111年專技高考-電機工程技師-工程數學詳解

111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：$$假設另一解y_2= y_1u(x) = xu(x) \Rightarrow y_2'=u+ xu' \Rightarrow y_2''= 2u'+xu''\\ \Rightarrow (x^2-x)(2u'+xu'')-2x(u+ xu')+ 2xu=0 \Rightarrow u''={2\over x^2-x}u' =({2\over x-1}-{2\over x})u' \\ \Rightarrow \ln u'=\ln ({x-1\over x})^2 +C_1 \Rightarrow u'=C_2 ({x-1\over x})^2 \Rightarrow u = C_2\int ({x-1\over x})^2\,dx  =C_2(x-2\ln x-{1\over x})+ C_3 \\ \Rightarrow y_2= C_2(x^2-2x\ln x-1)+ C_3x \Rightarrow 通解y=C_1^*y_1+ C_2^*y_2  =C_3(x^2-2x\ln x-1)+ C_4x \\ \Rightarrow \bbox[red,2pt]{y=C_1(x^2-2x\ln x-1)+ C_2x,C_1 \text{ and }C_2為常數}$$

解答：$$假設\cases{\mathcal L\{f(t)\} =F(s) \\ \mathcal L^{-1}\{ F(s)\}=f(t) },則 \mathcal L\{tf(t)\} =-{d\over ds} F(s) \Rightarrow tf(t)= \mathcal L^{-1}\{-{d\over ds} F(s)\} \\\Rightarrow f(t)={1\over t} \mathcal L^{-1}\{-{d\over ds} F(s)\}  \Rightarrow  \mathcal L^{-1} \{\ln{s-a\over s-b}\} =\mathcal L^{-1}\{\ln (s-a)\} -\mathcal L^{-1}\{\ln (s-b)\} \\={1\over t}\mathcal L^{-1}\{-{d\over ds}\ln (s-a)\} -{1\over t}\mathcal L^{-1}\{-{d\over ds}\ln (s-b)\}  ={1\over t}\mathcal L^{-1}\{-{1\over s-a}\} -{1\over t}\mathcal L^{-1}\{-{1\over s-b} \} \\=-{1\over t}e^{at}+{1\over t}e^{bt} =\bbox[red,2pt]{e^{bt}-e^{at}\over t}$$

解答：$$\mathbf{(一)}\;令f(z)= {2z+1\over z+3} \Rightarrow \oint_C {2z+1\over (z+3)(z-1)}\,dz = \oint {f(z)\over z-1}\,dz = 2\pi i\cdot f(1)= 2\pi i \cdot  {3\over 4} =\bbox[red,2pt]{{3\over 2}\pi i}\\ \mathbf{(二)}\;令f(z)= {2z+1\over z+3} \Rightarrow f'(z)= {2\over z+3} -{2z+1\over (z+3)^2}\Rightarrow \oint_C {2z+1\over (z+3)(z-1)^2}\,dz = \oint_C {f(z)\over (z-1)^2}\,dz \\ \qquad =2\pi i\cdot f'(1) = 2\pi i \cdot \left( {1\over 2}-{3\over 16} \right) =\bbox[red, 2pt]{{5\over 8}\pi i}$$

解答：$$\mathbf{(一)}\;A=\begin{bmatrix} 1 & 1\\ 1& 3\end{bmatrix} \Rightarrow B=AA^t = \begin{bmatrix} 2 & 4\\ 4& 10\end{bmatrix} \Rightarrow \det(B-\lambda I)=0 \Rightarrow \lambda^2-12\lambda +4=0 \\\qquad \Rightarrow \lambda =  6\pm 4\sqrt 2  \Rightarrow A的奇異值為\sqrt{\lambda} = \bbox[red, 2pt]{2\pm \sqrt 2} \\\mathbf{(二)}\; \det(A-\lambda I)=0 \Rightarrow \lambda^2-4\lambda +2 =0 \Rightarrow \lambda = 2\pm \sqrt 2 \Rightarrow \mathbf x^T A\mathbf x= \mathbf x^T \lambda \mathbf x \\ \qquad =\lambda \mathbf x^T \mathbf x =\lambda  \Rightarrow \bbox[red, 2pt]{\cases{\mathbf x^T A\mathbf x 最大值為2+\sqrt 2\\ \mathbf x^T A\mathbf x最小值為2-\sqrt 2}}$$

解答：$$\mathbf{(一)}\;散度= \mathbf{div  }\vec F = \nabla\cdot \vec F ={\partial \over \partial x}F_1 +{\partial \over \partial y}F_2 +{\partial \over \partial z}F_3 ={\partial \over \partial x}x^2y^3 \sin z +{\partial \over \partial y}x^2 y^2z^2 +{\partial \over \partial z} 4\cos(xyz) \\= \bbox[red,2pt]{2xy^3\sin z+2x^2yz^2 -4xy\sin(xyz)} \\旋度=\text{curl }\vec F =\nabla \times \vec F = ({\partial \over \partial y}F_3-{\partial \over \partial z}F_2, {\partial \over \partial z}F_1-{\partial \over \partial x}F_3, {\partial \over \partial x}F_2-{\partial \over \partial y}F_1) \\ = \bbox[red, 2pt]{(-4xz \sin(xyz)-2x^2y^2 z,x^2y^3 \cos z+4yz \sin(xyz),2xy^2z^2-3x^2y^2 \sin z) } \\\mathbf{(二)}\; \cases{x(t)=t\\ y(t)= -2t+1\\ z(t)= 5t+2} ,t\in [0,1] \Rightarrow \cases{x'(t)= 1\\ y'(t)=-2\\ z'(t)=5} \\ \Rightarrow \int 3x^2dx +(2yz)dy+ (y^2) dz = \int_0^1 3t^2 + 2(-2t+1)(5t+2)(-2) + (-2t+1)^2 5\; dt \\ =\int_0^1 63t^2-24t-3\,dt = \left. \left[ 21t^3- 12t^2 -3t\right]\right|_0^1 =\bbox[red, 2pt]{6}$$

解答：$$\mathbf{(一)}\;E(X) = \int xp(x)\,dx =\int_0^1 x\cdot x\,dx +\int_1^2 (2-x)x\,dx ={1\over 3} +{2\over 3}= \bbox[red, 2pt]1 \\ \mathbf{(二)}\;E(X^2) = \int x^2p(x)\,dx = \int_0^1 x^3\,dx +\int_1^2 x^2(2-x)\,dx = {1\over 4}+{11\over 12} ={7\over 6} \\ \qquad \Rightarrow Var(X) = E(X^2)-(E(X))^2 = {7\over 6}-1^2 = \bbox[red, 2pt]{1\over 6}\\ \mathbf{(三)}\; E(X^3) =\int x^3p(x)\,dx = \int_0^1 x^4\,dx +\int_1^2 x^3(2-x)\,dx = {1\over 5}+ {13\over 10} =\bbox[red, 2pt]{3\over 2}$$

 ========================== END ========================

解題僅供參考，其他國考試題及詳解