx =-e^{-2x}({1\over 2}x^2 \ln x-{1\over 4}x^2) +xe^{-2x}(x\ln x-x) \\ \Rightarrow y_p={1\over 2}x^2 e^{-2x}\ln x -{3\over 4}x^2e^{-2x} \Rightarrow y= y_h+y_p \Rightarrow \bbox[red, 2pt]{y= e^{-2x}(c_1+ c_2x) +{1\over 2}x^2 e^{-2x}\ln x -{3\over 4}x^2e^{-2x}}$$

解答: $$\begin{bmatrix}\frac{d x_1(t)}{dx} \\\frac{d  x_2(t)}{dx}\end{bmatrix} = \begin{bmatrix} \sigma& \omega \\ -\omega & \sigma \end{bmatrix} \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} \Rightarrow  \cases{x_1'(t)=\sigma x_1(t)+ \omega x_2(t) \\ x_2'(t) =-\omega x_1(t)+ \sigma x_2(t)}  \Rightarrow  \cases{L\{x_1'(t)\}= L\{\sigma x_1(t)+ \omega x_2(t) \}\\ L\{x_2'(t) \}= L\{-\omega x_1(t)+ \sigma x_2(t) \}} \\ \Rightarrow \cases{sX_1(s)-X_1(0) = \sigma X_1(s) + \omega X_2(s) \\ sX_2(s)-X_2(0) =-\omega X_1(s)+ \sigma X_2(s)} \Rightarrow \cases{(s-\sigma)X_1(s)= \omega X_2(s)+ A\\ (s-\sigma) X_2(s)= -\omega X_1(s)+ B}  \\ \Rightarrow X_1(s)={B\over \omega}-{s-\sigma\over \omega}X_2(s) ={B\over \omega}-{s-\sigma\over \omega}\left( {s-\sigma\over \omega}X_1(s)-{A\over \omega}\right) \\ \Rightarrow \left(1+{(s-\sigma)^2 \over \omega^2} \right)X_1(s)= {B\over \omega}+{A(s-\sigma) \over \omega^2} \Rightarrow X_1(s)= {1\over \omega^2+ (s-\sigma)^2} (B\omega+ A(s-\sigma))\\ \Rightarrow x_1(t) =L^{-1} \{X_1(s)\} = Be^{\sigma t} \sin(\omega t) +Ae^{\sigma t} \cos(\omega t)\\ \Rightarrow x_1'(t) =(A\sigma+ B\omega)e^{\sigma t} \cos(\omega t)+ (B\sigma -A\omega)e^{\sigma t} \sin(\omega t) \\ \Rightarrow x_2(t) ={1\over \omega}(x_1'(t)- \sigma x_1(t)) =B e^{\sigma t} \cos(\omega t)-Ae^{\sigma t} \sin(\omega t) \\ \Rightarrow \bbox[red, 2pt]{\cases{x_1(t)= e^{\sigma t}(A \cos(\omega t)+ B  \sin(\omega t)) \\ x_2(t) =e^{\sigma t}(B\cos (\omega t)-A\sin(\omega t))}}$$

解答: $$\textbf{(a) } \text{Let }A= \begin{bmatrix} 0& 0 & 0 \\0& 0 & 0 \\0& 0 & 0 \end{bmatrix}, \text{ then }-A^T=A. \text{ That is, }A \text{ is a 3x3 skew-symmetrix matrix.} \\ \text{Suppose }A \in S, \text{ then }A=-A^T \Rightarrow cA=-(cA)^T \Rightarrow cA \in S.\\\text{Suppose }A ,B\in S, \text{ then }A=-A^T \text{ and }B=-B^T \Rightarrow A+B=-A^T-B^T=-(A+B)^T \\ \Rightarrow A+B \in S. \\\text{Therefore, }V \text{ is a subspace of }V. \bbox[red, 2pt]{QED.} \\\textbf{(b) } A\in S \Rightarrow A= \begin{bmatrix} 0& a_{1,2} & a_{1,3} \\-a_{1,2}& 0 & a_{2,3} \\-a_{1,3}& -a_{2,3} & 0 \end{bmatrix} =a_{1,2} \begin{bmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0& 0 &0 \end{bmatrix} +a_{1,3} \begin{bmatrix}0 & 0 & 1\\ 0 & 0 & 0\\ -1& 0 &0 \end{bmatrix} +a_{2,3} \begin{bmatrix}0 & 0 & 0\\ 0 & 0 & 1\\ 0& -1 &0 \end{bmatrix} \\, \text{ then } \bbox[red, 2pt]{\left\{ \begin{bmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0& 0 &0 \end{bmatrix}, \begin{bmatrix}0 & 0 & 1\\ 0 & 0 & 0\\ -1& 0 &0 \end{bmatrix} ,\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0& -1 &0 \end{bmatrix} \right\}} \text{ is a basis }$$

解答: $$\textbf{(a) }A=\begin{bmatrix} 2 & -4 & 4 \\0 & 4 & -1 \\0 & -1 & 4\end{bmatrix} \Rightarrow \det(A-\lambda I) =-\lambda^3+10 \lambda^2-31\lambda+30 = -(\lambda-2) (\lambda-3) (\lambda-5)=0 \\\quad \Rightarrow \bbox[red, 2pt] {\text{eigenvalues: }2,3,5} \\ \textbf{(b) }\lambda_1=2 \Rightarrow (A-\lambda_1 I) v=0  \Rightarrow \begin{bmatrix} 0 & -4 & 4 \\ 0 & 2 & -1 \\ 0 & -1 & 2\end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_2=0\\ x_3=0} \\\qquad \Rightarrow v= x_1 \begin{pmatrix}1\\0\\0 \end{pmatrix}, \text{choose }v_1= \begin{pmatrix} 1\\0\\0 \end{pmatrix}\\ \lambda_2= 3 \Rightarrow (A-\lambda_2 I) v=0  \Rightarrow \begin{bmatrix} -1 & -4 & 4 \\ 0 & 1 & -1 \\0 & -1 & 1\end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3\end{bmatrix}=0 \Rightarrow \cases{x_1=0\\ x_2=x_3} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} 0\\1\\1 \end{pmatrix}, \text{choose }v_2=\begin{pmatrix}0 \\1\\1 \end{pmatrix} \\ \lambda_3= 5 \Rightarrow (A-\lambda_3 I) v=0  \Rightarrow \begin{bmatrix} -3 & -4 & 4 \\0 & -1 & -1 \\0 & -1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3\end{bmatrix}=0 \Rightarrow \cases{3x_1= 8x_3\\ x_2+x_3=0} \\\qquad \Rightarrow v= x_3 \begin{pmatrix} 8/3\\-1\\1 \end{pmatrix}, \text{choose }v_3 =\begin{pmatrix} 8/3 \\ -1\\1 \end{pmatrix} \\ \Rightarrow P=[v_1 v_2 v_3] \Rightarrow \bbox[red, 2pt]{P= \begin{bmatrix} 1 & 0 & \frac{8}{3} \\0 & 1 & -1 \\0 & 1 & 1\end{bmatrix}}$$

解答: $$\cases{4x_1-2x_2+ 3x_3=0\\ 3x_1-4x_2+ 2x_3=0 \\ 6x_1+2 x_2+5x_3 =0} \Rightarrow \begin{bmatrix} 4& -2& 3\\ 3& -4 & 2\\ 6& 2 & 5 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=0 \\ \text{Let }A= \begin{bmatrix} 4& -2& 3\\ 3& -4 & 2\\ 6& 2 & 5 \end{bmatrix}, \text{ then }rref(A)= \begin{bmatrix} 1 & 0 & \frac{4}{5}\\0 & 1 & \frac{1}{10}\\0 & 0 & 0 \end{bmatrix} \Rightarrow \cases{x_1+ {4\over 5}x_3 =0\\ x_2+{1\over 10}x_3=0} \\ \Rightarrow \text{solutions of }A\mathbf x=0: \{k(-{4\over 5},-{1\over 10},1) \mid k\in \mathbb R\} \\ \Rightarrow \cases{\text{ a basis: }\bbox[red, 2pt]{\{(8,1,-10)\}} \\ \text{nullity}(A)=\bbox[red, 2pt]1}$$

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解題僅供參考，碩士班歷年試題及詳解