114年台北科大車輛工程碩士班-工程數學詳解

國立臺北科技大學114學年度碩士班招生考試

系所組別:1301、1302、1303 車輛工程系碩士班
第一節 工程數學 試題

解答: $$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^m-2\\ \Rightarrow x^2y''+ 1.5xy'-0.5y= m(m-1)x^m+1.5 mx^m -0.5x^m = (m^2+0.5m-0.5)x^m=0\\ \Rightarrow m^2+0.5m-0.5 =0 \Rightarrow 0.5(2m-1)(m+1)=0 \Rightarrow m=1/2,-1\\ \Rightarrow \bbox[red, 2pt] {y=c_1 \sqrt x+{c_2\over x}}$$

解答: $$y''+0.4y'+9.04y=0 \Rightarrow \lambda^2+0.4\lambda+9.04 =0 \Rightarrow \lambda=-0.2\pm 3 i\\ \Rightarrow y=e^{-0.2x} (c_1 \cos 3x+ c_2\sin 3x) \Rightarrow y(0)=c_1=0 \Rightarrow y=c_2e^{-0.2x} \sin 3x \\\Rightarrow y'=-0.2c_2 e^{-0.2x} \sin 3x+ 3c_2e^{-0.2x} \cos 3x \Rightarrow y'(0)=3c_2=3 \Rightarrow c_2=1 \Rightarrow \bbox[red, 2pt]{y=e^{-0.2x} \sin 3x}$$

解答: $$v={1\over y} \Rightarrow v'=-{1\over y^2}y' \Rightarrow y'= -y^2v' \Rightarrow -y^2v'=Ay-By^2 \Rightarrow -v'={A\over y}-B =Av-B\\ \Rightarrow v'+Av=B \Rightarrow e^{Ax}v'+ Ae^{Ax}v=Be^{Ax} \Rightarrow \left( e^{Ax}v\right)'=Be^{Ax} \\\Rightarrow e^{Ax}v =\int Be^{Ax}\,dx ={B\over A}e^{Ax}+c_1 \Rightarrow v={1\over y}={B\over A}+c_1e^{-Ax} \Rightarrow \bbox[red, 2pt] {y={1\over {B\over A}+c_1e^{-Ax}}}$$

解答: $$L\{y''\}+ L\{y'\} +9L\{y\}=0 \Rightarrow s^2Y(s)-sy(0)-y'(0)+ sY(s)-y(0)+9Y(s)=0 \\ \Rightarrow (s^2+s+9)Y(s)=0.16(s+1) \Rightarrow Y(s)= {0.16(s+1) \over s^2+s+9} \Rightarrow y(t)= L^{-1}\{Y(s)\} \\ \Rightarrow y(t) =0.16L^{-1}\left\{{s+1/2\over (s+1/2)^2+35/4} \right\} +{0.16\over 2} L^{-1}\left\{ {1\over (s+1/2)^2+35/4}\right\} \\ =0.16 e^{-t/2} \cos(\sqrt{35}t/2)+ 0.08 e^{-t/2} \cdot {2\over \sqrt{35}}\sin(\sqrt{35}t/2) \\ \Rightarrow \bbox[red, 2pt]{ y(t)=e^{-t/2}\left( \cos{\sqrt{35}t\over 2}+ {0.16\over \sqrt{35}} \sin {\sqrt{35}t\over 2}\right)}$$

解答: $$B=\begin{bmatrix} 1& 0& 2\\ 2& -1& 3\\ 4& 1& 8\end{bmatrix} \\ \textbf{(a) }\det(B) =-8+4+8-3= \bbox[red, 2pt]1 \\\text{(b) } [B\mid I] =\left[ \begin{array}{rrr|rrr}1 & 0 & 2 & 1 & 0 & 0\\2 & -1 & 3 & 0 & 1 & 0\\4 & 1 & 8 & 0 & 0 & 1 \end{array} \right] \xrightarrow{R_2-2R_1\to R_2, R_3-4R_1\to R_3} \left[ \begin{array}{rrr|rrr}1 & 0 & 2 & 1 & 0 & 0\\0 & -1 & -1 & -2 & 1 & 0\\0 & 1 & 0 & -4 & 0 & 1 \end{array} \right] \\ \xrightarrow{R_2+R_3 \to R_3}  \left[ \begin{array}{rrr|rrr}1 & 0 & 2 & 1 & 0 & 0\\0 & -1 & -1 & -2 & 1 & 0\\0 & 0 & -1 & -6 & 1 & 1\end{array} \right] \xrightarrow{-R_2\to R_2, -R_3\to R_3}  \left[ \begin{array}{rrr|rrr}1 & 0 & 2 & 1 & 0 & 0\\0 & 1 & 1 & 2 & -1 & 0\\0 & 0 & 1 & 6 & -1 & -1\end{array} \right] \\ \xrightarrow{R_1-2R_3\to R_1, R_2-R_3\to R_2}  \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & -11 & 2 & 2\\0 & 1 & 0 & -4 & 0 & 1\\0 & 0 & 1 & 6 & -1 & -1\end{array} \right] \Rightarrow B^{-1} = \bbox[red, 2pt]{\begin{bmatrix}   -11 & 2 & 2\\  -4 & 0 & 1\\  6 & -1 & -1 \end{bmatrix}}$$
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解題僅供參考，其他碩士班試題及詳解