113年成大船舶機電碩士班-工程數學詳解

 國立成功大學113學年度碩士班招生考試

系所：系統及船舶機電工程學系
科目：工程數學

解答:$$f(x)=2|x| \Rightarrow f(x)=f(-x) \Rightarrow f \text{ is even} \Rightarrow b_n=0\\ a_0={1\over 2\pi}\int_{-\pi}^{\pi} 2|x|\,dx ={2\over \pi}\int_0^\pi x\,dx= \pi\\ a_n={1\over \pi} \int_{\pi}^\pi 2|x| \cos(nx)\,dx ={4\over \pi}\int_0^\pi x\cos(nx)\,dx ={4\over \pi} \left.\left[ {x\over n}\sin(nx)+ {1\over n^2}\cos(nx)\right] \right|_0^\pi \\\qquad ={4\over \pi}({1\over n^2}((-1)^n-1)) ={4\over n^2\pi}\left( (-1)^n-1\right) \\ \Rightarrow \bbox[red, 2pt]{f(x)=\pi+\sum_{n=1}^\infty {4\over n^2\pi}\left( (-1)^n-1\right) \cos(nx)} \\ \Rightarrow f(0)=0=\pi -{8\over \pi}({1\over 1^2}+ {1\over 3^2}+ {1\over 5^2}+ \cdots) \Rightarrow {1\over 1^2}+ {1\over 3^2}+ {1\over 5^2}+ \cdots = \bbox[red, 2pt]{\pi^2\over 8}$$

解答:$$u_t+u_x=u \Rightarrow {dt\over dt}u_t+{dx\over dt}u_x={du\over dt} \Rightarrow \cases{{dt\over dt}=1\\ {dx\over dt}=1\\ {du\over dt} =u} \Rightarrow \cases{x-t=C\\ u=u_0e^t} \\ \Rightarrow u(x,t) =f(x-t)e^t \Rightarrow u(0,t)=3e^{-t}=f(-t)e^t \Rightarrow f(-t)=3e^{-2t} \Rightarrow f(t)=3e^{2t}\\ \Rightarrow f(x-t)= 3e^{2(x-t)}\Rightarrow u(x,t)= 3e^{2(x-t)}e^t \Rightarrow \bbox[red, 2pt]{u(x,t)=3e^{2x-t}}$$

解答:$$\text{By divergence theorem, }\iint_S \vec F\cdot \vec n\,dA =\iiint_R\nabla\cdot \vec F\,dV=\iiint_R (1+1+4z^3)\,dV \\=\iiint_R 2\,dV+4 \iiint_R z^3\,dV =2\cdot {4\over 3}\pi\cdot 4^3+4\cdot 0 = \bbox[red, 2pt]{512\pi\over 3}$$

解答:$$L\{tf(t)\} =-{d\over ds}F(s)=-{d\over ds}\left(\ln{s^2+1\over (s-1)^4} \right)= -{d\over ds}\left(\ln(s^2+1)-\ln(s-1)^4 \right) \\=- \left( {2s\over s^2+1}-{ 4(s-1)^3\over (s-1)^4}\right) =- \left( {2s\over s^2+1}-{ 4 \over s-1 }\right) ={4\over s-1}-{2s\over s^2+1} \\ \Rightarrow tf(t)=L^{-1}\left\{ {4\over s-1} \right\}-L^{-1}\left\{ {2s\over s^2+1} \right\} =4e^t-2\cos(t) \Rightarrow f(t)= \bbox[red, 2pt]{4e^t-2\cos(t)\over t}$$

解答:$$y''-2y'+y=0 \Rightarrow \lambda^2-2\lambda+1=0 \Rightarrow \lambda=1,1 \Rightarrow y_h=c_1e^x+ c_2xe^x\\  y_p=e^x(Ax^2+Bx^3 +Cx^4) \Rightarrow y_p'= e^x(2Ax+(A+ 3B)x^2+ (B+4C)x^3+Cx^4) \\ \Rightarrow y_p''= e^x(2A+ (4A+6B)x+ (A+6B+12C)x^2+ (B+8C)x^3+Cx^4) \\ \Rightarrow y_p''-2y_p'+y_p= e^x(2A+6Bx+12Cx^2) =6x^2e^x \Rightarrow \cases{A=B=0\\ C=1/2} \\ \Rightarrow y_p={1\over 2}x^4e^x \Rightarrow y= y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^x+ c_2xe^x +{1\over 2}x^4e^x}$$

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解題僅供參考，碩士班歷年試題及詳解