111年成大船舶機電碩士班-工程數學詳解

國立成功大學111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: $$\cases{y_1'=2y_1-4y_2+u(t-2)e^t\\ y_2'=y_1-3y_2-u(t-2)e^t} \Rightarrow \cases{L\{y_1' \}= L\{2y_1-4y_2+u(t-2)e^t\} \\L\{ y_2'\} =L\{ y_1-3y_2-u(t-2)e^t \}} \\ \Rightarrow \cases{sY_1(s)-3 =2Y_1(s)-4Y_2(s)+ \cfrac{e^{-2(s-1)}}{ s-1}  \\ sY_2(s) =Y_1(s)-3Y_2(s)-  \cfrac{e^{-2(s-1)}}{ s-1}  }  \Rightarrow \cases{Y_1(s)={3\over s-2}-{4\over s-2}Y_2(s)+ \cfrac{e^{-2(s-1)}}{ (s-1)(s-2)} \\ Y_2(s)={1\over s+3}Y_1(s)-\cfrac{e^{-2(s-1)}}{ (s-1)(s+3) }} \\ \Rightarrow \cases{Y_1(s)= \cfrac{ 3(s+3)}{(s+2)(s-1)} + \cfrac{(s +7)(s-3)}{  (s-1)^2(s +2)(s+3)} e^{-2(s-1)} \\ Y_2(s) =\cfrac{3(s-3)}{(s+2) (s+3) (s-1)} -\cfrac{ (s-3)^2}{ (s-1)^2(s+2)(s +3)} e^{-2(s-1)}}   \Rightarrow \cases{y_1(t) = L^{-1} \{Y_1(s) \} \\y_2(t)= L^{-1} \{Y_2(s)} \\ \Rightarrow \bbox[red, 2pt]{\cases{y_1(t) = 2e^t +e^{-2t} + \cfrac{u(t-2) }{18} e^2(-24te^{t-2}+27e^{-3(t-2)} +71e^{t-2}-50e^{-2(t-2)}) \\ y_2(t)=-{1\over 2}e^t+5e^{-2t}-{9\over 2}e^{-3t}- \cfrac{u(t-2)e^2}{ 36}(12te^{t-2}+100e^{-2(t-2)} -81 e^{-3(t-2)} -43e^{t-2})}}$$

解答: $$f(t)=\begin{cases}2& 0\lt t\lt 1\\ t^2/2 & 1\lt t\lt \pi/2\\ \sin(t)& t\gt \pi/2 \end{cases}\\ \Rightarrow f(t)=2(u(t)-u(t-1))+{1\over 2}t^2(u(t-1)-u(t-{\pi\over 2})) +\sin(t)u(t-{\pi \over 2}) \\\qquad = \bbox[red, 2pt]{2u(t)+{1\over 2}(t^2-4)u(t-1)+(\sin t-{1\over 2}t^2)u(t-{\pi\over 2})}$$

解答: $$先求齊次解, 令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow x^2y''-xy'-3y=m(m-1)x^m-mx^m-3x^m =(m^2-2m-3)x^m=0\\ \Rightarrow m^2-2m-3=0 \Rightarrow (m-3)(m+1)=0 \Rightarrow m=-1,3 \Rightarrow y_h=c_1x^{-1} +c_2x^3\\ y_p=Ax^2+Bx+C \Rightarrow y_p'=2Ax+B \Rightarrow y_p''=2A \\ \Rightarrow x^2y''-xy'-3y=2Ax^2-2Ax^2-Bx-3Ax^2-3Bx-3C \\=-3Ax^2-4Bx-3C =4x^2 \Rightarrow \cases{A=-4/3\\ B=C=0} \Rightarrow y_p=-{4\over 3}x^2 \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y={c_1\over x}+ c_2x^{3}-{4\over 3}x^2}$$

解答: $$4x_1^2+6x_1x_2-4x_2^2=[x_1,x_2] \begin{bmatrix}4 & 3 \\3 & -4 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix} \equiv \textbf x^t A \textbf x\\ 又A＝  \begin{bmatrix}3/\sqrt{10} & -1/\sqrt{10} \\1/\sqrt{10} & 3/\sqrt{10} \end{bmatrix} \begin{bmatrix} 5 & 0 \\0 & -5 \end{bmatrix} \begin{bmatrix}3/\sqrt{10} & 1/\sqrt{10} \\-1/\sqrt{10} & 3/\sqrt{10} \end{bmatrix} ＝QDQ^T \\ \Rightarrow  \textbf x^t A \textbf x =\equiv \textbf x^t QDQ^t \textbf x =(Q^t\textbf x)^tD(Q^t \textbf x) \equiv (\textbf{x}')^t D\textbf{x}' \\ \Rightarrow 4x_1^2+6x_1x_2-4x_2^2=5x_1'^2-5x_2'^2=5 \Rightarrow \bbox[red,2pt]{x_1'^2-x_2'^2=1為一雙曲線}$$

解答: $$S:r=[u,v,3u-2v] \Rightarrow z-3x+2y=0 \Rightarrow NdS=[-3,2,1] \Rightarrow F\cdot NdS =3x^2+2y^2 \\ \Rightarrow \text{Flux integral =}\int_{-3}^3 \int_0^2 3x^2+2y^2\,dxdy =\int_{-3}^3 8+4y^2\,dy =2\int_0^38+4y^2\,dy \\=2\times 60=\bbox[red, 2pt]{120}$$

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