113年中山光電碩士班-工程數學詳解

國立中山大學113學年度碩士班招生考試

科目名稱:工程數學【光電系碩士班】

 

解答: $$y'+2y=4e^{-2x} \Rightarrow e^{2x}y'+2e^{2x}y=4 \Rightarrow (e^{2x}y)'=4 \Rightarrow e^{2x}y=4x+c_1 \\ \Rightarrow \bbox[red, 2pt]{y=4xe^{-2x}+c_1e^{-2x}}$$

解答: $$\cos(x+y)dx +(3y^2+2y+ \cos(x+y))dy=0 \Rightarrow \cases{P(x,y)=\cos(x+y)\\ Q(x,y)=3y^2+2y+\cos(x+y)} \\ \Rightarrow P_y=-\sin(x+y)= Q_x \Rightarrow \text{exact} \Rightarrow \Phi(x,y)=\int P\,dx =\int Q\,dy \\ \Rightarrow \Phi=\int \cos(x+y)\,dx = \int (3y^2+2y+ \cos(x+y))dy \\\Rightarrow \Phi=\sin(x+y)+ \phi(y) = y^3+y^2+\sin(x+y)+ \rho(x) \\ \Rightarrow \Phi=\bbox[red, 2pt]{\sin(x+y) +y^2 +y^2+c_1=0}$$

解答: $$y''+3y+2=0 \Rightarrow \lambda^2+ 3\lambda+2=0 \Rightarrow (\lambda+2)(\lambda+1)=0 \Rightarrow \lambda=-1,-2 \\ \Rightarrow \bbox[red, 2pt]{y=c_1e^{-x}+ c_2e^{-2x}}$$

解答: $${dy\over dx}=(x+1)e^{-x}y^2 \Rightarrow {1\over y^2}dy=(x+1)e^{-x}dx \Rightarrow -{1\over y}=-xe^{-x}-2e^{-x}+c_1 \\ \Rightarrow \bbox[red, 2pt]{y={1\over xe^{-x}+2e^{-x}+c_2}}$$

解答: $$\cos(wt) ={1\over 2}(e^{iwt}+e^{-iwt}) \Rightarrow L\{f(t)\} =\int_0^\infty e^{at}\cos(wt)e^{-st}\,dt ={1\over 2 }\int_0^\infty \left(e^{(iw+a-s)t} + e^{(-iw+a-s)t}\right) \,dt\\ = {1\over 2}\left. \left[ {1\over iw+a-s}e^{(-iw+a-s)t} + {1\over -iw+a-s}e^{(-iw+a-s)t}\right] \right|_0^\infty \\={1\over 2}  \left[ -{1\over iw+a-s} - {1\over -iw+a-s} \right]  =\bbox[red, 2pt]{s-a\over (a-s)^2+w^2}$$

解答: $$L\{y''\}-L\{y\}=L\{t\} \Rightarrow s^2Y(s)-s-1-Y(s)={1\over s^2} \Rightarrow Y(s)={1\over s^2(s^2-1)}+{s+1\over s^2-1} \\ \Rightarrow Y(s)={1\over s^2(s^2-1)}+{1\over s-1}=-{1\over s^2}-{1\over 2(s+1)}+{3\over 2(s-1)}\\ \Rightarrow y(t)= L^{-1}\{Y(s)\}  \Rightarrow \bbox[red, 2pt]{y(t)=-t-{1\over 2}e^{-t}+{3\over 2}e^t}$$

解答: $$\cases{x_1-x_2+x_3 =0\\ -x_1+x_2-x_3=0\\ 10x_2+25x_3=90\\ 20x_1+10x_2=80} =\begin{bmatrix}1 & -1& 1 \\-1 & 1& -1\\ 0& 10& 25\\ 20& 10& 0 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix}=\begin{bmatrix}0 \\0\\ 90\\ 80 \end{bmatrix} \\ \text{augmented matrix:} \left[ \begin{array}{rrr|r}1 & -1 & 1 & 0\\-1 & 1 & -1 & 0\\0 & 10 & 25 & 90\\20 & 10 & 0 & 80\end{array} \right] \xrightarrow{R_1 +R_2\to R_2,R_4-20R_1\to R_4}  \left[ \begin{array}{rrr|r}1 & -1 & 1 & 0\\0 & 0 & 0 & 0\\0 & 10 & 25 & 90\\0 & 30 & -20 & 80\end{array} \right]  \\ \xrightarrow{R_3/10\to R_3} \left[ \begin{array}{rrr|r}1 & -1 & 1 & 0\\0 & 0 & 0 & 0\\0 & 1 & \frac{5}{2} & 9\\0 & 30 & -20 & 80 \end{array} \right] \xrightarrow{R_1+R_3\to R_1, R_4-30R_3 \to R_4} \left[ \begin{array}{rrr|r}1 & 0 & \frac{7}{2} & 9\\0 & 0 & 0 & 0\\0 & 1 & \frac{5}{2} & 9\\0 & 0 & -95 & -190 \end{array} \right] \\ \xrightarrow{R_4 /(-95)\to R_4}  \left[ \begin{array}{rrr|r}1 & 0 & \frac{7}{2} & 9\\0 & 0 & 0 & 0\\0 & 1 & \frac{5}{2} & 9\\0 & 0 & 1 & 2 \end{array} \right] \xrightarrow{R_1-(7/2)R_4\to R_2, R_3-(5/2)R_4\to R_3}  \left[ \begin{array}{rrr|r}1 & 0 & 0 & 2\\0 & 0 & 0 & 0\\0 & 1 & 0 & 4\\0 & 0 & 1 & 2 \end{array} \right] \\ \Rightarrow\bbox[red, 2pt]{ \cases{x_1=2\\ x_2=4\\ x_3=2}}$$

解答: $$[A\mid I]= \left[ \begin{array}{rrr|rrr}-1 & 1 & 2 & 1 & 0 & 0\\3 & -1 & 1 & 0 & 1 & 0\\-1 & 3 & 4 & 0 & 0 & 1\end{array} \right] \xrightarrow{-R_1\to R_1}\left[ \begin{array}{rrr|rrr} 1 & -1 & -2 & -1 & 0 & 0\\3 & -1 & 1 & 0 & 1 & 0\\-1 & 3 & 4 & 0 & 0 & 1 \end{array} \right]\\ \xrightarrow{R_2-3R_1\to R_2, R_1+R_3\to R_3} \left[ \begin{array}{rrr|rrr} 1 & -1 & -2 & -1 & 0 & 0\\0 & 2 & 7 & 3 & 1 & 0\\0 & 2 & 2 & -1 & 0 & 1\end{array} \right] \xrightarrow{R_2/2 \to R_2}\left[ \begin{array}{rrr|rrr} 1 & -1 & -2 & -1 & 0 & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 2 & 2 & -1 & 0 & 1 \end{array} \right] \\  \xrightarrow{R_1+R_2\to R_1, R_3-2R_2\to R_2}\left[ \begin{array}{rrr|rrr} 1 & 0 & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & -5 & -4 & -1 & 1 \end{array} \right] \xrightarrow{R_3/(-5) \to R_3}\left[ \begin{array}{rrr|rrr} 1 & 0 & \frac{3}{2} & \frac{1}{2} & \frac{1}{2} & 0\\0 & 1 & \frac{7}{2} & \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right]\\ \xrightarrow{R_1-1.5R_3\to R_1, R_2-3.5R_3\to R_2}\left[ \begin{array}{rrr|rrr} 1 & 0 & 0 & - \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\0 & 1 & 0 & - \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\0 & 0 & 1 & \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{array} \right] \Rightarrow \bbox[red, 2pt]{A^{-1}=\left[ \begin{array}{rrr|rrr}- \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\- \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\\frac{4}{5} & \frac{1}{5} & - \frac{1}{5} \end{array} \right]}$$

解答: $$\mathcal F(f(x))=\int_{-\infty}^\infty f(x)e^{-i\omega t}\,dt =\int_{-1}^1 e^{-i\omega t}\,dt =\left.\left[ {1\over -i\omega} e^{-i\omega t} \right] \right|_{-1}^1 = {1\over -i\omega}\left( e^{-i\omega}-e^{i\omega}\right) \\={1\over -i\omega}(-2i\sin \omega)= \bbox[red, 2pt]{2\sin \omega\over \omega}$$

解答: $$ex. digital signal processing...$$

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解題僅供參考,其他歷年試題及詳解