113年台師大機電碩士班-工程數學詳解

國立臺灣師範大學113學年度碩士班招生考試

程目: 工程數學
適用系所: 機電工程學系 

解答:$$xy'=y^2+y \Rightarrow {1\over y^2+y} dy ={1\over x}dx \Rightarrow \ln {y\over y+1} =\ln x+c_1 \Rightarrow y={c_2x \over 1-c_2x} \\ \Rightarrow y(2)= {2c_2\over 1-2c_2} =4 \Rightarrow c_2= {2\over 5} \Rightarrow y={(2/5) x\over 1-(2/5)x} \Rightarrow \bbox[red, 2pt]{y={2x\over 5-2x}}$$

解答:$$(x^2D^2+ 2xD-6I)y= x^2y''+2xy'-6y=0 \\ y=x^m \Rightarrow \cases{y'=mx^{m-1} \\ y''=m(m-1)x^{m-2}} \Rightarrow (m^2-m+2m-6)x^m=  (m^2+m-6)x^m =0\\ \Rightarrow (m+3)(m-2)x^m=0 \Rightarrow m=2,-3 \Rightarrow y=c_1x^2 +c_2x^{-3} \Rightarrow y'=2c_1x -3c_2x^{-4} \\ \Rightarrow \cases{ y(1) =c_1 +c_2= 0.5\\ y'(1)=2c_1-3c_2=1.5} \Rightarrow \cases{c_1=0.6\\ c_2= -0.1} \Rightarrow \bbox[red, 2pt]{y=0.6x^2-0.1x^{-3}}$$

解答:$$\cases{y_1'=-2y_1+3y_2\\ y_2' = 4y_1-y_2} \Rightarrow \mathbf y'=\begin{bmatrix} -2 & 3 \\4 & -1 \end{bmatrix} \begin{bmatrix}y_1 \\ y_2 \end{bmatrix} \equiv A\mathbf y\\ A=\begin{bmatrix} -2 & 3 \\4 & -1 \end{bmatrix} = \begin{bmatrix}3/4 & -1 \\1 & 1 \end{bmatrix} \begin{bmatrix}2 & 0 \\0 & -5 \end{bmatrix} \begin{bmatrix}4/7 & 4/7 \\-4/7 & 3/7 \end{bmatrix} \\ \Rightarrow e^{At}= \begin{bmatrix}3/4 & -1 \\1 & 1 \end{bmatrix} \begin{bmatrix}e^{2t} & 0 \\0 & e^{-5t} \end{bmatrix} \begin{bmatrix}4/7 & 4/7 \\-4/7 & 3/7 \end{bmatrix} = \begin{bmatrix}{4\over 7}e^{-5t}+{3\over 7}e^{2t} & -{3\over 7}e^{-5t}+{3\over 7}e^{2t} \\ -{4\over 7}e^{-5t}+{4\over 7}e^{2t} & {3\over 7}e^{-5t}+{4\over 7}e^{2t} \end{bmatrix} \\ \Rightarrow \mathbf y= e^{At} \begin{bmatrix}c_1 \\c_2 \end{bmatrix} \Rightarrow \begin{bmatrix}4 \\3 \end{bmatrix} =e^{0} \begin{bmatrix}c_1 \\c_2 \end{bmatrix} =\begin{bmatrix}c_1 \\c_2 \end{bmatrix} \Rightarrow \mathbf y=  \begin{bmatrix}{4\over 7}e^{-5t}+{3\over 7}e^{2t} & -{3\over 7}e^{-5t}+{3\over 7}e^{2t} \\ -{4\over 7}e^{-5t}+{4\over 7}e^{2t} & {3\over 7}e^{-5t}+{4\over 7}e^{2t} \end{bmatrix}  \begin{bmatrix}4 \\3 \end{bmatrix} \\ \Rightarrow \mathbf y= \bbox[red, 2pt]{ \begin{bmatrix}y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix}e^{-5t}+ 3e^{2t} \\ -e^{-5t}  +4e^{2t} \end{bmatrix}}$$

解答:$$\cases{2x+2y+3z=4\\ 2x+3y+az=5\\ 3x+4y+ 5z=b} \Rightarrow \begin{bmatrix}2 & 2 & 3\\ 2 & 3 & a \\3 & 4& 5 \end{bmatrix} \begin{bmatrix}x \\y\\ z \end{bmatrix} = \begin{bmatrix}4 \\5 \\b \end{bmatrix} \equiv A\mathbf x= \mathbf b\\ \det(A)=-2a+7\\ \textbf{(a)}\; a=7/2 \Rightarrow \text{rref }[A\mid b]= \begin{bmatrix}2b-14 & 0 & 2b-14 & 0 \\ 0 & 2b-14 & b-7 & 0 \\0 & 0 & 0 & 2b-14 \end{bmatrix} \\ \qquad \text{No solution: } \bbox[red, 2pt]{a=7/2,b\ne 7} \\ \textbf{(b)}\; \text{A unique solution: }\bbox[red, 2pt]{a\ne 7/2} \\ \textbf{(c)}\; \text{Infinitely many solutions: }\bbox[red, 2pt]{a=7/2,b=7}$$

解答:$$\textbf{(a)}\; f=x^4+y^2+z \Rightarrow \nabla f=(f_x,f_y,f_z)= (4x^3,2y,1) \Rightarrow \nabla f(4,-1,3)= \bbox[red, 2pt]{(256,-2,1)} \\\textbf{(b)}\; \nabla^2 f =\frac{\partial }{\partial x}f_x + \frac{\partial }{\partial y}f_y + \frac{\partial }{\partial z}f_z  =12x^2+2+0= \bbox[red, 2pt]{12x^2+2} \\ \textbf{(c)}\; \vec v\cdot \nabla f =((x+y)^2,z^2,2yz) \cdot (4x^3,2y,1)= \bbox[red, 2pt]{4x^3(x+y)^2 +2yz^2+ 2yz} \\\textbf{(d)}\; \text{curl }\vec v =\begin{vmatrix}\vec i & \vec j & \vec k \\\frac{\partial  }{\partial x} & \frac{\partial  }{\partial y} & \frac{\partial  }{\partial z} \\ (x+y)^2 & z^2 & 2yz \end{vmatrix} = \bbox[red, 2pt]{-2(x+y) \vec k}  \\ \textbf{(e)} \;\cases{ \nabla f(3,0,2)=(108,0,1) \\ \vec v(3,0,2)= (9,4,0)}\Rightarrow D_v f(3,0,2) =(108,0,1) \cdot {(9,4,0)\over |(9,4,0)|} =\bbox[red, 2pt]{972\over \sqrt{97}}$$

解答:$$\textbf{(a)}\; f(-x)=-f(x) \Rightarrow f(x) \text{ is odd} \Rightarrow a_n=0\\ \qquad b_n= {1\over \pi} \left( \int_{-\pi}^0 -\sin(nx) dx+\int_0^\pi \sin(nx)\,dx \right) ={1\over \pi} \left( \left. \left[ {1\over n}\cos(nx) \right] \right|_{-\pi}^0 + \left. \left[ -{1\over n}\cos(nx) \right] \right|_0^\pi\right) \\={1\over \pi} \cdot {2\over n}(1-(-1)^n) ={2\over n\pi}(1-(-1)^n) \Rightarrow \bbox[red, 2pt]{f(x)= \sum_{n=1}^\infty {2\over n\pi}(1-(-1)^n) \sin(nx)} \\ \textbf{(b)}\; f(x)={4\over \pi} \sin x+ {4\over 3\pi} \sin(3x) +{4\over 5\pi} \sin(5x)+\cdots \\\qquad \Rightarrow f({\pi \over 2})= 1={4\over \pi}-{4\over 3\pi}+{4\over 5\pi}-\cdots ={4\over \pi}(1-{1\over3}+ {1\over 5}-{1\over 7}+\cdots )\\ \qquad \Rightarrow 1-{1\over3}+ {1\over 5}-{1\over 7}+\cdots =\bbox[red, 2pt]{\pi\over 4}$$

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