108年台綜大轉學考-工程數學D04詳解

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：D04

解答：

(a)

$$\mathbf{(b)}\;f(x)=\begin{cases} x+\pi, &-\pi \lt x\lt 0 \\ -x+\pi,& 0\lt x\lt \pi\end{cases} \Rightarrow f(x)為偶函數\Rightarrow b_n=0，只需計算a_n;\\ a_0= {1\over 2\pi} \int_{-\pi}^\pi f(x)\,dx ={1\over 2\pi}\left( \int_{-\pi}^0 x+\pi\,dx + \int_0^{\pi}  -x+\pi\,dx \right) ={1\over 2\pi}\times \pi^2 ={\pi\over 2}\\ a_n ={1\over \pi}\int_{-\pi}^\pi f(x)\cos(nx) \,dx ={1\over \pi}\left( \int_{-\pi}^0 (x+\pi)\cos(nx),dx + \int_0^{\pi}  (-x+\pi)\cos(nx) \,dx \right)\\ ={1\over \pi}\left(\left.\left[ {x\over n}\sin(nx) +{1\over n^2}\cos(nx) +{\pi\over n}\sin(nx) \right] \right|_{-\pi}^0 + \left.\left[ -{x\over n} \sin(nx) -{1\over n^2}\cos(nx) +{\pi\over n}\sin(nx) \right] \right|_{0}^{\pi} \right)\\ ={1\over \pi}\cdot {2\over n^2}(1-(-1)^n) =\cases{{4\over n^2\pi},n是奇數\\[1ex] 0,n是偶數} \Rightarrow f(x)=a_0 +\sum_{n=1}^ \infty a_n\cos(nx) \\ \Rightarrow \bbox[red, 2pt]{f(x)={\pi \over 2}+ {2\over \pi}\sum_{n=1}^ \infty {1\over n^2}(1-(-1)^n)\cos(nx)}$$

解答：$$先求齊次解，即y''+\omega^2 y=0 \Rightarrow 特徵方程式:\lambda^2+\omega^2=0 \Rightarrow \lambda = \pm \omega i\\ \Rightarrow y_h=A\cos(\omega t)+B\sin(\omega t) \\ 由於r(t)=\sin t，因此需考慮\omega是否等於1;由題意知\omega=0.5,1.2,2,10,皆不等於1\\ 因此y_p= C\sin t \Rightarrow y_p''=-C\sin t \Rightarrow -C\sin t+\omega^2C\sin t=\sin t \Rightarrow C(\omega^2-1) =1\\ \Rightarrow C = 1/(\omega^2-1) \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=A\cos(\omega t) +B\sin(\omega t)+ {1\over \omega^2-1} \sin t,A 與B皆為常數}$$

解答：$$y'={2x\over y} \Rightarrow \int ydy = \int 2x\,dx \Rightarrow {1\over 2}y^2 = x^2+ C\\ y(1)=5 \Rightarrow {25\over 2} =1+C \Rightarrow C ={23\over 2} \Rightarrow \bbox[red, 2pt]{y^2 = 2x^2+ 23}$$

解答：$$f=2e^{xyz} \Rightarrow \cases{f_x=2yze^{xyz} \\ f_y =2xz e^{xyz}\\ f_z= 2xye^{xyz}} \Rightarrow \cases{f_{xx} =2y^2 z^2 e^{xyz} \\ f_{yy} =2x^2 z^2 e^{xyz}\\ f_{zz}= 2x^2y^2 e^{xyz}} \\ \Rightarrow \nabla^2 f= f_{xx}+ f_{yy}+ f_{zz} = \bbox[red, 2pt]{ 2e^{xyz} (x^2y^2 +y^2z^2 +z^2x^2)}$$

解答：$$假設\cases{\vec u=\begin{bmatrix} a\\b\\ c\end{bmatrix} \\[1ex] \vec v= \begin{bmatrix} d\\e\\ f\end{bmatrix}} \Rightarrow \vec u\times \vec v=\begin{bmatrix} bf-ce\\ cd-af \\ ae-bd\end{bmatrix} \Rightarrow (\vec u\times \vec v)' =\begin{bmatrix} (bf-ce)'\\ (cd-af)' \\ (ae-bd)'\end{bmatrix} =\begin{bmatrix} b'f+bf'-c'e-ce' \\ c'd+cd' -a'f-af' \\ a'e +ae' -b'd-bd'\end{bmatrix} \cdots(1)\\ 而\cases{\vec u'\times \vec v =\begin{bmatrix} a'\\b'\\ c'\end{bmatrix} \times \begin{bmatrix} d\\e\\ f\end{bmatrix} = \begin{bmatrix} b'f-c'e\\c'd-a'f\\ a'e-b'd\end{bmatrix} \\[1ex] \vec u\times \vec v'=\begin{bmatrix} a\\b\\ c\end{bmatrix} \times \begin{bmatrix} d'\\e' \\ f'\end{bmatrix} =\begin{bmatrix} bf'-ce'\\cd'-af' \\ ae'-bd'\end{bmatrix}} \Rightarrow \vec u'\times \vec v+\vec u\times \vec v'=\begin{bmatrix} b'f+bf'-c'e-ce' \\ c'd+cd' -a'f-af' \\ a'e +ae' -b'd-bd'\end{bmatrix} \cdots(2)\\ (1)=(2) \Rightarrow \vec u\times \vec v= \vec u'\times \vec v+\vec u\times \vec v'，\bbox[red,2pt]{故得證}$$

解答：$$A=\left[\begin{matrix} -1  & -1 & 0 \\-1 & -1  & 0 \\0 & 0 & 1 \end{matrix}\right] \Rightarrow \det(A-\lambda I)=0 \Rightarrow -\lambda(\lambda+2)( \lambda-1)=0 \Rightarrow 特徵值\cases{ \lambda_1=0 \\\lambda_2= -2 \\\lambda_3 =1 } \\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)\mathbf x= \left[\begin{matrix} -1  & -1 & 0 \\-1 & -1  & 0 \\0 & 0 & 1 \end{matrix}\right] \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_1+x_2=0 \\ x_3=0},取v_1=\begin{bmatrix} -1\\ 1\\0\end{bmatrix} \\\lambda_2=-2 \Rightarrow (A-\lambda_1 I)\mathbf x= \left[\begin{matrix} 1  & -1 & 0 \\-1 & 1  & 0 \\0 & 0 & 3 \end{matrix}\right] \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_1=x_2 \\ x_3=0},取v_2=\begin{bmatrix} 1\\ 1\\0\end{bmatrix} \\\lambda_3= 1 \Rightarrow (A-\lambda_1 I)\mathbf x= \left[\begin{matrix} -2  & -1 & 0 \\-1 & -2  & 0 \\0 & 0 & 0 \end{matrix}\right] \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow x_1=x_2=0,取v_3=\begin{bmatrix} 0\\ 0\\1\end{bmatrix} \\ \Rightarrow P=[v_1\;v_2\; v_3]=  \left[ \begin{matrix}-1 & 1 & 0 \\1 & 1 & 0 \\0 & 0 & 1\end{matrix}\right] \Rightarrow P^{-1} =\left[\begin{matrix} \frac{-1}{2} & \frac{1}{2} & 0 \\\frac{1}{2} & \frac{1}{2} & 0 \\0 & 0 & 1\end{matrix}\right] \\ \Rightarrow A= P \begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 &\lambda_2 & 0\\ 0 & 0 & \lambda_3 \end{bmatrix}  P^{-1} =\bbox[red, 2pt]{ \left[ \begin{matrix}-1 & 1 & 0 \\1 & 1 & 0 \\0 & 0 & 1\end{matrix}\right] \begin{bmatrix} 0 & 0 & 0 \\ 0 &-2 & 0\\ 0 & 0 & 1 \end{bmatrix} \left[\begin{matrix}\frac{-1}{2} & \frac{1}{2} & 0 \\\frac{1}{2} & \frac{1}{2} & 0 \\0 & 0 & 1\end{matrix}\right]}$$

解答：$$\cases{y_1'= y_2+2-u(t-1)\\ y_2'= -y_1+1-u(t-1)} \Rightarrow \begin{bmatrix} y_1'\\ y_2'\end{bmatrix} = \begin{bmatrix} 0 & 1\\ -1& 0\end{bmatrix}  \begin{bmatrix} y_1\\ y_2\end{bmatrix} + \begin{bmatrix} 2-u(t-1)\\ 1-u(t-1) \end{bmatrix} \equiv y'=Ay+ \mathbf b\\ A= \begin{bmatrix} 0 & 1\\ -1& 0\end{bmatrix} \Rightarrow \det(A-\lambda I)=\lambda^2+1=0 \Rightarrow \lambda=\pm i\\  \lambda_1=i \Rightarrow (A-\lambda_1 I)\mathbf x=\begin{bmatrix} -i & 1\\ -1& -i\end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow x_1=-ix_2,取v_1=\begin{bmatrix} -i\\ 1\end{bmatrix}\\ \lambda_2=-i \Rightarrow (A-\lambda_2 I)\mathbf x=\begin{bmatrix} i & 1\\ -1& i\end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow x_1= ix_2,取v_2=\begin{bmatrix} i\\ 1\end{bmatrix} \\ 因此y'=Ay 的齊次解為\Rightarrow \begin{bmatrix} y_1\\ y_2\end{bmatrix} =C_1e^{it}\begin{bmatrix} -i\\ 1\end{bmatrix} +C_2e^{-it} \begin{bmatrix} i\\ 1\end{bmatrix} \Rightarrow \cases{y_1=-C_1ie^{it} +C_2ie^{-it}\\ y_2= C_1e^{it}+ C_2e^{-it}} \\ \Rightarrow \cases{y_1'=  C_1e^{it}+ C_2e^{-it}+2-u(t-1)\\ y_2'=C_1 ie^{it} -C_2ie^{-it} +1-u(t-1)} \Rightarrow   \cases{\mathcal L\{y_1'\}= \mathcal L\{ C_1e^{it}+ C_2e^{-it}+2-u(t-1) \}\\ \mathcal L\{y_2'\}= \mathcal L\{C_1 ie^{it} -C_2ie^{-it} +1-u(t-1)\} } \\ \Rightarrow \cases{sY_1(s)-y_1(0) ={C_1\over s-i} +{C_2\over s+i} +{2\over s}-e^{-s}/s \\ sY_2(s)-y_2(0)= {C_1i\over s-i}-{C_2 i\over s+i}+ {1\over s}-e^{-s}/s} \Rightarrow \cases{Y_1(s) = {C_1\over s(s-i)} +{C_2\over s(s+i)} +{2\over s^2}+{2\over s}-e^{-s}/s^2 \\ Y_2(s)= {C_1i\over s(s-i)}-{C_2 i\over s(s+i)}+ {1\over s^2}+ {1\over s}-e^{-s}/s^2} \\ \Rightarrow  \cases{Y_1(s) = {C_1\over i}({1\over s-i}-{1\over s}) +{C_2\over i}({1\over s }-{1\over s+i}) +{2\over s^2}+{2\over s}-e^{-s}/s^2 \\ Y_2(s)= C_1({1\over s-i}-{1\over s})-C_2 ({1\over s}-{1\over s+i})+ {1\over s^2}+ {1\over s}-e^{-s}/s^2}  \\ \Rightarrow \cases{y_1= {C_1\over i}(e^{it}-1 ) +{C_2\over i}(1-e^{-it}) +2t+ 2-(t-1)u(t-1) \\ y_2= C_1(e^{it}-1)-C_2 (1-e^{-it})+ t+ 1-(t-1)u(t-1)}$$

解答：$$A= \left[\begin{matrix}1 & 0 & 3\\1 & 1 & 1\\0 & 1 & 0\end{matrix}\right] \Rightarrow rref(A)=\left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right] \Rightarrow Rank(A)=3 \Rightarrow \bbox[red,2pt]{線性獨立}$$

解答：$$(2,0, a_1)\cdot (1,0,8) = 2+8a_1=0 \Rightarrow a_1= \bbox[red,2pt]{-{1\over 4}}$$

解答：$$\cases{3x+y+ 2z=1\\ x-y+3z=-3 \\ y-2z=1} \Rightarrow \begin{bmatrix}3 & 1 &2 \\ 1 & -1 & 3\\ 0 & 1 &-2 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 1\\ -3\\1\end{bmatrix} \equiv A\mathbf x=\mathbf b \\ \Rightarrow \cases{A_x= \begin{bmatrix}1 & 1 &2 \\ -3 & -1 & 3\\ 1 & 1 &-2 \end{bmatrix} \\[1ex] A_y= \begin{bmatrix}3 & 1 &2 \\ 1 & -3 & 3\\ 0 & 1 &-2 \end{bmatrix} \\[1ex] A_z =\begin{bmatrix}3 & 1 &1 \\ 1 & -1 & -3\\ 0 & 1 &1 \end{bmatrix}} \Rightarrow \cases{\det(A_x)= -8 \\ \det(A_y) =13 \\\det(A_z) =6 \\\det(A)=1} \Rightarrow \cases{x=\det(A_x)/\det(A)= -8\\ y=\det(A_y) /\det(A) = 13\\ z= \det(A_z) / \det(A) = 6}\\ \Rightarrow \bbox[red,2pt]{\cases{x=  -8\\ y=  13\\ z=   6}}$$

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