113年專技高考電機工程技師-工程數學詳解

113年專門職業及技術人員高等考試

等 別：高等考試
類 科：電機工程技師
科 目：工程數學（包括線性代數、微分方程、複變函數與機率）

解答: $$y''-4y'+5y=0 \Rightarrow \lambda^2-4\lambda+5=0 \Rightarrow \lambda=2\pm i \Rightarrow y_h= e^{2x} (c_1 \cos(x)+c_2 \sin(x))\\ \text{Using variations of parameters, let }\cases{y_1=\cos(x)e^{2x}\\ y_2= \sin(x)e^{2x}}, \text{ then }W=\begin{vmatrix}y_1 & y_2\\ y_1' & y_2' \end{vmatrix} =e^{4x} \\ \Rightarrow y_p =-\cos(x)e^{2x} \int{ \sin(x)e^{2x}\cdot e^{2x} \csc x\over e^{4x}}\,dx + \sin(x)e^{2x} \int {\cos(x)e^{2x} \cdot e^{2x} \csc x \over e^{4x}}\,dx \\ \qquad =-\cos(x)e^{2x} \int  1\,dx + \sin(x)e^{2x} \int \cot (x)\,dx   =-x\cos(x)e^{2x}  + \sin(x)e^{2x} \ln(\sin x) \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=e^{2x}(c_1 \cos x+c_2 \sin x-x\cos x+ \sin x\ln (\sin x))}$$

解答: $$L\{f(t )\} ={1\over 2\beta^3} L\{ \sin(\beta t)\} -{1\over 2\beta^2}L\{t \cos(\beta t)\} \\ \Rightarrow F(s)={1\over 2\beta^3}\cdot {\beta\over s^2+\beta^2} -{1\over 2\beta^2} \cdot {s^2-\beta^2 \over (s^2+\beta^2)^2} \Rightarrow \bbox[red, 2pt]{F(s)={1\over (s^2+\beta^2)^2}}$$

解答:

$$\text{Let }f(z)={\sqrt 2\over 1+16z^4}= {\sqrt 2/16\over z^4-{1\over 16}} ={\sqrt 2\over 16 (z-{1\over 2}e^{\pi i/4}) (z-{1\over 2}e^{3\pi i/4}) (z-{1\over 2}e^{5\pi i/4}) (z-{1\over 2}e^{7\pi i/4}) }, \\ \text{and  let }C \text{ be the closed contour consisting of the interval }[-R, R] \text{ on x-axis and the semicircle }C_R\\ \text{ of radius }R\gt 1.\\ \oint_C f(z)\,dz= \int_{-R}^R {\sqrt 2\over 1+16x^4}\,dx + \int_{C_R} f(z)\,dz= I_1+I_2 = 2\pi i(\text{Res}(f,{1\over 2}e^{\pi i/4}) + \text{Res} ( f,{1\over 2}e^{2\pi i/4})), \\ =2\pi i\left({1\over -4+4i} +{1\over 4+4i} \right)=2\pi i\left(-{1\over 4}i \right)={\pi \over 2} \\\lim_{R\to \infty} \int_{C_R}f(z) \,dz=0 \Rightarrow \lim_{R\to \infty} \int_{-R}^R {\sqrt 2\over 1+16x^4}\,dx =\int_{-\infty}^\infty {\sqrt 2\over 1+16x^4}\,dx = \bbox[red, 2pt]{\pi\over 2}$$

解答: $$\textbf{(一) }P為對稱矩陣 \Rightarrow P= \begin{bmatrix} a & c\\ c& b\end{bmatrix} \Rightarrow PA+A^TP =\begin{bmatrix} a & c\\ c& b\end{bmatrix} \begin{bmatrix} 0 & 1\\ -5& -6\end{bmatrix} +\begin{bmatrix} 0 & -5\\ 1& -6\end{bmatrix} \begin{bmatrix} a & c\\ c& b\end{bmatrix} =-I_2 \\ \Rightarrow \begin{bmatrix} -10c & a-5b-6c\\ a-5b-6c& 2c-12b \end{bmatrix} =\begin{bmatrix} -1 & 0\\ 0& -1\end{bmatrix} \Rightarrow \cases{c=1/10\\ a-5b=6c\\ 2c+1=12b} \Rightarrow \cases{a=11/10\\ b=c=1/10} \\\quad \Rightarrow \bbox[red, 2pt]{P=\begin{bmatrix}11/ 10& 1/10\\ 1/10& 1/10 \end{bmatrix}} \\\textbf{(二) }P=\begin{bmatrix}11/ 10& 1/10\\ 1/10& 1/10 \end{bmatrix} \Rightarrow \det(P-\lambda I) =\lambda^2-{6\over 5}\lambda +{1\over 10} =0 \\ \quad \Rightarrow 特徵值\lambda =\bbox[red, 2pt]{{3\over 5}\pm {\sqrt{26} \over 10}}$$


解答: $$\textbf{(一) } r(t) =[3t,4t^2,8t^4] \Rightarrow 切線向量r'(t)= \bbox[red, 2pt]{[3,8t,32t^3]}\\ \Rightarrow 單位切線向量{r'(t) \over ||r'(t)||} ={1 \over \sqrt{3^2+(8t)^2+ (32t^3)^2}}[3,8t,32t^3]   = \bbox[red, 2pt]{{1\over \sqrt{1024t^6 +64t^2+9}}[ 3,8t,32t^3]} \\\textbf{(二) }\text{By Gauss Divergence Theorem, } \iint_S F\cdot ndA = \iiint \text{div }F\,dV =\iiint (7+3-1)\,dV \\\qquad =9\times {4\over 3}\pi\cdot 3^3= \bbox[red, 2pt]{324\pi}$$

 

解答: $$\textbf{(一) } \iint P\,dxdy=1 \Rightarrow \int_0^\infty \int_0^\infty ke^{-x-y/2}\,dxdy =\int_0^\infty ke^{-y/2}\, dy= \left. \left[ -2ke^{-y/2} \right] \right|_0^\infty =2k=1\\ \qquad \Rightarrow \bbox[red, 2pt]{k={1\over 2}} \\\textbf{(二) } p(y) =\int_0^\infty {1\over 2}e^{-x-y/2}\,dx= {1\over 2}e^{-y/2} \Rightarrow E\{Y\} =\int_0^\infty y\cdot {1\over 2}e^{-y/2}\,dy = \left. \left[ -(y+2)e^{-y/2}\right] \right|_0^\infty =\bbox[red, 2pt]2 \\ \textbf{(三) }E\{X^2Y^2\} =\int_0^\infty \int_0^\infty {1\over 2}x^3y^2e^{-x-y/2} \,dxdy = \int_0^\infty {1\over 2}y^2 \left. \left[ -(x^3+3x^2+6x+6)e^{-x-y/2} \right] \right|_0^\infty\, dy \\\qquad = \int_0^\infty3y^2e^{-y/2}\, dy = 3\left. \left[ -2(y^2+4y+8)e^{-y/2} \right] \right|_0^\infty =3\times 16= \bbox[red, 2pt]{48}$$

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解題僅供參考，高普考歷年試題及詳解