110年北科大機電整合碩士班乙組-工程數學詳解

國立臺北科技大學110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
科目:工程數學

解答: $$\text{integrating factor }I(x)=e^{\int 1\,dx}=e^x \Rightarrow I(x)y'+I(x)y={1\over 2}I(x)(e^x-e^{-x}) \\ \Rightarrow e^xy'+e^xy={1\over 2}(e^{2x}-1) \Rightarrow (e^xy)'= {1\over 2}(e^{2x}-1) \Rightarrow e^x y={1\over 2}({1\over 2}e^{2x}-x)+c_1 \\ \Rightarrow \bbox[red, 2pt]{y={1\over 4}e^x-{1\over 2}xe^{-x}+c_1e^{-x}}$$

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

解答: $$y''+y=0 \Rightarrow y_h= c_1\cos x+c_2\sin x\\ y_p=Ax\cos x+Bx\sin x \Rightarrow y_p'=A\cos x-Ax\sin x+B\sin x+Bx\cos x \\ \Rightarrow y_p''=-2A\sin x -Ax\cos x+2B\cos x -Bx\sin x \\ \Rightarrow y_p''+y_p=-2A\sin x+2B\cos x=\cos x \Rightarrow \cases{A=0\\ B=1/2} \Rightarrow y_p={1\over 2}x\sin x\\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=  c_1\cos x+c_2\sin x+ {1\over 2}x\sin x}$$

解答: $$\text{Let }\cases{y_1=\cos x\\ y_2=\sin x} \Rightarrow W=\begin{vmatrix} y_1&y_2 \\ y_1'& y_2' \end{vmatrix} =\begin{vmatrix} \cos x&\sin x \\-\sin x& \cos x \end{vmatrix} =1 \\ \Rightarrow y_p=-\cos x\int \sin x \cdot \cos x\,dx + \sin x\int \cos x\cdot \cos x\,dx={1\over 4}\cos x+{1\over 2}x\sin x \\ \Rightarrow y=y_h+y_p =c_1\cos x+c_2\sin x+{1\over 4}\cos x+{1\over 2}x\sin x \\ \Rightarrow \bbox[red, 2pt]{y=c_3\cos x+c_2\sin x+{1\over 2}x\sin x}$$

解答: $$L\{ y''\}+8L\{ y'\}+16 L\{y\}=L\{ t^2e^{-4t}\} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0)+8(sY(s)-y(0))+ 16Y(s)=\frac{\text{d}^2 }{\text{d}t^2} \left( {1\over s+4}\right) \\ \Rightarrow (s^2+8s+16)Y(s)-s-4 ={2\over (s+4)^3} \Rightarrow Y(s)={2\over (s+4)^5}+{1\over s+4} \\ \Rightarrow y(t)=L^{-1}\left\{{2\over (s+4)^5} \right\} +L^{-1}\left\{{1\over s+4} \right\} \Rightarrow \bbox[red, 2pt]{y(t)={1\over 12}e^{-4t}t^4 +e^{-4t}}$$

解答: $$L\{ y''\}+4 L\{y\}=L\{ f(t)\} =L\{u(t-\pi) 3\cos(t)\}\\ \Rightarrow s^2Y(s)-sy(0)-y'(0)+4Y(s)= -{3s\over s^2+1}e^{-\pi s} \\ \Rightarrow (s^2+4)Y(s)-s-1= -{3s\over s^2+1}e^{-\pi s} \Rightarrow Y(s)= -{3s\over (s^2+1)(s^2+4)}e^{-\pi s}+ {s+1\over s^2+4} \\ \Rightarrow y(t)=L^{-1}\{Y(s) \} =L^{-1}\left\{ -{3s\over (s^2+1)(s^2+4)}e^{-\pi s} \right\} +L^{-1}\left\{{s+1\over s^2+4} \right\} \\=u(t-\pi)(-\cos(t-\pi)+ \cos(2(t-\pi))+\cos(2t)+{1\over 2}\sin(2t) \\ \Rightarrow \bbox[red, 2pt]{ y(t)=u(t-\pi)(\cos t+\cos(2t))+\cos(2t)+{1\over 2}\sin(2t)}$$

解答: $$y(x,t)=X(x)T(t) \Rightarrow \cases{\frac{\partial^2 y}{\partial t^2}=c^2 \frac{\partial^2 y}{\partial x^2} \Rightarrow XT''=c^2X''T\\ BC \cases{y(0,t)=0 \\ y(L,t)=0} \Rightarrow \cases{X(0)T(t)=0\\ X(L)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0}} \\ XT''=c^2X''T \Rightarrow {X'' \over X}={T'' \over c^2T} =k\\ \textbf{Case 1: k=0}\\ \qquad X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow BC\cases{X(0)=c_2=0\\ X(L)=c_1L+c_2=0} \Rightarrow c_1=c_2=0 \Rightarrow X=0\\ \textbf{Cases 2: }\mathbf{k=\mu^2 \gt 0(\mu\gt 0)}\\\qquad X''-\mu^2X=0 \Rightarrow X=c_1e^{\mu x}+ c_2e^{-\mu x} \Rightarrow BC\cases{X(0)=c_1+ c_2=0 \Rightarrow c_2=-c_1\\ X(L)= c_1e^{\mu L}+ c_2e^{-\mu L}=0} \\\qquad \Rightarrow c_1e^{\mu L}- c_1e^{-\mu L}=0 \Rightarrow c_1(e^{2\mu L)}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0\\ \textbf{Case 3: }\mathbf{k=-\mu^2 \lt 0}\\\qquad X''+\mu^2 X=0 \Rightarrow X=c_1\cos(\mu x)+ c_2\sin(\mu x) \Rightarrow BC\cases{X(0)=c_1=0\\ X(L)=c_2\sin(\mu L)=0} \\ \qquad \Rightarrow \mu L= n\pi \Rightarrow \mu={n\pi\over L} \Rightarrow X_n= \sin({n\pi x\over L}), n\in \mathbb N\\ \qquad \Rightarrow T''-kc^2T=0 \Rightarrow T''+\mu^2c^2T=0 \Rightarrow T= c_3\cos(\mu c t)+ c_4\sin(\mu ct)\\ \Rightarrow T'=-c_3\mu c \sin(\mu c t) +c_4 \mu c\cos(\mu c t)\\\qquad  \frac{\partial y}{\partial t}(x,0)=0 \Rightarrow X(x)T'(0) =0 \Rightarrow T'(0)=0 \Rightarrow c_4=0 \Rightarrow T= c_3\cos(\mu c t)\\ \Rightarrow T_n= \cos({nc\pi t\over L}) \Rightarrow y_n(x,t) =X_n(x)T_n(t) = \sin({n\pi x\over L}) \cos({nc\pi t\over L}), n\in \mathbb N\\ \Rightarrow y(x,t)= \sum_{n=1}^\infty a_n \sin({n\pi x\over L}) \cos({nc\pi t\over L}) \Rightarrow y(x,0)=\sum_{n=1}^\infty a_n \sin({n\pi x\over L}) =f(x) \\ \Rightarrow a_n= {2\over L} \int_0^L f(x) \cos({n\pi x\over L})\,dx \\ \Rightarrow \bbox[red, 2pt]{y(x,t)= \sum_{n=1}^\infty a_n \sin({n\pi x\over L}) \cos({nc\pi t\over L}), \text{where }a_n= {2\over L} \int_0^L f(x) \cos({n\pi x\over L})\,dx }$$

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解題僅供參考,其他歷年試題及詳解