111年北科大車輛工程碩士班-工程數學詳解

國立臺北科技大學111學年度碩士班招生考試

系所組別: 1311 、 1312 、 1313 、 1321 、 1322 、 1323 車輛工程系碩士班甲 、乙組
第一節 工程數學 試題

解答:$$\textbf{(a)} \; v=x+y \Rightarrow v'=1+y' \Rightarrow y'=\sqrt{x+y} \Rightarrow v'-1=\sqrt v\\ \Rightarrow {1\over \sqrt v+1}dv =1\,dx \Rightarrow 2(\sqrt v+1)-2\ln(\sqrt v+1)=x+c_1 \\\Rightarrow  \bbox[red, 2pt]{\sqrt{x+y}+1 - \ln(\sqrt{x+y}+1)={1\over 2}x+c_2}\\ \textbf{(b)} \;\text{Intergation factor }I(x)=e^{\int \tan x\,dx}=\sec x \Rightarrow y'\sec x+ y\tan x\sec x  =\sec x\sin(2x) \\ \Rightarrow (y\sec x)'=2\sin x \Rightarrow y\sec x=\int 2\sin x\,dx =-2\cos x+c_1 \Rightarrow \bbox[red, 2pt]{y=-2\cos^2 x+ c_1\cos x}$$

解答:$$\textbf{ (a)}\;y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda +2=0 \Rightarrow (\lambda-2)(\lambda-1)=0 \Rightarrow \lambda=1,2 \\ \quad \Rightarrow \text{Homogeneous solution } \bbox[red, 2pt]{y_h=c_1e^x +c_2e^{2x}} \\ \textbf{(b)}\; y_p=Axe^x \Rightarrow y_p'= Ae^x+ Axe^x \Rightarrow 2Ae^x+ Axe^x \\\quad \Rightarrow   y_p''-3y_p'+2y_p= -Ae^x = e^x \Rightarrow A=-1 \Rightarrow \text{Particular solution } \bbox[red, 2pt]{y_p=-xe^x} \\\textbf{(c)}\; \text{General solution }y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y_h=c_1e^x +c_2e^{2x}-xe^x}$$

解答:$$y=\sum_{n=0}^\infty a_nx^n \Rightarrow y'=\sum_{n=0}^\infty na_nx^{n-1} \Rightarrow y''=\sum_{n=0}^\infty n(n-1)a_nx^{n-2} \\ \Rightarrow y''+xy'+y= \sum_{n=0}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=0}^\infty (n+1)a_nx^{n}\\  =\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n+ \sum_{n=0}^\infty (n+1)a_nx^{n}=0 \\  \Rightarrow (n+2)(n+1)a_{n+2}+(n+1)a_n=0 \Rightarrow a_{n+2} =-{ 1\over n+2}a_n,n\ge 0\\ \Rightarrow \begin{cases} a_{2n}=(-1)^n  {1\over  \prod_{k=1}^n (2k)} a_0 \\ a_{2n+1}=(-1)^n  {1\over  \prod_{k=1}^n (2k+1) }a_1\end{cases},n\ge 1 \\ \Rightarrow \bbox[red, 2pt]{y=a_0+a_1x+ \sum_{n=1}^\infty (-1)^n \left( {1\over  \prod_{k=1}^n (2k)} a_0 x^{2n} +  {1\over  \prod_{k=1}^n (2k+1) }a_1 x^{2n+1}\right)}$$

解答:$$L\{ y''\}+ 3L\{ y'\}+ 2L\{ y\} =L\{ u(x+1) \} \Rightarrow s^2Y(s)-1+3sY(s)+2Y(s)={e^s\over s} \\ \Rightarrow Y(s)={1\over s^2+3s+2}\left({e^s\over s}+1 \right) \Rightarrow y(x)=L^{-1} \{Y(s)\} =L^{-1}\left\{ {1\over (s+2)(s+1)}\left({e^s\over s}+1 \right) \right\} \\  =L^{-1}\left\{ {1\over (s+2)(s+1)}\left({e^s\over s}+1 \right) \right\} =u(x+1)L^{-1}\left\{ {1\over s(s+2)(s+1)}  \right\}(x+1)+L^{-1}\left\{ {1\over (s+2)(s+1)}\right\} \\ \Rightarrow \bbox[red, 2pt]{y(x)=u(x+1)\left({1\over 2}-e^{x+1} +{1\over 2}e^{-2(x+1)} \right) +e^{-x}-e^{-2x}}$$

解答:$$A=\begin{bmatrix}1 &1  & 1 \\0 & 0 & 1 \\1 & 1 & 0\end{bmatrix} \Rightarrow \det(A-\lambda I) = -\lambda(\lambda+1)(\lambda-2) =0 \Rightarrow \lambda=0,-1,2 \\ \lambda_1=0 \Rightarrow (A- \lambda_1 I)v =0 \Rightarrow \begin{bmatrix}1 &1  & 1 \\0 & 0 & 1 \\1 & 1 & 0\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{x_1+x_2=0 \\ x_3=0} \Rightarrow v=\begin{bmatrix}-x_2 \\x_2 \\ 0 \end{bmatrix} \\\qquad,取v_1= \begin{bmatrix}-1 \\1 \\ 0 \end{bmatrix} \\ \lambda_2=-1\Rightarrow (A- \lambda_2 I)v =0 \Rightarrow \begin{bmatrix}2 &1  & 1 \\0 & 1 & 1 \\1 & 1 & 1\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{x_1 =0 \\ x_2+x_3=0} \Rightarrow   v=\begin{bmatrix}0 \\-x_2 \\ x_3 \end{bmatrix} \\\qquad,取v_2= \begin{bmatrix}0 \\-1 \\ 1 \end{bmatrix} \\ \lambda_3=2 \Rightarrow (A- \lambda_3 I)v =0 \Rightarrow \begin{bmatrix}-1 &1  & 1 \\0 & -2 & 1 \\1 & 1 & -2\end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow \cases{2x_1 =3x_3 \\ 2x_2=x_3} \Rightarrow   v=\begin{bmatrix}3x_3/2 \\x_3/2 \\ x_3 \end{bmatrix}\\\qquad ,取v_3= \begin{bmatrix}3/2 \\ 1/2 \\ 1 \end{bmatrix} \\ \Rightarrow \text{eigenvalues: }\bbox[red, 2pt]{0,-1,2}\text{ and eigenvectors: }\bbox[red, 2pt]{\begin{bmatrix}-1 \\1 \\ 0 \end{bmatrix} , \begin{bmatrix}0 \\-1 \\ 1 \end{bmatrix}, \begin{bmatrix}3/2 \\ 1/2 \\ 1 \end{bmatrix} }$$
 

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解題僅供參考,其他歷年試題及詳解