111年台綜大轉學考-工程數學D04詳解

 臺灣綜合大學系統111學年度學士班轉學生聯合招生考試

科目：工程數學
類組代碼：D04

解答：$$\cases{f_1(x)= e^x \Rightarrow \cases{f_1(-x)=e^{-x} \\-f_1(x)= -e^x} \Rightarrow \cases{f_1(-x)\ne f_1(x)\\ f_1(-x)\ne -f_1(x)} \Rightarrow e^x 不是偶函數也不是奇函數\\ f_2(x)= e^{x^2} \Rightarrow \cases{f_2(-x)=e^{x^2}\\ -f_2(x)=-e^{x^2}} \Rightarrow f_2(x)=f_2(-x) \Rightarrow e^{x^2} 為偶函數\\ f_3(x)= \tan x \Rightarrow \cases{f_3(-x)= -\tan x\\ -f_3(x)=-\tan x} \Rightarrow f_3(-x)=-f_3(x) \Rightarrow \tan x為奇函數\\ f_4(x)= \sinh x \Rightarrow \cases{f_4(-x)= -\sinh x\\ -f_4(x)= -\sinh x} \Rightarrow f_4(-x)=-f_4(x) \Rightarrow \sinh x為奇函數} \\ \Rightarrow \bbox[red,2pt]{\cases{ e^x 不是偶函數也不是奇函數\\ e^{x^2} 為偶函數 \\\tan x為奇函數 \\\sinh x為奇函數}}$$

解答：$$\cases{f=x+y-z \Rightarrow \cases{f_x=1\\ f_y=1\\ f_z=-1}\\ g=xyz \Rightarrow \cases{g_x=yz\\ g_y= xz\\ g_z= xy}} \Rightarrow \cases{{\partial \over \partial x}fg = f_xg+ fg_x =xyz+(x+y-z)yz \\{\partial \over \partial y}fg = f_yg +fg_y = xyz + (x+y-z)xz\\ {\partial \over \partial z}fg = f_zg+ fg_z = -xyz+ (x+y-z)xy} \\ \Rightarrow grad(fg)= ({\partial \over \partial x}fg, {\partial \over \partial y}fg,{\partial \over \partial z}fg) =(yz(2x+y-z), xz(x+2y-z), xy(x+y-2z)) \\ \Rightarrow div(grad(fg))={\partial \over \partial x}yz(2x+y-z) +{\partial \over \partial y}xz(x+2y-z) +{\partial \over \partial z}xy(x+y-2z) \\ =2yz+ 2xz-2xy =\bbox[red,2pt]{2(yz+xz-xy)}$$

解答：$$A=\left(\begin{matrix}-1 & -1 & 0 \\-1 & -1 & 0 \\0 & 0 & 2\end{matrix}\right) \Rightarrow \det(A-\lambda I)=0 \Rightarrow -\lambda(\lambda-2)(\lambda+2)=0 \Rightarrow \cases{\lambda_1=0\\ \lambda_2=2 \\ \lambda_3=-2}\\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}-1 & -1 & 0 \\-1 & -1 & 0 \\0 & 0 & 2\end{matrix}\right)\left(\begin{matrix}x_1 \\x_2 \\ x_3\end{matrix}\right)=0 \Rightarrow \cases{x_1=-x_2\\ x_3=0},取v_1=\left(\begin{matrix}-1 \\1 \\ 0\end{matrix}\right)\\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}-3 & -1 & 0 \\-1 & -3 & 0 \\0 & 0 & 0\end{matrix}\right)\left(\begin{matrix}x_1 \\x_2 \\ x_3\end{matrix}\right)=0 \Rightarrow \cases{x_1=0\\ x_2=0},取v_2=\left(\begin{matrix}0 \\0 \\ 1\end{matrix}\right)\\ \lambda_3=-2 \Rightarrow (A-\lambda_3 I)\mathbf x=0 \Rightarrow \left(\begin{matrix}1 & -1 & 0 \\-1 & 1 & 0 \\0 & 0 & 4\end{matrix} \right)\left(\begin{matrix}x_1 \\x_2 \\ x_3\end{matrix}\right)=0 \Rightarrow \cases{x_1=x_2\\ x_3=0},取v_3=\left(\begin{matrix}1 \\1 \\ 0\end{matrix}\right)\\ 取P=(v_1\;v_2\;v_3) 及D=\left(\begin{matrix}\lambda_1 & 0 & 0 \\0 & \lambda_2 & 0 \\0 & 0 & \lambda_3\end{matrix}\right) \Rightarrow A=PDP^{-1}\\ 因此\text{eigenbasis}=\bbox[red,2pt]{\left\{ \left(\begin{matrix}-1 \\1 \\ 0 \end{matrix} \right), \left(\begin{matrix}0 \\0 \\ 1\end{matrix}\right), \left(\begin{matrix}1 \\1 \\ 0\end{matrix}\right)\right\}},\\對角化A=\bbox[red,2pt]{\left(\begin{matrix}-1 & 0 & 1 \\1 & 0 & 1 \\0 & 1 & 0\end{matrix}\right) \left(\begin{matrix}0 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & -2\end{matrix}\right) \left(\begin{matrix}\frac{-1}{2} & \frac{1}{2} & 0 \\0 & 0 & 1 \\\frac{1}{2} & \frac{1}{2} & 0\end{matrix}\right)}$$

解答：$$\cases{y_1'- 2y_1+3y_2=0 \\ y_2'-y_1+2y_2=0} \Rightarrow \cases{\mathcal L\{y_1'- 2y_1+3y_2\}=0 \\ \mathcal L\{y_2'-y_1+2y_2\} =0} \Rightarrow \cases{sY_1(s)-y_1(0)-2Y_1(s)+3Y_2(s)=0 \\ sY_2(s)-y_2(0)-Y_1(s)+2 Y_2(s)=0} \\\Rightarrow \cases{(s-2)Y_1(s) +3Y_2(s)=1  \cdots(1)\\ -Y_1(s)+(s+2) Y_2(s)=0 \cdots(2)},由(2)得Y_1(s)= (s+2) Y_2(s) 代入(1) \\\Rightarrow (s^2-4)Y_2(s)+ 3Y_2(s)=1 \Rightarrow Y_1(s)= {1\over s^2-1} \Rightarrow y_1(t)= \mathcal L^{-1}\left\{ {1\over s^2-1} \right\} ={1\over 2}(e^t-e^{-t})\\ 再將Y_1(s)= {1\over s^2-1}代入(2) \Rightarrow Y_2(s)= {1\over (s+2)(s^2-1)} \Rightarrow y_2(t)= \mathcal L^{-1}\left\{ {1\over (s+2)(s^2-1)}\right\} \\={1\over 3}e^{-2t} -{1\over 2}e^{-t} +{1\over 6}e^t \\ \Rightarrow \bbox[red,2pt]{y_1(t)= {1\over 2}(e^t-e^{-t}),y_2(t)= {1\over 3}e^{-2t} -{1\over 2}e^{-t} +{1\over 6}e^t}$$

解答：$$y=a_0+ a_1x +a_2x^2+ a_3x^3+a_4x^4 + \cdots \Rightarrow y'=a_1+ 2a_2x + 3a_3x^2 +4a_4 x^3 +\cdots\\ \Rightarrow 3xy'=3a_1x +6a_2x^2 +9a_3x^3 +12a_4x^4+\cdots \\ \Rightarrow (3x-1)y'= -a_1+ (3a_1-2a_2)x +(6a_2-3a_3)x^2 +(9a_3-4a_4)x^3+ \cdots\\ \Rightarrow y''=2a_2 + 6a_3x + 12a_4x^2 +  20a_5x^3+\cdots \\ \Rightarrow x(x-1)y'' =-2a_2 x+(2a_2-6a_3)x^2 +(6a_3-12a_4)x^3 +(12a_4-20a_5)x^4+\cdots\\ \Rightarrow x(x-1)y''+(3x-1)y'+y= (a_0-a_1)+ (4a_1-4a_2)x + (9a_2-9a_3)x^2 +(16a_3-16a_4)x^3+\cdots =0 \\ \Rightarrow y=C(1+x+x^2 + \cdots) ={C_1\over 1-x}\\ 由於原微分方程為二階，因有另一解設為y_2={C_2\over 1-x}P(x) \Rightarrow y_2'={C_2\over 1-x}P'+ {C_2\over (1-x)^2}P \\\Rightarrow y_2''={C_2\over 1-x}P'' +{2C_2\over (1-x)^2}P' +{2C_2\over (1-x)^3}P 將y_2,y_2'及y_2''代回原式  \Rightarrow  xP''+P'=0 \Rightarrow P=\ln x\\ \Rightarrow \bbox[red, 2pt]{y={C_1\over 1-x}+{C_2\over 1-x}\ln x}$$

解答：$$\mathbf{(a)}\;令\cases{M(x,y)=e^{x+y}+ ye^y\\ N(x,y)= xe^y-1}，則(e^{x+y}+ ye^y)dx +(xe^y-1)dy=0 相當於Mdx+Ndy=0\\ 由於\cases{M_y= e^{x+y}+e^y+ ye^y\\ N_x=e^y} \Rightarrow M_y\ne N_x \Rightarrow 非恰當(Not\; Exact)\\ 令積分因子\mu(y) \Rightarrow \cases{{\partial \over \partial y}M\mu =M_y\mu +M\mu_y= (e^{x+y}+e^y+ ye^y)\mu +(e^{x+y}+ ye^y)\mu_y\\ {\partial \over \partial x}N\mu = N_x\mu =e^y\mu} \\ \Rightarrow (e^{x+y}+e^y+ ye^y)\mu +(e^{x+y}+ ye^y)\mu_y=e^y\mu \Rightarrow \mu+\mu_y=0 \Rightarrow \mu=e^{-y}\\ \Rightarrow \cases{M\mu= e^x+y\\ N\mu = x-e^{-y}}\Rightarrow \Psi(x,y)=\int (e^x+y)\,dx = \int x-e^{-y}\,dy \Rightarrow \Psi =e^x+xy+ \phi(y)=xy+e^{-y}+ \varphi(x)\\ \Rightarrow \Psi=e^x+xy+ e^{-y}=C，再將初始值y(0)=-3代入\Rightarrow 1+e^{3} =C\\ \Rightarrow \bbox[red,2pt]{e^x+xy+ e^{-y}= 1+e^{3}}$$

解答：$$\lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n\to \infty} \left|{x^{2n+3}\over (2n+3)!} \cdot { (2n+1)!\over x^{2n+1}} \right| = x^2\lim_{n\to \infty} {1\over (2n+3)(2n+2)} =0\\ \Rightarrow 收斂半徑= \bbox[red, 2pt]{\infty}$$

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