112年雲科大碩士班-工程數學(1)詳解

國立雲林科技大學112學年度碩士班招生考試

系所: 電機系
科目: 工程數學(1)

解答: $$\mathbf{(1)}\; y'=5\sin(3x) \Rightarrow \int 1\,dy= \int 5\sin(3x)\,dx \Rightarrow \bbox[red, 2pt]{y=-{5\over 3}\cos (3x)+C}\\ \mathbf{(2)}\; y'+y=e^{5x} \Rightarrow 積分因子I(x)=e^{\int 1\,dx}=e^x \Rightarrow y'e^x+ ye^x=e^{6x} \\ \Rightarrow (ye^x)'=e^{6x} \Rightarrow ye^x =\int e^{6x}\,dx = {1\over 6}e^{6x}+C \Rightarrow \bbox[red, 2pt]{y={1\over 6}e^{5x}+Ce^{-x}} \\\mathbf{(3)}\; u=x+y+3 \Rightarrow u'=1+y' \Rightarrow u'-1=u^2 \Rightarrow \int {1\over u^2+1}du=\int 1\,dx \\ \Rightarrow \arctan(u)= \arctan(x+y+3)=x+c \Rightarrow x+y+3=\tan(x+c) \\ \Rightarrow \bbox[red,2pt]{y=\tan(x+c)-x-3}$$

解答: $$\mathbf{(1)}\; F(s)=L[f(t)] =L[te^{4t}] +L[e^{2t}\sin t]= \bbox[red, 2pt]{{1\over (s-4)^2}+{1\over (s-2)^2+1}}\\ \mathbf{(2)}\; f(t)=L^{-1}[F(s)] =L^{-1}[{e^{-2s}\over s(s-1)}] =L^{-1}[e^{-2s}({1\over s-1}-{1\over s})] =\bbox[red, 2pt]{u(t-2)(e^{t-2}-1)}\\ \mathbf{(3)}\;F(s)= L[f(t)]=L[\sin t]\cdot L[\cos t]={1\over s^2+1}\cdot {s\over s^2+1} =\bbox[red,2pt] {s\over (s^2+1)^2}$$

解答: $$L\{y'' \}-L\{y' \}=L\{e^t\cos t \} \Rightarrow [s^2Y(s)-sy(0)-y'(0)]-[sY(s)-y(0)]= {s-1\over (s-1)^2+1} \\ \Rightarrow (s^2-s)Y(s)= {s-1\over (s-1)^2+1} \Rightarrow Y(s) = {1\over s((s-1)^2+1)}\\={1\over 2}\cdot {1\over s}-{1\over 2}\cdot {s-1\over (s-1)^2+1}+{1\over 2}\cdot {1\over (s-1)^2+1} \\ \Rightarrow y(t)={1\over 2}L^{-1}\{ {1\over s}\}-{1\over 2}L^{-1}\{ {s-1\over (s-1)^2+1}\}+{1\over 2}L^{-1}\{{1\over (s-1)^2+1} \}\\ \Rightarrow \bbox[red, 2pt]{y(t)= {1\over 2}\left( u(t)-e^t\cos t+e^t\sin t\right)}$$

解答: $$先求齊次解:2x^2y''+5xy'+y=0, 令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow 2m(m-1)x^m+5mx^m+x^m=0 \Rightarrow (2m^2+3m+1)x^m=0 \\ \Rightarrow 2m^2+3m+1=0 \Rightarrow (2m+1)(m+1)=0 \Rightarrow m=-1/2,-1 \Rightarrow y_h=c_1x^{-1/2}+c_2x^{-1}\\ 接著令y_p=ax^2+bx+c \Rightarrow y_p'=2ax+b \Rightarrow y_p''=2a \\ \Rightarrow 4ax^2+10ax^2+5bx+ ax^2+bx+c=x^2-x \Rightarrow \cases{15a=1\\ 6b=-1\\c=0} \Rightarrow \cases{a=1/15\\ b=-1/6\\ c=0} \\ \Rightarrow y_p={1\over 15}x^2-{1\over 6}x \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y={1\over 15}x^2-{1\over 6}x+{c_1\over \sqrt x}+{c_2\over x}}$$

解答: $$\mathbf{(a)}\;\text{zero matrix }\begin{bmatrix} 0 & 0\\0 & 0 \end{bmatrix} \in V\\ A=\begin{bmatrix} a & c\\c & b \end{bmatrix} \in V \Rightarrow \alpha A=\begin{bmatrix} \alpha a & \alpha c\\\alpha c & \alpha b \end{bmatrix} \in V\\ \cases{A=\begin{bmatrix} a1 & c1\\c1 & b1 \end{bmatrix} \in V \\B=\begin{bmatrix} a2 & c2\\c2 & b2 \end{bmatrix} \in V} \Rightarrow A+B=A=\begin{bmatrix} a1+a2 & c1+c2\\c1+c2 & b1+b2 \end{bmatrix} \in V\\ 因此V\text{ is a vector space}$$

解答: $$A=\left[\begin{matrix}1 & 2 & 2 & -1 \\3 & 6 & 5 & 0 \\1 & 2 & 1 & 2\end{matrix}\right] \xrightarrow{-3R_1+R_2\to R_2,-R_1+R_3\to R_3} \begin{bmatrix}1 & 2 & 2 & -1 \\0 & 0 & -1 & 3 \\0 & 0 & -1 & 3\end{bmatrix} \xrightarrow{-R_2+R_3 \to R_3}\begin{bmatrix}1 & 2 & 2 & -1 \\0 & 0 & -1 & 3 \\0 & 0 & 0 & 0\end{bmatrix} \\ \xrightarrow{2R_2+R_1\to R_1, -R_2\to R_2} \begin{bmatrix}1 & 2 & 0 & 5 \\0 & 0 & 1 & -3 \\0 & 0 & 0 & 0\end{bmatrix}\Rightarrow \bbox[red, 2pt]{Rank(A)=2}, \\\bbox[red, 2pt]{\cases{\text{bases for the row space of A} =\{(1,2,0,5), (0,0,1,-3)\} \\\text{bases for the column space of A} =\{\begin{pmatrix} 1\\3\\1\end{pmatrix},\begin{pmatrix} 2\\5\\1\end{pmatrix} \}}}$$

解答: $$A=\left[\begin{matrix}1 & 1 & -1 & -1 \\3 & 2& 0 & 1 \\1 & 0 & 1 & 0\end{matrix}\right] \xrightarrow{-3R_1+R_2\to R_2,-R_1+R_3\to R_3} \begin{bmatrix}1 & 1& -1& -1\\ 0& -1& 3& 4\\ 0& -1& 2& 1 \end{bmatrix} \\ \xrightarrow{-R_2+R_3\to R_3}\begin{bmatrix}1 & 1& -1& -1\\ 0& -1& 3& 4\\ 0& 0& -1& -3 \end{bmatrix} \xrightarrow{R_2+R_1\to R_1}\begin{bmatrix}1 & 0& 2& 3\\ 0& -1& 3& 4\\ 0& 0& -1& -3 \end{bmatrix}  \xrightarrow{3R_3+R_2\to R_2}\begin{bmatrix}1 & 0& 2& 3\\ 0& -1& 0& -5\\ 0& 0& -1& -3 \end{bmatrix} \\ \xrightarrow{-R_2,-R_3} \begin{bmatrix}1 & 0& 2& 3\\ 0& 1& 0& 5\\ 0& 0& 1& 3 \end{bmatrix} \Rightarrow \text{the basis of the row space of }A=\{(1,0,2,3), (0,1,0,5), (0,0,1,3)\}\\ 假設\cases{a_1=(0,0,1,3) \\a_2=(0,1,0,5)\\ a_3=(1,0,2,3)}, 取u_1=a_1=(0,0,1,3) \Rightarrow e_1={u_1\over \Vert u_1\Vert}=(0,0,{1\over \sqrt{10}}, {3\over \sqrt{10}}) \\ u_2= a_2-(a_2\cdot e_1)e_1= (0,1,0,5)-{15\over \sqrt{10}}(0,0,{1\over \sqrt{10}}, {3\over \sqrt{10}}) =(0,1,-{3\over 2},{1\over 2}) \\ \Rightarrow e_2={u_2\over \Vert u_2\Vert} =(0,{\sqrt 2\over \sqrt 7},-{3\over \sqrt{14}},{1\over \sqrt{14}})\\ u_3=a_3-(a_3\cdot e_1)e_1-(a_3\cdot e_2)e_2 =(1,0,2,3)-{11\over \sqrt{10}}(0,0,{1\over \sqrt{10}}, {3\over \sqrt{10}})+{3\over \sqrt{14}}(0,{\sqrt 2\over \sqrt 7},-{3\over \sqrt{14}},{1\over \sqrt{14}})\\ =(1,{3\over 7},{9\over 35},-{3\over 35}) \\\Rightarrow  \text{orthogonal basis} =\{u_1,u_2,u_3\} =\bbox[red,2pt]{\{(0,0,1,3), (0,1,-{3\over 2},{1\over 2}), (1,{3\over 7},{9\over 35},-{3\over 35})\}}$$

解答: $$\cases{X=\begin{bmatrix}1 &-1\\ -1& 1\end{bmatrix}\\[1em]Y=[1,-1]\\[1em] Z=[-2]} \Rightarrow  \cases{X^2=\begin{bmatrix}2 &-2\\ -2& 2\end{bmatrix}\\ YX=[2,-2]\\ ZY=[-2,2]} \Rightarrow \cases{X^4=\begin{bmatrix}8 &-8\\ -8& 8\end{bmatrix}\\ YX+ZY=[0,0]}\\\Rightarrow A^2=\begin{bmatrix}X & 0\\ Y & Z \end{bmatrix}\begin{bmatrix}X & 0\\ Y & Z \end{bmatrix} =\begin{bmatrix}X^2 & 0\\ YX+ZY & Z^2 \end{bmatrix} =\begin{bmatrix}X^2 & 0\\ 0 & Z^2 \end{bmatrix} \Rightarrow A^8=\begin{bmatrix}X^8 & 0\\ 0 & Z^8 \end{bmatrix}\\=\bbox[red, 2pt]{\begin{bmatrix}128& -128 & 0\\ -128& 128 & 0\\ 0 & 0& 256 \end{bmatrix}}$$