臺灣綜合大學系統108學年度學士班轉學生聯合招生考試
科目名稱：工程數學
類組代碼：D37
解答：$$先求齊次解:y''-4y'+4y=0 \Rightarrow \lambda^2-4\lambda+ 4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow \lambda=2\\ \Rightarrow y_h =C_1e^{2x} +C_2xe^{2x};\\ 令\cases{y_1=e^{2x}\\ y_2= xe^{2x}\\ r(x)=(12x^2-6x)e^{2x}} \Rightarrow \cases{y_1'=2e^{2x}\\ y_2'=e^{2x} +2xe^{2x}} \Rightarrow W=\begin{vmatrix}y_1 & y_2\\ y_1' & y_2' \end{vmatrix} =\begin{vmatrix}e^{2x} & xe^{2x} \\ 2e^{2x} & e^{2x}+2xe^{2x} \end{vmatrix} =e^{4x}\\ \Rightarrow y_p = -y_1\int {y_2r(x) \over W}\,dx + y_2\int {y_1r(x)\over W}\,dx = -e^{2x}\int 12x^3-6x^2\,dx +xe^{2x} \int 12x^2-6x\,dx\\ =-e^{2x}(3x^4-2x^3)+xe^{2x}(4x^3-3x^2) =e^{2x}(x^4-x^3) \\\Rightarrow y=y_h+y_p =C_1e^{2x} +C_2xe^{2x} +e^{2x}(x^4-x^3) \\\Rightarrow y'=(2C_1+C_2) e^{2x} +2C_2xe^{2x}+e^{2x}(2x^4+2x^3-3x^2) \\將\cases{y(0) =1\\ y'(0)=0} 代入 \Rightarrow \cases{1=C_1\\ 0= 2C_1+C_2} \Rightarrow C_2=-2 \Rightarrow y= e^{2x}-2xe^{2x} +e^{2x}(x^4-x^3) \\ \Rightarrow \bbox[red,2pt]{y=e^{2x}(x^4-x^3-2x+1)}$$
解答：$$\cases{\mathcal L\{ x''\} =s^2X(s)-sx(0)-x'(0) = s^2X(s)-s\\\mathcal L\{x'\} = sX(s)-x(0)=sX(s)-1\\ \mathcal L\{y''\} = s^2Y(s)-sy(0)-y'(0)=s^2Y(s)+s -5\\ \mathcal L\{y'\} = sY(s)-y(0)= sY(s)+1} \\ \Rightarrow \cases{ \mathcal L\{ x''+x'+y'\} =0 \\\mathcal L\{y''+y'-4x'\} =0 } \Rightarrow \cases{ (s+1)X(s)+ Y(s)=s \cdots(1)\\ (s+1)Y(s)-4X(s)=-1 \cdots(2)}\\ 由(1) \Rightarrow Y(s)=s-(s+1)X(s)代入(2) \Rightarrow X(s)= { s(s+1)+1 \over (s+1)^2 +4} \Rightarrow Y(s)= {3s-1 \over (s+1)^2+4} \\ \Rightarrow \cases{x(t) = \mathcal L^{-1}\{X(s)\} \\y(t) = \mathcal L^{-1}\{Y(s)\}} \Rightarrow \bbox[red, 2pt]{\cases{x(t)=\delta(t)-e^{-t}\cos (2t)-{3\over 2}e^{-t} \sin(2t) \\ y(t)=3e^{-t}\cos(2t)-2e^{-t}\sin (2t)}}$$
解答：$$假設X=PDP^{-1}，其中D=\begin{bmatrix} \lambda_1& 0\\ 0 &\lambda_2\end{bmatrix} \Rightarrow X^2=PD^2P^{-1} \\\Rightarrow X^2-5X +6I =P(D^2-5D+6I)P^{-1} = \begin{bmatrix} -4 & -5\\ 8 & 10 \end{bmatrix}= \left[\begin{matrix}\frac{-5}{4} & \frac{-1}{2} \\1 & 1\end{matrix}\right] \left[\begin{matrix}0 & 0 \\0 & 6\end{matrix}\right]\left[ \begin{matrix}
\frac{-4}{3} & \frac{-2}{3} \\\frac{4}{3} & \frac{5}{3}\end{matrix}\right]\\ \Rightarrow \cases{\lambda_1^2 -5\lambda_1+6 = 0\\ \lambda_2^2 -5\lambda_2+6 =6} \Rightarrow \cases{\lambda_1=2,3\\ \lambda_2=0,5} \Rightarrow X= \begin{bmatrix}\frac{-5}{4} & \frac{-1}{2} \\1 & 1\end{bmatrix} \begin{bmatrix} \lambda_1& 0\\ 0 & \lambda_2 \end{bmatrix} \begin{bmatrix} \frac{-4}{3} & \frac{-2}{3} \\\frac{4}{3} & \frac{5}{3}\end{bmatrix} \\
\Rightarrow \bbox[red, 2pt]{X=\begin{bmatrix} \frac{10}{3} & \frac{5}{3} \\ \frac{-8}{3} & \frac{-4}{3} \end{bmatrix}, \begin{bmatrix} 0 & \frac{-5}{2} \\4 & 7\end{bmatrix} ,\begin{bmatrix} 5 & \frac{5}{2} \\-4 & -2 \end{bmatrix}, \begin{bmatrix} \frac{5}{3} & \frac{-5}{3} \\\frac{8}{3} & \frac{19}{3} \end{bmatrix} }$$
解答：$$取\cases{x(t)=2\cos t\\ y(t)=2\sin t\\ z(t)=1} \Rightarrow \cases{x'(t)= -2\sin t\\y'(t)=2\cos t\\ z'(t)=0} \\\Rightarrow 左式=\oint_C \vec F\cdot d\vec r = \int_0^{2\pi} (10\sin t,-10\cos t,3)\cdot (-2\sin t,2\cos t,0)\,dt\\ =\int_0^{2\pi} -20\sin^2 t-20\cos^2 t\,dt =\int_0^{2\pi } -20\,dt =-40\pi\\ \vec F=(5y,-5x,3) \Rightarrow \text{curl }\vec F =({\partial \over \partial x} ,{\partial \over \partial y} ,{\partial \over \partial z} )\times (5y,-5x,3) =(0,0,-10) \\ \vec r=(x,y,1) \Rightarrow \cases{\vec r_x=(1,0,0)\\ \vec r_y=(0,1,0)} \Rightarrow \vec r_x\times \vec r_y=(0,0,1)\\ 因此右式:\iint_S (\text{curl }\vec F)\cdot \mathbf n\,dS = \iint_S (0,0,-10)\cdot (0,0,1)\,dA =\iint_S-10\,dA = -10\times 4\pi=-40\pi\\ 因此左式=右式=-40\pi，\bbox[red,2pt]{故得證}$$
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