國立雲林科技大學110學年度碩士班招生考試
系所: 電子系
科目:工程數學
解答: $$\textbf{(a)}\; y'+2y+5=0 \Rightarrow e^{2x}y'+2e^{2x}y=-5e^{2x} \Rightarrow (e^{2x}y)'=-5e^{2x} \\ \quad \Rightarrow e^{2x}y= \int -5e^{2x}\,dx = -{5\over 2}e^{2x}+c_1 \Rightarrow \bbox[red, 2pt]{y=-{5\over 2}+c_1e^{-2x}} \\ \textbf{(b)}\; y''-6y'+10y=0 \Rightarrow \lambda^2-6\lambda+10=0 \Rightarrow \lambda=3\pm i \Rightarrow \bbox[red, 2pt]{y=e^{3x}(c_1\cos x+ c_2\sin x)} \\ \textbf{(c)}\; \text{Let }y=x^m, \text{ then } y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \quad \Rightarrow x^2y''-xy'-2y= m(m-1)x^m-mx^m-2x^m = (m^2-2m-2)x^m=0 \\ \quad \Rightarrow m^2-2m-2=0 \Rightarrow m=1\pm \sqrt 3 \Rightarrow \bbox[red, 2pt]{y=c_1x^{(1+\sqrt 3)} +c_2 x^{(1-\sqrt 3)}}$$
解答: $$\cases{P(x,y)=e^{x+y}+ ye^y\\ Q(x,y)= xe^y-1} \Rightarrow \cases{P_y =e^{x+y} +e^y+ye^y\\ Q_x=e^y} \Rightarrow -{P_y-Q_x\over P} =-1 \\ \Rightarrow u'=-u \Rightarrow \text{ integration factor }u(y)=\bbox[red, 2pt]{e^{-y}} \\ \Rightarrow \cases{uP=e^x+y\\ uQ=x-e^{-y}} \Rightarrow \cases{(uP)_y=1\\ (uQ)_x=1} \Rightarrow \text{Exact} \Rightarrow \Phi(x,y)= \int uP\,dx = \int uQ\,dy\\ \Rightarrow \Phi=\int( e^x+y)\,dx =\int (x-e^{-y})\,dy\Rightarrow e^x+ xy+\phi(y)=xy+e^{-y}+\rho(x) \\ \Rightarrow \Phi= \bbox[red, 2pt]{e^x+xy+e^{-y}+c_1=0}$$
解答: $$y''-6y'+9y=0 \Rightarrow \lambda^2-6\lambda+9=0 \Rightarrow (\lambda-3)^2=0 \Rightarrow \lambda=3 \Rightarrow y_h=c_1e^{3x}+ c_2xe^{3x}\\ \text{Applying variation of parameters, let }\cases{y_1=e^{3x}\\ y_2=xe^{3x}} \Rightarrow W=\begin{vmatrix} y_1& y_2\\ y_1'& y_2'\end{vmatrix} =e^{6x} \\ \Rightarrow y_p =-e^{3x}\int { xe^{3x}\cdot {e^{3x}\over x^2}\over e^{6x}} \, dx + xe^{3x} \int {e^{3x}\cdot {e^{3x}\over x^2} \over e^{6x}} \,dx =-e^{3x} \int {1\over x}\,dx+ xe^{3x} \int {1\over x^2}\,dx\\ =-e^{3x}\ln x-e^{3x} \Rightarrow y=y_h +y_p \Rightarrow \bbox[red, 2pt]{y=c_3e^{3x}+ c_2xe^{3x}-e^{3x}\ln x}$$
解答: $$\textbf{(a)}\; L\{5\}-3L\{e^{-4t}\}+2L\{ t^3\} ={5\over s}-3\cdot {1\over s+4}+2\cdot {6\over s^4} =\bbox[red, 2pt]{{5\over s}-{3\over s+4}+{12\over s^4}} \\ \textbf{(b)}\; L^{-1}\left\{{1\over (s+2)(s+3)} \right\} = L^{-1}\left\{{1\over s+2} -{1\over s+3}\right\} = \bbox[red, 2pt]{e^{-2t}-e^{-3t}}$$
解答: $$\cases{x+2y+3z=a\\ x+3y+8z=b\\ x+2y+2z=c} \Rightarrow \begin{bmatrix}1 & 2& 3 \\1 & 3& 8\\ 1& 2& 2 \end{bmatrix} \begin{bmatrix} x \\y\\z \end{bmatrix} =\begin{bmatrix}a \\ b \\c \end{bmatrix} \Rightarrow \begin{bmatrix} x \\y\\z \end{bmatrix} = \begin{bmatrix}1 & 2& 3 \\1 & 3& 8\\ 1& 2& 2 \end{bmatrix}^{-1} \begin{bmatrix}a \\ b \\c \end{bmatrix} \\=\begin{bmatrix} 10 & -2 & -7 \\-6 & 1 & 5 \\1 & 0 & -1 \end{bmatrix} \begin{bmatrix} a \\ b \\c \end{bmatrix}= \begin{bmatrix} 10a-2b-7c \\ -6a+ b+5c \\ a-c \end{bmatrix} \Rightarrow \bbox[red, 2pt]{\cases{x= 10a-2b-7c \\y =-6a+b+5c \\z=a-c}}$$
解答: $$\textbf{(a)}\; \bbox[red, 2pt]{v_1=(1,1), v_2=(1,-1)} \\ \textbf{(b)} \; T= \bbox[red, 2pt]{\begin{bmatrix}1 & 1 \\1 & -1 \end{bmatrix}} \\\textbf{(c)} T(x,y) 相當於將點(x,y)投影到直線x＋y＝0，也因此不可逆\\\textbf{(d)}\; A=\begin{bmatrix} \frac{1}{2} & \frac{-1}{2} \\\frac{-1}{2} & \frac{1}{2} \end{bmatrix} \Rightarrow A^2=\begin{bmatrix}\frac{1}{2} & \frac{-1}{2} \\\frac{-1}{2} & \frac{1}{2} \end{bmatrix} \begin{bmatrix}\frac{1}{2} & \frac{-1}{2} \\\frac{-1}{2} & \frac{1}{2} \end{bmatrix} =\begin{bmatrix}\frac{1}{2} & \frac{-1}{2} \\\frac{-1}{2} & \frac{1}{2} \end{bmatrix} =A \\ \Rightarrow A^2=A \Rightarrow \lim_{n \rightarrow \infty}A^n=A, \bbox[red, 2pt]{Q.E.D.}$$
解答: $$\textbf{(a)}\; \begin{array}{c}& X& Y &X^2 & XY &Y^2 \\\hline & 0 & 3 & 0 & 0 & 9\\ & 1& 3& 1& 3 & 9\\ & 1& 6 & 1& 6 & 36\\\hdashline \sum & 2 & 12& 2 & 9 & 54\end{array} \Rightarrow c_1={n\sum XY-(\sum X)(\sum Y)\over n\sum X^2-(\sum X)^2} ={3\cdot 9-2\cdot 12\over 3\cdot 2-2^2}={3\over 2}\\ \cases{\bar x =\sum X/n=2/3\\ \bar y=\sum Y/n=4} \Rightarrow \bar y=c_0+c_1\bar x \Rightarrow 4=c_0+ {3\over 2}\cdot {2\over 3} \Rightarrow c_0=3 \Rightarrow \bbox[red, 2pt]{f=3+{3\over 2}t} \\\textbf{(b)}\; \cases{f(0)=3\\ f(1)=9/2} \Rightarrow \text{least square error }=(3-3)^2+(3-{9\over 2})^2 +(6-{9\over 2})^2=\bbox[red, 2pt]{9\over 2}$$
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