115年公務人員高等考試二級考試試題
等 別：高考二級
類 科：醫學工程
科 目：工程數學
解答:$$y''+2y'+2y=0 \Rightarrow r^2+2r+2=0 \Rightarrow r=-1\pm i \Rightarrow y_h= e^{-x} (C_1\cos x+C_2\sin x) \\ 取\cases{y_1= e^{-x}\cos x\\ y_2=e^{-x}\sin x} \Rightarrow W(y_1, y_2) = \begin{vmatrix} e^{-x}\cos x & e^{-x}\sin x \\ -e^{-x}(\cos x + \sin x) & e^{-x}(\cos x - \sin x) \end{vmatrix} =e^{-2x} \\ 利用變數變換法(\text{Variation of Parameters}):\\ y_p= -e^{-x} \cos x\int{ e^{-x}\sin x\cdot 4e^{-x} \sec^3 x\over e^{-2x}}\,dx + e^{-x}\sin x \int {e^{-x}\cos x\cdot 4e^{-x} \sec^3 x\over e^{-2x}}\,dx \\\qquad =-e^{-x} \cos x \int 4\sin x \sec^3 x\,dx +e^{-x}\sin x \int 4\cos x\sec^3 x\,dx \\=-e^{-x} \cos x\cdot \left( 2\sec^2 x \right)+ e^{-x}\sin x\cdot 4\tan x = -2e^{-x}\sec x + 4e^{-x}\sin x \tan x \\ y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y(x) = e^{-x}(C_1 \cos x + C_2 \sin x) - 2e^{-x}\sec x + 4e^{-x}\sin x \tan x }$$
解答:$$\mathcal L\left\{ y(t) \right\}+ \mathcal L\left\{ \int_{\tau=0}^t y(\tau) \cosh(t-\tau)\, d\tau\right\}=\mathcal L\left\{ 2t \right\}+ \mathcal L\left\{2e^t \right\} \\ \Rightarrow Y(s)+ Y(s)\cdot \mathcal{L}\{ \cosh(t)\} ={2\over s^2}+{2\over s-1} \Rightarrow Y(s) + Y(s) \cdot \frac{s}{s^2 - 1} = \frac{2}{s^2} + \frac{2}{s - 1} \\ \Rightarrow Y(s) = \frac{2(s^2 - 1)}{s^2(s - 1)} = \frac{2(s - 1)(s + 1)}{s^2(s - 1)} ={2\over s}+{2\over s^2} \\\Rightarrow y(t)= \mathcal{L}^{-1}\left\{\frac{2}{s} + \frac{2}{s^2}\right\} = 2 + 2t \Rightarrow \bbox[red, 2pt] {y(t)=2+2t}$$
解答:$$\textbf{(一) }圓C: |z|=2 \Rightarrow \cases{z=\pm 1在圓C內部\\ z=-3在圓C外部} \\\quad f (z) = \frac{10z}{(z^2 - 1)(z + 3)} = \frac{10z}{(z - 1)(z + 1)(z + 3)} \Rightarrow \text{Res}(f , 1) = \left. {10z\over (z+1)(z+3)}\right|_{z=1} ={5\over 4} \\ \quad \Rightarrow \text{Res}(f , -1) = \left. {10z\over (z-1)(z+3)}\right|_{z=-1} = {5\over 2} \Rightarrow \oint_C f (z)\,dz = 2\pi i \left( {5\over 4}+{5\over 2} \right)= \bbox[red, 2pt]{15\pi i\over 2} \\ \textbf{(二) } 依據\text{Argument Principle: } \oint_C{d\over dz}\ln f(z)\,dz = \oint_C{f'(z)\over f(z)}\,dz = 2\pi i(N-P), 其中\cases{N:零點總數\\ P:極點總數} \\ 零點: 10z=0 \Rightarrow z=0在圓C內部 \Rightarrow N=1\\ 極點: (z-1)(z+1)(z+3)=0 \Rightarrow \cases{z=\pm 1在圓C內部\\ z=-3在圓C外部} \Rightarrow P=2 \Rightarrow 2 \pi i(N-P)=2\pi i(1-2) \\=\bbox[red, 2pt]{-2\pi i}$$
解答:$$\textbf{(一) } \det(A-\lambda I) =0 \Rightarrow \begin{vmatrix} -\lambda & 1 \\ -2 & -3-\lambda \end{vmatrix} = 0 \Rightarrow \lambda^2+3\lambda +2=0 \Rightarrow (\lambda+2)( \lambda+1)=0 \\ \quad \Rightarrow \lambda_1 = -1, \lambda_2 = -2 \Rightarrow D= \begin{bmatrix}\lambda_1& 0\\ 0& \lambda_2 \end{bmatrix} = \begin{bmatrix}-1& 0\\0& -2 \end{bmatrix} \\ \lambda_1=-1 \Rightarrow (A-\lambda_1 I)v= \begin{bmatrix} 1 & 1 \\ -2 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies x_1 + y_1 = 0 \Rightarrow 取v_1= \begin{bmatrix}1\\-1 \end{bmatrix} \\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I)v= \begin{bmatrix} 2 & 1 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies 2x_2 + y_2 = 0 \Rightarrow 取v_2= \begin{bmatrix} 1\\-2 \end{bmatrix} \\ \Rightarrow P = \begin{bmatrix}v_1 & v_2 \end{bmatrix} \Rightarrow A=PDP^{-1} \Rightarrow \bbox[red, 2pt]{A= \begin{bmatrix} 1 & 1 \\ -1 & -2 \end{bmatrix} \begin{bmatrix}-1& 0\\ 0& -2 \end{bmatrix} \begin{bmatrix}2& 1\\-1& -1 \end{bmatrix}} \\ \textbf{(二) } e^{At} = \begin{bmatrix} 1 & 1 \\ -1 & -2 \end{bmatrix} \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-2t} \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -1 & -1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} 2e^{-t} - e^{-2t} & e^{-t} - e^{-2t} \\ -2e^{-t} + 2e^{-2t} & -e^{-t} + 2e^{-2t} \end{bmatrix}} $$
解答:$$\textbf{(一) } \vec{F} = 9x\vec{i} + 5y\vec{j} - 4z\vec{k} \Rightarrow 散度= \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(9x) + \frac{\partial}{\partial y}(5y) + \frac{\partial}{\partial z}(-4z) = 9 + 5 - 4 = 10 \\\quad 旋度= \nabla \times \vec{F} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 9x & 5y & -4z \end{vmatrix} =0 \Rightarrow \bbox[red, 2pt]{\cases{散度=10\\ 旋度=0}} \\\textbf{(二) }封閉曲面 S 包圍了一個圓柱體區域 V，其範圍為 x^2 + y^2 \le 4 且 $-2 \le z \le 2 \\\Rightarrow 體積V=2^2\pi(2-(-2))= 16\pi\\ 依高斯散度定理: \iint_S \vec{F} \cdot \vec{n} \, dA = \iiint_V (\nabla \cdot \vec{F}) \, dV = \iiint_V 10 \, dV = 10 \times 16\pi = \bbox[red, 2pt]{160 \pi}$$
解答:$$\textbf{(一) } \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} k e^{-x - \frac{y}{2} - \frac{z}{3}} \,dx \,dy \,dz = 1 \Rightarrow k \left( \int_{0}^{\infty} e^{-x} \,dx \right) \left( \int_{0}^{\infty} e^{-\frac{y}{2}} \,dy \right) \left( \int_{0}^{\infty} e^{-\frac{z}{3}} \,dz \right) = 1 \\\quad k\cdot 1\cdot 2\cdot 3=1 \Rightarrow k= \bbox[red, 2pt]{1\over 6} \\ \textbf{(二) } p(x, y, z) = \left(e^{-x}\right) \left(\frac{1}{2} e^{-\frac{y}{2}}\right) \left(\frac{1}{3} e^{-\frac{z}{3}}\right) \Rightarrow \cases{X\sim Exp(1) \\ Y\sim Exp(1/2)\\ Z\sim Exp(1/3)}\\ \quad Z\sim Exp(1/3) \Rightarrow E\{Z\}={1\over 1/3}= \bbox[red, 2pt]{3} \\ \textbf{(三) } \cases{ E\{X^3\} = \int_{0}^{\infty} x^3 e^{-x} \,dx = \Gamma(4) = 3! = 6 \\ E\{Y^2\} = \int_{0}^{\infty} y^2 \cdot \frac{1}{2} e^{-\frac{y}{2}} \,dy = \frac{2!}{(1/2)^2} = \frac{2}{1/4} = 8 } \Rightarrow E\{X^3 Y^2 Z\} = E\{X^3\} \cdot E\{Y^2\} \cdot E\{Z\} \\\quad =6\cdot 8\cdot 3= \bbox[red, 2pt]{144} \\註: 指數分配: X\sim Exp(\lambda) \Rightarrow f(x;\lambda)= \lambda e^{-\lambda x} \Rightarrow E(X)={1\over \lambda} \\ \text{Gamma 函數: }\Gamma(x)= \int_0^\infty t^{x-1}e^{-t}\,dt \Rightarrow \Gamma(n)= (n-1)! \\ 取t={y\over 2} \Rightarrow 2dt=dy \Rightarrow \int_0^\infty y^2\cdot {1\over 2}e^{-y/2}\,dy= \int_0^\infty 4t^2\cdot {1\over 2}e^{-t} (2\,dt) =4\Gamma(3)=4\cdot 2!=8$$
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