114年台北科大機電碩士班甲組-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:1110 機械工程系機電整合碩士班甲組
第一節 工程數學 試題

解答: $$v={1\over y^3} \Rightarrow v'= -{3\over y^4}y' \Rightarrow y'=-{y^4v' \over 3} \Rightarrow -{y^4v' \over 3}+{1\over 3}y={1\over 3}(1-2x)y^4 \Rightarrow -v'+ v=1-2x \\ \Rightarrow v'-v=2x-1 \Rightarrow v'e^{-x}-ve^{-x}=(2x-1)e^{-x} \Rightarrow (ve^{-x})'=(2x-1)e^{-x} \\ \Rightarrow ve^{-x}= \int (2x-1)e^{-x}\,dx = -e^{-x}(2x+1)+c_1 \Rightarrow v={1\over y^3}=-(2x+1)+ c_1e^x \\ \Rightarrow \bbox[red, 2pt]{y=\sqrt[3]{1\over c_1e^x-2x-1}}$$

解答: $$y''+2y'+ 2y=0 \Rightarrow \lambda^2+2\lambda +2=0 \Rightarrow \lambda=-1\pm i \Rightarrow y_h= e^{-x} (c_1 \cos x+c_2\sin x) \\ y_p=A\cos 3x+ B\sin 3x \Rightarrow y_p'=-3A\sin 3x+ 3B\cos 3x \Rightarrow y_p''=-9A\cos 3x-9B\sin 3x \\ \Rightarrow y_p''+2y_p'+ 2y_p =(-7A+6B) \cos 3x+(-6A-7B)\sin 3x = 3.5\sin 3x-3\cos 3x \\ \Rightarrow \cases{-7A+6B=-3\\ -6A-7B=3.5} \Rightarrow \cases{A=0 \\ B=-0.5} \Rightarrow y_p=-0.5\sin 3x \Rightarrow y=y_h+y_p \\ \Rightarrow y= e^{-x} (c_1 \cos x+c_2\sin x)-0.5 \sin 3x \Rightarrow y(0)=c_1=0 \Rightarrow y=c_2 e^{-x}\sin x-0.5 \sin 3x \\ \Rightarrow y'=-c_2e^{-x}\sin x+ c_2e^{-x}  \cos x-1.5\cos 3x \Rightarrow y'(0)=c_2-1.5=0.5 \Rightarrow c_2=2\\ \Rightarrow \bbox[red, 2pt]{y=2e^{-x}\sin x-0.5\sin 3x}$$

解答: $$[A\mid I] =\left[ \begin{array}{rrr|rrr}0 & 0 & 1 & 1 & 0 & 0\\0 & \frac{1}{2} & 1 & 0 & 1 & 0\\\frac{1}{3} & 0 & 0 & 0 & 0 & 1 \end{array} \right] \xrightarrow{2R_2\to R_2, 3R_3\to R_3} \left[ \begin{array}{rrr|rrr} 0 & 0 & 1 & 1 & 0 & 0\\0 & 1 & 2 & 0 & 2 & 0\\1 & 0 & 0 & 0 & 0 & 3\end{array} \right] \xrightarrow{R_1 \leftrightarrow R_3} \\\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & 0 & 0 & 3\\0 & 1 & 2 & 0 & 2 & 0\\0 & 0 & 1 & 1 & 0 & 0\end{array} \right] \xrightarrow{R_2-2R_3\to R_2}\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & 0 & 0 & 3\\0 & 1 & 0 & -2 & 2 & 0\\0 & 0 & 1 & 1 & 0 & 0\end{array} \right]  \Rightarrow A^{-1} = \bbox[red, 2pt]{\left[ \begin{array}{rrr|rrr} 0& 0& 3\\ -2 & 2& 0\\ 1& 0& 0\end{array} \right]  }$$

解答: $$L\{y''\}+ 2L\{y'\}+ 2L\{y\}=L\{\delta(t-3)\} \Rightarrow s^2Y(s) +2sY(s)+2Y(s) =e^{-3s} \\ \Rightarrow Y(s)={e^{-3s} \over s^2+2s+2} \Rightarrow y(t) =L^{-1}\{Y(s)\} =L^{-1} \left\{ {e^{-3s} \over s^2+2s+2}\right\} =L^{-1} \left\{ {e^{-3s} \over (s+1)^2+1^2}\right\} \\ \Rightarrow \bbox[red, 2pt]{y(t)=u(t-3)e^{-(t-3)} \sin(t-3)}$$

解答: $$y(x,t) =X(x)T(t) \Rightarrow XT''=c^2X''T \Rightarrow {T''\over c^2T}={X'' \over X} =k\\ B.C.: \cases{y(0,t) =X(0)T(t)=0\\ y(L,t)=X(L)T(t) =0} \Rightarrow \cases{X(0)=0\\ X(L)=0}, y(x,0) =X(x)T(0) =0 \Rightarrow T(0)=0 \\ \textbf{Case I }k=0 \Rightarrow X=c_1x+c_2 \Rightarrow \cases{X(0)=c_2=0\\ X(L)=c_1L+c_2=0} \Rightarrow \cases{c_1=0\\ c_2=0} \Rightarrow X=0 \Rightarrow y=0\\ \textbf{Case II }k=\rho^2 \gt 0 \Rightarrow X''-\rho^2X=0 \Rightarrow X=c_1e^{\rho x} +c_2e^{-\rho x} \Rightarrow \cases{X(0)=c_1+c_2 =0\\ X(L)= c_1e^{\rho L}+ c_2e^{-\rho L} =0} \\\qquad \Rightarrow c_1e^{\rho L}-c_1e^{-\rho L}=0 \Rightarrow c_1(e^{2\rho L}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0 \Rightarrow y=0\\ \textbf{Case III }k=-\rho^2 \lt 0 \Rightarrow X''+\rho^2X=0 \Rightarrow X=c_1 \cos(\rho x)+ c_2\sin (\rho x) \Rightarrow X(0)=c_1=0 \\\qquad \Rightarrow X=c_2\sin(\rho x) \Rightarrow X(L)=c_2 \sin (\rho L)=0 \Rightarrow \sin(\rho L)=0 \Rightarrow  \rho L=n\pi \Rightarrow \rho={n\pi\over L}\\ \qquad \Rightarrow X_n= \sin({n\pi x \over L}),n=1,2,\dots \\ T''+\rho^2 c^2 T=0 \Rightarrow T= c_1 \cos (\rho c t)+ c_2 \sin(\rho ct) \Rightarrow T(0) =c_1=0 \Rightarrow T=c_2\sin(\rho c t) \\ \Rightarrow T_n= \sin({n\pi c t\over L}), n=1,2, \dots \\ \Rightarrow y_n= \sin({n\pi x\over L}) \sin({n\pi ct \over L}) \Rightarrow y= \sum_{n=1}^\infty a_n \sin({n\pi x\over L}) \sin({n\pi ct \over L}) \\ \Rightarrow {\partial y\over \partial t}(x,t) = \sum_{n=1}^\infty a_n {n\pi c\over L}\sin({n\pi x\over L}) \cos({n\pi ct \over L}) \Rightarrow {\partial y\over \partial t}(x,0) =\sum_{n=1}^\infty a_n {n\pi c\over L}\sin({n\pi x\over L}) =A \\ \Rightarrow a_n {n\pi c\over L}={2\over L} \int_0^L A \sin({n\pi x\over L})\,dx ={2A\over L} \cdot {L\over n\pi}(1-(-1)^n) \Rightarrow a_n={2AL \over n^2\pi^2 c}(1-(-1)^n) \\ \Rightarrow \bbox[red, 2pt]{y(x,t) =\sum_{n=1}^\infty {2AL \over n^2\pi^2 c}(1-(-1)^n)\sin({n\pi x\over L}) \sin({n\pi ct \over L})}$$

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解題僅供參考，其他碩士班試題及詳解