113年成大土木碩士班-工程數學

 國立成功大學113學年度碩士班招生考試試題

系所：土木工程學系
考試科目：工程數學

解答： $$\textbf{(a)}\;  y''+4y'-5y=0 \Rightarrow \lambda^2+4\lambda-5=0 \Rightarrow (\lambda+5)(\lambda -1)=0 \\\quad \Rightarrow \lambda=1,-5 \Rightarrow y_h=c_1e^t+ c_2e^{-5t} \\ \textbf{(b)}\; \text{Applying variations of parameters, let } \cases{y_1=e^t\\ y_2=e^{-5t}}  \Rightarrow W= \begin{vmatrix}y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} =-6e^{-4t} \\ \quad \Rightarrow y_p=-e^t \int{e^{-5t}\cdot 10te^{-t} \over -6e^{-4t}}\,dt +e^{-5t} \int{e^t\cdot 10te^{-t} \over -6e^{-4t}}\,dt ={5\over 3}e^t \int te^{-2t}\,dt -{5\over 3} e^{-5t}\int te^{4t}\,dt \\\quad ={5\over 3}e^t \cdot \left(-{1\over 4}e^{-2t}(2t+1) \right) -{5\over 3} e^{-5t} \left( {1\over 16}e^{4t}(4t-1)\right)\Rightarrow \bbox[red, 2pt]{y_p =-{5\over 16}(4t+1)e^{-t}} \\\textbf{(c)}\; y=y_h+y_p \Rightarrow \bbox[red, 2pt] {y=c_1e^t+ c_2e^{-5t}-{5\over 16}(4t+1)e^{-t}}$$

解答： $$\textbf{(a)} \;A=\begin{bmatrix}8 & 3 \\-3 & 2 \end{bmatrix} \Rightarrow \det(A-\lambda I)=(8-\lambda)(2-\lambda)+9 = \lambda^2-10\lambda+ 25 =0 \\ \quad \Rightarrow (\lambda-5)^2=0 \Rightarrow \lambda=5 \Rightarrow A\text{ has eigenvalue } \bbox[red, 2pt]{\lambda_1=5} \text{ of multiplicity two.} \\ \textbf{(b)}\; c_0+c_1\lambda 有兩個變數c_0及c_1，而\lambda_1是重根，無法完整求出兩變數\\\textbf{(c)}\; \cases{f(A)=A^n \\ r(A)=c_1A+c_0I} \Rightarrow \cases{f(\lambda)= \lambda^n \\ r(\lambda)=c_1\lambda+c_0} \Rightarrow  \cases{f(\lambda)=r(\lambda) \\ f'(\lambda)= n\lambda^{n-1} =r'(\lambda) =c_1} \\\Rightarrow \cases{f(5)=r(5)\\ f'(5)=r'(5)} \Rightarrow \cases{5^n=5c_1+c_0\\ n\cdot 5^{n-1}= c_1} \Rightarrow c_0=(1-n)5^n \\\Rightarrow A^n =n\cdot 5^{n-1} A+(1-n)5^n I = n\cdot 5^{n-1} \begin{bmatrix}8 & 3 \\-3 & 2 \end{bmatrix}+(1-n)5^n \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} \\=\begin{bmatrix}8n5^{n-1}+(1-n)5^n & 3n5^{n-1} \\-3n5^{n-1} & 2n5^{n-1} +(1-n)5^n\end{bmatrix} \Rightarrow \bbox[red, 2pt]{A^n=\begin{bmatrix}(3n+5)5^{n-1} & 3n5^{n-1} \\-3n5^{n-1} & (5-3n)5^{n-1} \end{bmatrix}} \\\textbf{(d)}\; n=9 \Rightarrow \bbox[red, 2pt]{A^9 = \begin{bmatrix} 32\cdot 5^{8} & 27\cdot 5^{8} \\-27\cdot 5^{8} & -22\cdot 5^{8} \end{bmatrix}}$$

解答： $$\textbf{(a)}\; g\mathbf v= yze^{x^2+y^2+z^2} \mathbf i+ xz e^{x^2+y^2+z^2} \mathbf j+ xye^{x^2+y^2+z^2} \mathbf k \\\qquad \Rightarrow \text{div}(g\mathbf v) = \frac{\partial }{\partial x} yze^{x^2+y^2+z^2}+\frac{\partial }{\partial y} xze^{x^2+y^2+z^2}+\frac{\partial }{\partial z} xye^{x^2+y^2+z^2} \\\qquad =2xyz e^{x^2+y^2+z^2}+ 2xyz e^{x^2+y^2+z^2}+2xyz e^{x^2 +y^2+z^2} = \bbox[red, 2pt]{6xyz e^{x^2+y^2+z^2}} \\\textbf{(b)}\;  \nabla^2 g= \frac{ \partial^2 }{\partial x^2}g + \frac{\partial^2 }{\partial y^2}g + \frac{\partial^2 }{\partial z^2}g =\frac{\partial }{\partial x} 2xe^{x^2+y^2+z^2} + \frac{\partial }{\partial y}2ye^{x^2+y^2+z^2} + \frac{\partial }{\partial z} 2ze^{x^2+y^2+z^2}  \\\qquad=2e^{x^2+y^2+z^2} +4x^2e^{x^2+y^2+z^2} +2e^{x^2+ y^2+z^2} +4y^2e^{x^2+y^2+z^2} +2e^{x^2+y^2+z^2} +4z^2e^{x^2+y^2+z^2}\\\qquad = \bbox[red, 2pt]{(6+4(x^2+y^2 +z^2)) e^{x^2+y^2+z^2} } \\\textbf{(c)}\; \text{grad }g =\left( \frac{\partial }{\partial x}g,   \frac{\partial }{\partial y}g,  \frac{\partial }{\partial z}g \right) = \left( 2xe^{x^2+ y^2+z^2},    2ye^{x^2+y^2+z^2},   2z e^{x^2+y^2+z^2} \right) \\ \qquad \Rightarrow \text{curl(grad }g)=\begin{vmatrix}\mathbf i & \mathbf j& \mathbf k \\{\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ 2xe^{x^2+y^2+z^2}& 2ye^{x^2+y^2+z^2} & 2ze^{x^2+ y^2+z^2}\end{vmatrix} = \bbox[red, 2pt]{ \mathbf {0}} \\\textbf{(d)}\; \text{curl }\mathbf v= \begin{vmatrix} \mathbf i & \mathbf j& \mathbf k \\{\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ yz& xz & xy \end{vmatrix} =0 \Rightarrow \text{div(curl }\mathbf v)  = \bbox[red, 2pt]{ \mathbf {0}} \\\textbf{(e)}\; \mathbf v\cdot \mathbf v =y^2z^2 +x^2z^2+ x^2y^2\\\qquad  \Rightarrow \text{grad}(\mathbf v\cdot \mathbf v) = \bbox[red, 2pt]{2x(y^2+z^2)\mathbf i+2y(x^2+z^2)\mathbf j +2z(x^2+y^2) \mathbf k}$$

解答： $$A(\alpha)= \int_{-\infty}^\infty f(x)\cos (\alpha x)\,dx =\int_0^3 \cos(\alpha x)\,dx= \left. \left[ {1\over \alpha}\sin(\alpha x) \right] \right|_0^3 ={1\over \alpha}\sin(3\alpha)\\ B(\alpha)=\int_{-\infty}^\infty f(x)\sin(\alpha x)\,dx =\int_0^3 \sin(\alpha x)\,dx = \left. \left[ -{1\over \alpha} \cos(\alpha x) \right] \right|_0^3 = {1\over \alpha}(1-\cos(3\alpha) \\ \Rightarrow f(x)= {1\over \pi} \int_0^\infty \left( A(\alpha) \cos(\alpha x)+ B(\alpha)\sin(\alpha x) \right) \,d\alpha \\={1\over \pi} \int_0^\infty \left( {1\over \alpha} \sin(3\alpha) \cos(\alpha x)+ {1\over \alpha}(1-\cos(3\alpha) \sin(\alpha x) \right) \,d\alpha \\\qquad =\bbox[red, 2pt]{{1\over \pi} \int_0^\infty {1\over \alpha} \left(  1+ \sin(3\alpha-\alpha x)  \right) \,d\alpha}$$

解答： $$A=\begin{bmatrix}3 & 3 & -6 \\ 2 & -2 & 4 \\1 & 1 & -2 \end{bmatrix} \xrightarrow{R_2-(\color{red}{2/3})R_1\to R_2} \begin{bmatrix}3 & 3 & -6 \\ 0 & -4 & 8 \\1 & 1 & -2 \end{bmatrix} \xrightarrow{R_3-(\color{red}{1/3})R_1\to R_2} \begin{bmatrix}3 & 3 & -6 \\ 0 & -4 & 8 \\0 & 0 & 0 \end{bmatrix} \\ \Rightarrow A= \begin{bmatrix}1 & 0 &0 \\\color{red}{2/3} & 1 &0 \\ \color{red}{1/3}& 0 & 1 \end{bmatrix}  \begin{bmatrix}3 & 3 & -6 \\ 0 & -4 & 8 \\0 & 0 & 0 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{L= \begin{bmatrix}1 & 0 &0 \\ {2/3} & 1 &0 \\  {1/3}& 0 & 1 \end{bmatrix} ,U=  \begin{bmatrix}3 & 3 & -6 \\ 0 & -4 & 8 \\0 & 0 & 0 \end{bmatrix} }$$
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