115年雲林科大電子碩士班-工程數學詳解

國立雲林科技大學115學年度碩士班招生考試

系所：電子系 科目：工程數學(2)

解答:$$xy'=x^2+3y \Rightarrow y'-{3\over x}y=x \Rightarrow x^{-3}y'-{3\over x^4}y=x^{-2} \Rightarrow \left( x^{-3}y \right)'=x^{-2} \\ \Rightarrow x^{-3}y= \int x^{-2} \,dx =-x^{-1}+C \Rightarrow \bbox[red, 2pt]{y=-x^2+Cx^3}$$

解答:$$y''+4y=0 \Rightarrow r^2+4=0 \Rightarrow r=\pm 2i \Rightarrow y_h= c_1\cos(2x)+ c_2\sin(2x) \\ y_p=A\cos x+B\sin x \Rightarrow y_p'=-A\sin x+B\cos x \Rightarrow y_p''= -A\cos x-B\sin x\\ \Rightarrow y_p''+4y_p=3A\cos x+3B\sin x=2\sin x \Rightarrow \cases{A=0 \\ B=2/3} \Rightarrow y_p={2\over 3}\sin x \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1\cos(2x)+ c_2\sin(2x) +{2\over 3} \sin x}$$

解答:$${dy\over dx} =-{2xy+\cos x\over x^2+ \sin y} \Rightarrow (2xy+\cos x)dx+ (x^2+\sin y)dy=0 \\ \cases{M(x,y)= 2xy+ \cos x\\ N(x,y)=x^2+\sin y} \Rightarrow \cases{M_y=2x\\ N_x=2x} \Rightarrow M_y =N_x \Rightarrow \text{Exact} \\ \Rightarrow \Phi(x,y) = \int (2xy+\cos x)dx= \int(x^2+\sin y)dy \Rightarrow x^2y+\sin x+ \phi(y) =x^2y-\cos y+\rho(x) \\ \Rightarrow \Phi(x,y)= \bbox[red, 2pt]{x^2y+\sin x-\cos y=C}$$

解答:$$\textbf{(a) }\cases{y_1'= y_1-3y_2\\ y_2'=4y_1-6y_2} \Rightarrow   \begin{bmatrix}y_1'\\ y_2' \end{bmatrix}= \begin{bmatrix}1&-3\\ 4&-6 \end{bmatrix} \begin{bmatrix}y_1\\ y_2 \end{bmatrix} \\A =\begin{bmatrix}1&-3\\ 4&-6 \end{bmatrix} \Rightarrow \text{eigenvalues }\cases {\lambda_1=-2 \\\lambda_2= -3} \Rightarrow \text{eigenvectors}  \cases{v_1= \begin{bmatrix}1\\1 \end{bmatrix} \\ v_2= \begin{bmatrix}3\\4 \end{bmatrix}} \\ \Rightarrow \begin{bmatrix}y_1\\ y_2 \end{bmatrix}=c_1e^{\lambda_1 t}v_1+ c_2e^{\lambda_2 t} v_2 =c_1e^{-2t} \begin{bmatrix}1\\ 1 \end{bmatrix}+ c_2 e^{-3t} \begin{bmatrix}3\\4 \end{bmatrix} \Rightarrow \begin{bmatrix}y_1(0)  \\ y_2(0)\end{bmatrix} = \begin{bmatrix}2\\1 \end{bmatrix} = c_1 \begin{bmatrix}1\\1 \end{bmatrix} +c_2 \begin{bmatrix}3\\4 \end{bmatrix} \\ \Rightarrow \cases{c_1+3c_2=2\\ c_1+4c_2=1} \Rightarrow \cases{c_1=5\\ c_2=-1} \Rightarrow \bbox[red, 2pt] {\cases{y_1=5e^{-2t}-3e^{-3t} \\y_2=5e^{-2t}-4e^{-3t}}} \\ \textbf{(b) }$$

$$\textbf{(c) } \text{Both eigenvalues are strictly real, distinct, and negative. And }\\ \lim_{t\to \infty} \mathbf y=(0,0) \Rightarrow \text{ The system is } \bbox[red, 2pt]{\text{asymptotically stable.}}$$

解答:$$\textbf{(a) } A+B= \begin{bmatrix}3&1&0\\ 0& 2& 3 \end{bmatrix} + \begin{bmatrix}1&0& -1\\ 1& 2& 3 \end{bmatrix} = \begin{bmatrix}4& 1&-1\\ 1&4&6 \end{bmatrix} \Rightarrow (A+B)^T= \bbox[red, 2pt]{\begin{bmatrix}4& 1\\1 &4\\-1 &6\end{bmatrix}} \\ \textbf{(b )} \cases{2A-B= \begin{bmatrix}6&2 &0\\ 0& 4& 6 \end{bmatrix} -\begin{bmatrix}1&0& -1\\ 1& 2& 3 \end{bmatrix} = \begin{bmatrix}5& 2& 1\\-1& 2& 3 \end{bmatrix} \\[1ex] C+D= \begin{bmatrix}0& 1\\1& 0\\2&-1 \end{bmatrix} + \begin{bmatrix}2& 2\\-1& 4\\0&1 \end{bmatrix} = \begin{bmatrix}2& 3\\0& 4\\2& 0 \end{bmatrix}} \Rightarrow (2A-B)(C+D) = \bbox[red, 2pt]{ \begin{bmatrix}12&23\\4& 5 \end{bmatrix}} \\\textbf{(c) }2X-(A-B)=0 \Rightarrow X={1\over 2}(A-B) ={1\over 2} \begin{bmatrix}2& 1& 1\\-1& 0&0 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}1& 1/2& 1/2\\ -1/2& 0 & 0 \end{bmatrix}}$$

解答:$$\textbf{(a) } \begin{vmatrix} \sin n\theta & -\cos n\theta\\ \cos n\theta& \sin n\theta \end{vmatrix} =\sin^2n\theta+\cos^2n\theta= \bbox[red, 2pt]1 \\ \textbf{(b) } \begin{vmatrix} -2& 0& 0\\ 3& 1& 0\\1& 3& 5 \end{vmatrix} =-2 \begin{vmatrix} 1&0\\3& 5 \end{vmatrix} =-2\cdot 5=\bbox[red, 2pt]{-10} \\\textbf{(c) } \begin{vmatrix} 2& 1& 1& 1\\ 1& 2& 1& 1\\1& 1& 2& 1\\ 1& 1& 1& 2\end{vmatrix} =2 \begin{vmatrix} 2& 1& 1\\1& 2& 1\\1& 1& 2 \end{vmatrix} - \begin{vmatrix} 1& 1& 1\\ 1& 2& 1 \\1& 1& 2\end{vmatrix} + \begin{vmatrix} 1& 2& 1\\1& 1& 1\\ 1& 1& 2 \end{vmatrix} - \begin{vmatrix} 1& 2& 1\\1& 1& 2\\1& 1& 1 \end{vmatrix} =2\cdot 4-1-1-1= \bbox[red, 2pt]5$$

解答:$$\begin{bmatrix}1\\2 \end{bmatrix} = \begin{bmatrix}\cos \theta &-\sin \theta\\\sin \theta& \cos \theta \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} \Rightarrow \begin{bmatrix}x\\ y \end{bmatrix} =\begin{bmatrix}\cos \theta &-\sin \theta\\\sin \theta& \cos \theta \end{bmatrix}^{-1} \begin{bmatrix}1\\2 \end{bmatrix} =\begin{bmatrix}\cos \theta &\sin \theta\\ -\sin \theta& \cos \theta \end{bmatrix} \begin{bmatrix}1\\2 \end{bmatrix} \\= \bbox[red, 2pt]{\begin{bmatrix}\cos \theta-2\sin \theta\\ -\sin \theta+2\cos \theta \end{bmatrix}}$$

解答:$$ A = \begin{bmatrix} 2 & 0 & -2 \\ 0 & 4 & 0 \\ -2 & 0 & 5 \end{bmatrix} \Rightarrow \det(A-\lambda I) =   \begin{vmatrix} 2 - \lambda & 0 & -2 \\ 0 & 4 - \lambda & 0 \\ -2 & 0 & 5 - \lambda \end{vmatrix} =   (4 - \lambda) (\lambda - 1)(\lambda - 6) =0 \\ \Rightarrow \bbox[red, 2pt]{\text{The eigenvalues are: }1,4,6 }\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow   \begin{bmatrix} 1 & 0 & -2 \\ 0 & 3 & 0 \\ -2 & 0 & 4 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=2x_3\\ x_2=0} \\\qquad \Rightarrow v= x_3 \begin{bmatrix}2\\ 0\\ 1\end{bmatrix} , \text{ choosing }v_1= \begin{bmatrix}2\\0\\1 \end{bmatrix} \\ \lambda_2=4  \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow   \begin{bmatrix} -2 & 0 & -2 \\ 0 & 0 & 0 \\ -2 & 0 & 1 \end{bmatrix}  \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=-x_3 \\ x_3=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}0\\1\\ 0 \end{bmatrix}, \text{ choosing }v_2= \begin{bmatrix}0\\1\\0 \end{bmatrix} \\ \lambda_3=6 \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow      \begin{bmatrix} -4 & 0 & -2 \\ 0 & -2 & 0 \\ -2 & 0 & -1 \end{bmatrix}  \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_3=-2x_1\\ x_2=0} \\ \qquad \Rightarrow v= x_1 \begin{bmatrix}1\\0\\-2 \end{bmatrix}, \text{ choosing }v_3= \begin{bmatrix}1 \\0\\-2  \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{The eigenvectors are: }\begin{bmatrix}2\\0\\1 \end{bmatrix}, \begin{bmatrix}0\\1\\0 \end{bmatrix} ,\begin{bmatrix}1\\0\\-2  \end{bmatrix}}$$

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解題僅供參考，碩士班歷年試題及詳解