114年雲科大電子碩士班-工程數學詳解

 國立雲林科技大學114學年度碩士班招生考試

系所:電子系
科目:工程數學

解答:$$\textbf{(a) }xy'-3y=0 \Rightarrow xdy=3ydx \Rightarrow \int {1\over 3y}\,dy =\int {1\over x}\,dx  \Rightarrow {1\over 3}\ln(3y) =\ln x+c_1 \\\quad \Rightarrow \ln (3y) =\ln x^3 +c_2 =\ln c_3x^3 \Rightarrow y={c_3\over 3}x^3 \Rightarrow \bbox[red, 2pt]{y=c_4x^3} \\\textbf{(b) }y''-4y'+4y=0 \Rightarrow \lambda^2-4\lambda+4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow \lambda=2 \Rightarrow \bbox[red, 2pt]{y=c_1e^{2x}+ c_2xe^{2x}} \\\textbf{(c) }y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow y''+{1\over x}y'-{1\over x^2}y=0 \\\quad \Rightarrow x^2y''+xy'-y= m(m-1)x^m+mx^m-x^m =(m^2-1)x^m=0 \Rightarrow m^2-1=0\\\quad \Rightarrow m=\pm 1 \Rightarrow \bbox[red, 2pt]{y=c_1x+{c_2 \over x}} $$

解答:$$\cases{P(x,y)=x^2-4y\\ Q(x,y) =-x} \Rightarrow \cases{P_y=-4\\ Q_x=-1} \Rightarrow u'={P_y-Q_x\over Q}u ={3\over x}u \Rightarrow \int {1\over u}\,du =\int{3\over x}\,dx \\\Rightarrow \bbox[red, 2pt]{\text{ integration factor }u(x)= x^3} \Rightarrow \cases{uP=x^5-4x^3y\\ uQ=-x^4} \Rightarrow (uP)_y =-4x^3=(uQ)_x  \\ \Rightarrow \Phi(x,y)= \int (x^5-4x^3y)\,dx = \int -x^4\,dy \Rightarrow \Phi(x,y) = {1\over 6}x^6-x^4y +\phi(y) =-x^4y+\rho(x) \\ \Rightarrow {1\over 6}x^6-x^4y=c_1 \Rightarrow \bbox[red, 2pt]{y={1\over 6}x^2-{c_1\over x^4}}$$

解答:$$y''-4y'+3y=0 \Rightarrow \lambda^2-4\lambda+3=0 \Rightarrow (\lambda-3)(\lambda-1)=0 \Rightarrow \lambda=1,3 \Rightarrow y_h=c_1e^x+ c_2e^{3x}\\ y_p=A\cos 2x+B\sin 2x \Rightarrow y_p'=-2A\sin 2x+2B\cos 2x \Rightarrow y_p''=-4A\cos 2x-4B\sin 2x \\ \Rightarrow y_p''-4y_p'+3y_p =(-A-8B)\cos 2x+(8A-B)\sin 2x =\sin 2x \Rightarrow \cases{-A-8B=0\\ 8A-B=1} \\ \Rightarrow \cases{A=8/65\\ B=-1/65} \Rightarrow y_p={8\over 65} \cos 2x-{1\over 65}\sin 2x \Rightarrow y=y_h+y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1e^x+ c_2e^{3x}+ {8\over 65} \cos 2x-{1\over 65}\sin 2x}$$

解答:$$\textbf{(a) }L\{f(t)\} =L\{\cos(2t+\theta_0)\} = L\{\cos(2t)\cos \theta_0-\sin(2t) \sin \theta_0 \} = \bbox[red, 2pt]{{S\over S^2+4} \cos\theta_0-{2\over S^2+4} \sin \theta_0} \\\textbf{(b) }L^{-1}\{F(s)\} =L^{-1} \left\{{1\over (S+1)(S+2)}\right\}  =L^{-1} \left\{{1\over S+1}-{1\over S+2}\right\} = \bbox[red, 2pt]{e^{-t}-e^{-2t}}$$

解答:$$$$

解答:$$\textbf{(a) }\cases{\mathbf x=\begin{bmatrix} x_1\\ x_2\end{bmatrix} \\ \mathbf y= \begin{bmatrix} y_1\\ y_2\end{bmatrix}} \Rightarrow \cases{T(\mathbf x)=x_1-x_2 \\T(\mathbf y)=y_1-y_2\\ T(\mathbf x+\mathbf y) =(x_1+y_1)-(x_2+y_2)} \Rightarrow T(\mathbf x)+T(\mathbf y) =T(\mathbf x+\mathbf y) \cdots(1)\\\mathbf x=\begin{bmatrix} x_1\\ x_2\end{bmatrix} \Rightarrow a\mathbf x=\begin{bmatrix} ax_1\\ ax_2 \end{bmatrix} \Rightarrow T(a\mathbf x)=ax_1-ax_2=a(x_1-x_2) =aT(\mathbf x) \cdots(2)\\\text{By (1) and (2)}, T(\mathbf x) \text{ is linear. }\bbox[red, 2pt]{QED} \\\textbf{(b) }u=\begin{bmatrix} x\\ y\\ z\end{bmatrix} \in W \Rightarrow x+2y=z \Rightarrow u= \begin{bmatrix} x\\ y\\ x+2y\end{bmatrix} =x\begin{bmatrix} 1\\ 0\\ 1\end{bmatrix}+y\begin{bmatrix} 0\\ 1\\ 2\end{bmatrix} \Rightarrow W= span\left\{\begin{bmatrix} 1\\ 0\\ 1\end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 2\end{bmatrix} \right\} \\ \Rightarrow v= \begin{bmatrix} 1\\ 0\\ 1\end{bmatrix}\times\begin{bmatrix} 0\\ 1\\ 2\end{bmatrix} =\begin{bmatrix} -1\\ -2\\ 1\end{bmatrix} \Rightarrow v\in W^\bot \Rightarrow \bbox[red, 2pt]{W^\bot =span \left\{\begin{bmatrix} -1\\ -2\\ 1\end{bmatrix} \right\}}$$

解答:$$\textbf{(a) }A=\begin{bmatrix} -1& -1& 1\\0& -2& 1\\0 & 0& -1\end{bmatrix}  \Rightarrow \det(A-\lambda I)=-(\lambda+1)^2 (\lambda+2)=0 \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: -1,-2}} \\\textbf{(b) } \lambda_1=-1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 0 & -1 & 1 \\0 & -1 & 1 \\0 & 0 & 0\end{bmatrix} \begin{bmatrix} x_1\\x_2 \\x_3\end{bmatrix} =0 \Rightarrow x_2=x_3 \\\qquad \Rightarrow v= \begin{bmatrix} x_1\\x_3\\x_3\end{bmatrix} =x_1\begin{bmatrix} 1\\ 0\\ 0\end{bmatrix} +x_3 \begin{bmatrix} 0\\ 1\\1\end{bmatrix}, \text{choose }v_1=  \begin{bmatrix} 1\\ 0\\0\end{bmatrix},v_2= \begin{bmatrix} 0\\ 1\\1\end{bmatrix} \\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} 1 & -1 & 1 \\0 & 0 & 1 \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=x_2\\ x_3=0} \\ \qquad \Rightarrow v =\begin{bmatrix} x_2\\x_2\\0\end{bmatrix} =x_2\begin{bmatrix} 1\\1\\0\end{bmatrix} , \text{choose v_3= } \begin{bmatrix} 1\\1\\0\end{bmatrix}  \\ \Rightarrow V=\{v_1,v_2, v_3\}\Rightarrow \bbox[red, 2pt]{V = \left\{ \begin{bmatrix} 1\\ 0\\0\end{bmatrix},  \begin{bmatrix} 0\\ 1\\1\end{bmatrix}, \begin{bmatrix} 1\\1\\0\end{bmatrix}\right\}}$$

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