台北科大機械碩士乙-工程數學詳解

國立臺北科技大學113學年度碩士班招生考試

系所組別: 機械工程系機電整合碩士班乙組
第一節 工程數學

解答:$$\textbf{1.}\;xy'＋y＝0 \Rightarrow (xy)'=0 \Rightarrow xy=c_1 \Rightarrow \bbox[red, 2pt]{y={c_1\over x}} \\ \textbf{2.}\; \cases{P(x,y)=x^2+y^2 \\Q(x,y)= 2xy} \Rightarrow P_x=2x=Q_y \Rightarrow \text{exact} \\\quad \Rightarrow \Phi(x,y)=\int (x^2+y^2)\,dx = \int 2xy \,dy \Rightarrow \Phi={1\over 3}x^3+xy^2+ \phi(y)= xy^2+ \rho(x) \\\quad \Rightarrow \bbox[red, 2pt]{{1\over 3}x^3+xy^2+c_1=0}$$

解答:$$\textbf{1.}\; y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\\quad \Rightarrow x^2y''- xy+y = m(m-1)x^m-mx^m+x^m=(m^2-2m+1)x^m=0 \\\quad \Rightarrow m^2-2m+1=0 \Rightarrow (m-1)^2=0 \Rightarrow m=1 \Rightarrow \bbox[red, 2pt]{y=c_1x+ c_2x\ln x} \\\textbf{2.}\; y''-2y'+y=0 \Rightarrow \lambda^2-2\lambda+1=0 \Rightarrow (\lambda-1)^2=0 \Rightarrow \lambda=1 \Rightarrow y_h=c_1e^x +c_2xe^x \\\quad y_p=Ax^2e^x \Rightarrow y_p'=2Axe^x+ Ax^2e^x \Rightarrow y_p''= 2Ae^x+ 4Axe^x  + Ax^2e^x  \\ \quad \Rightarrow y_p''-2y_p'+y_p= 2Ae^x = e^x \Rightarrow A={1\over 2} \Rightarrow y_p={1\over 2}x^2e^x \Rightarrow y=y_h+y_p \\\quad \Rightarrow  \bbox[red, 2pt]{y=c_1e^x +c_2xe^x+{1\over 2}x^2e^x}$$

解答:$$\textbf{1.}\; L^{-1}\{F(s)\} =-{1\over t} L^{-1}\{F'(s)\} =-{1\over t} L^{-1}\{ {1\over s+2}-{1\over s+1}\} =-{1\over t}(e^{-2t}-e^{-t}) \\ \quad \Rightarrow f(t)= \bbox[red, 2pt]{{1\over t}(e^{-t}-e^{-2t})} \\\textbf{2.} \;L\{y''\}-16L\{y\}=L\{ u(t-2\pi)\} =s^2Y(s)-16Y(s)= {e^{-2\pi s}\over s} \Rightarrow Y(s)= {e^{-2\pi s}\over s(s^2-4)} \\\quad \Rightarrow y(t)=L^{-1}\{ Y(s)\} =L^{-1}\left\{{e^{-2\pi s}\over s(s^2-4)} \right\} =L^{-1}\left\{e^{-2\pi s}\left(-{1\over 4}\cdot {1\over s} +{1\over 8}\cdot {1\over s-2} +{1\over 8}\cdot  {1\over s+2}\right) \right\} \\ \quad \Rightarrow \bbox[red, 2pt]{y(t)=u(t-2\pi) \left(-{1\over 4}+{1\over 8}e^{-2(t-2\pi) } +{1\over 8}e^{2(t-2\pi)}\right)}$$

解答:$$\textbf{1.}\; A=\begin{bmatrix}3 & 4 \\-1 & 7 \end{bmatrix} \Rightarrow \det(A-\lambda I)=(\lambda-5)^2=0 \Rightarrow \lambda=5\\ \quad \Rightarrow (A-\lambda I)v=0 \Rightarrow \begin{bmatrix}-2 & 4 \\-1 & 2 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} =0 \Rightarrow x_1=2x_2 \Rightarrow v=x_2\begin{pmatrix}2 \\1\end{pmatrix} \\\Rightarrow \bbox[red, 2pt]{特徵值:5,特徵向量:\begin{pmatrix}2 \\1\end{pmatrix}} \\\textbf{2.}\; rank(A)=2但只有一個特徵向量,因此\bbox[red, 2pt]{A無法對角化} \\\textbf{3.}\; A=D+N= \begin{bmatrix}5 & 0\\0 & 5 \end{bmatrix} +\begin{bmatrix}-2 & 4 \\-1 & 2 \end{bmatrix} \Rightarrow N^2=0 \Rightarrow e^N=I +N =\begin{bmatrix}-1 & 4 \\-1 & 3 \end{bmatrix} \\\quad e^A=e^{D+N} =e^D\cdot e^N= \begin{bmatrix} e^5 & 0 \\0 & e^5 \end{bmatrix}  \begin{bmatrix}-1 & 4 \\-1 & 3 \end{bmatrix}= \bbox[red, 2pt]{ \begin{bmatrix}-e^5 & 4e^5 \\-e^5 & 3e^5 \end{bmatrix}}$$

解答:$$\cases{f(x,y,z)=x^2+y^2+z^2-2\\ g(x,y,z)= x^2+y^2-2-z} \Rightarrow \cases{\nabla f=(2x,2y,2z)\\ \nabla g=(2x,2y,-1)} \Rightarrow \cases{\vec u=\nabla f(1,-1,0)= (2,-2,0)\\ \vec v= \nabla g(1,-1,0)= (2,-2,-1)} \\ \Rightarrow \cos \theta ={\vec u\cdot \vec v\over |\vec u|| \vec v|} ={8\over \sqrt 8\cdot \sqrt 9} ={2\over 3}\sqrt 2 \Rightarrow \bbox[red, 2pt]{\theta=\cos^{-1} {2\over 3}\sqrt 2}$$

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