國立臺灣科技大學111學年度碩士班招生試題
系所組別：自動化及控制
科 目：工程數學
解答：$$\textbf{(1)}\;\text{Bernoulli D.E.} \Rightarrow v=y^{1-3}=y^{-2} \Rightarrow v'=-{2y' \over y^3} \Rightarrow y'=-{1 \over 2}y^3v'\\ \Rightarrow -{1 \over 2}y^3v'+{y \over x}=3x^2y^3 \Rightarrow -{1 \over 2}y^2v'+{1 \over x}=3x^2y^2 \Rightarrow -{1 \over 2}{v'\over v}+{1 \over x}={3x^2\over v} \\ \Rightarrow v'-{2\over x}v=-6x^2 \Rightarrow {v'\over x^2}-{2 \over x^3}v'=-6 \Rightarrow ({v\over x^2})'=-6 \Rightarrow {v\over x^2}=-6x+c_1 \\ \Rightarrow v={1\over y^2}=-6x^3+c_1x^2 \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{1\over -6x^3+c_1x^2}} \\\text{(2)}\; y''-3y'=0 \Rightarrow y_h=c_1+c_2e^{3x}\\ \text{Let }\cases{y_1=1\\ y_2=e^{3x}} \Rightarrow W=\begin{vmatrix}y_1& y_2\\ y_1'& y_2' \end{vmatrix}=3e^{3x} \Rightarrow y_p=-\int{e^{3x}\cdot 2e^{2x}\sin x\over 3e^{3x}}dx +e^{3x}\int {2e^{2x}\sin x\over 3e^{3x}}dx \\ \Rightarrow y_p=-{1\over 5}e^{2x}(\cos x+3\sin x) \Rightarrow y=y_h+y_p =c_1+c_2e^{3x}-{1\over 5}e^{2x}(\cos x+3\sin x) \\ \Rightarrow y'=3c_2e^{3x}-{2\over 5}e^{2x}(\cos x+3\sin x)-{1\over 5}e^{2x}(-\sin x+3\cos x) \\ \Rightarrow \cases{y(0)=c_1+c_2-{1\over 5}=1\\ y'(0)=3c_2-{2\over 5}-{3\over 5}=2} \Rightarrow \cases{c_1 ={1\over 5}\\c_2=1} \Rightarrow \bbox[red, 2pt]{y={1\over 5}+e^{3x}-{1\over 5}e^{2x}(\cos x+3\sin x)}$$
解答：$$\cases{x'-2y'+3z=0 \cdots(1)\\ x-4y'+3z'=t \cdots(2)\\ x-2y'+3z'=-1\cdots(3)}, (3)-(2) \Rightarrow 2y'=-1-t \Rightarrow 2L\{ y'\}=L\{ -1-t\} \\ \Rightarrow 2sY(s)=-{1\over s}-{1 \over s^2} \Rightarrow Y(s)=-{1\over 2s^2}-{1\over 2s^3} \Rightarrow y(t)=L^{-1}\{ Y(s)\} =-{1\over 2}t-{1\over 4}t^2 \\ \Rightarrow y'=-{1\over 2}-{1\over 2}t \Rightarrow 2y'=-1-t \Rightarrow \cases{x'+3z=-1-t\cdots(1')\\ x+3z'=-2-t \cdots(3')} \Rightarrow \cases{L\{ x'+3z\} =L\{ -1-t\} \\ L\{ x+3z'\}= L\{ -2-t\}} \\ \Rightarrow \cases{sX(s)+3Z(s)=-{1\over s}-{1\over s^2} \cdots(4)\\ X(s)+3sZ(s)=-{2\over s}-{1\over s^2} \cdots(5)} \Rightarrow s(5)-(4) =(3s^2-3)Z(s)=-2+{1\over s^2} \\ \Rightarrow Z(s)=-{2\over 3(s^2-1)}+ {1\over s^2(s^2-1)} \Rightarrow X(s)=-{2\over s}-{1\over s^2}+{2s\over (s^2-1)}- {3\over s(s^2-1)} \\ \Rightarrow \cases{x(t)=L^{-1}\{X(s)\} =-t+1-{1\over 2}(e^t+e^{-t})\\ z(t)=L^{-1}\{Z(s)\} =-t+{1\over 6}(e^t-e^{-t})} \Rightarrow \bbox[red, 2pt]{\cases{x(t)= -t+1-{1\over 2}(e^t+e^{-t})\\y(t) =-{1\over 2}t-{1\over 4}t^2 \\ z(t)=-t+{1\over 6}(e^t-e^{-t})}}$$
解答：$$\cases{y'-y/x=2x^2\\ y(1)=4} \Rightarrow \cases{x_0=1\\ y_0=4\\ f(x,y)=2x^2+y/x} \Rightarrow y_{i+1}=y_i+hf(x_i,y_i) \Rightarrow \cases{y_1=5.2\\ y_2 =6.64\\ y_3=8.37\\ y_4=10.44\\ y_5=12.9}\\ 又y'-{y\over x}=2x^2 \Rightarrow {y'\over x}-{y\over x^2}=2x \Rightarrow ({y\over x})'=2x \Rightarrow {y\over x}=x^2+ c_1\\ y(1)=4 \Rightarrow {4\over 1}=1+c_1 \Rightarrow c_1=3 \Rightarrow y=x^3+3x \Rightarrow \cases{y(1.2)=5.328\\ y(1.4)= 6.944\\ y(1.6)=8.896\\ y(1.8)=11.232\\ y(2)=14}\\\Rightarrow \bbox[red, 2pt]{\begin{array}{cc| rr} i & x_i & \text{approx }y_i & \text{exact }y_i\\ \hline 1 & 1.2 & 5.2 & 5.328\\ 2& 1.4 & 6.64 & 6.944\\ 3& 1.6& 8.37& 8.896\\ 4& 1.8 & 10.44 & 11.232\\ 5& 2 & 12.9 & 14 \end{array}}$$
\end{matrix}\right]\left[\begin{matrix}-1 & 0 \\0 & 6\end{matrix} \right] \left[ \begin{matrix} \frac{-3}{7} & \frac{4}{7} \\\frac{3}{7} & \frac{3}{7}\end{matrix} \right] \\ Y'=AY \Rightarrow Y= e^{At}Y_0 =\left[ \begin{matrix}-1 & \frac{4}{3} \\1 & 1\end{matrix} \right] \left[ \begin{matrix}e^{-t} & 0 \\0 & e^{6t}\end{matrix} \right] \left[ \begin{matrix} \frac{-3}{7} & \frac{4}{7} \\\frac{3}{7} & \frac{3}{7}\end{matrix} \right] \begin{bmatrix} 6\\ 1\end{bmatrix} \\=\left[ \begin{matrix}-1 & \frac{4}{3} \\1 & 1\end{matrix} \right] \left[\begin{matrix}e^{-t} & 0 \\0 & e^{6t}\end{matrix} \right] \begin{bmatrix} -2\\ 3\end{bmatrix}=\left[ \begin{matrix}-1 & \frac{4}{3} \\1 & 1\end{matrix} \right] \begin{bmatrix} -2e^{-t}\\ 3e^{6t}\end{bmatrix} =\begin{bmatrix} 2e^{-t} +4e^{6t}\\-2e^{-t}+ 3e^{6t}\end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{Y=\begin{bmatrix} 2e^{-t} +4e^{6t}\\-2e^{-t}+ 3e^{6t}\end{bmatrix} }$$
解答：$$f(-x)=f(x) \Rightarrow f\text{ is even} \Rightarrow b_n=0\\ a_0= {1\over 6}\int_{-3}^3 x^2 \,dx ={1\over 3} \int_{0}^3 x^2 \,dx =3\\ a_n={1\over 3}\int_{-3}^3 x^2 \cos{n\pi \over 3}x\,dx ={36\over n^2\pi^2}(-1)^n\\ \Rightarrow f(x)=a_0+\sum_{n=1}^\infty a_n \cos{n\pi \over 3}x \Rightarrow \bbox[red, 2pt]{f(x)=3+\sum_{n=1}^\infty {36\over n^2\pi^2}(-1)^n \cos{n\pi \over 3}x}$$
解答：$$o=\{(1,1),(1,2),\dots,(1,6), (2,1),(2,2),\dots, (2,6), (3,1),(3,2),\dots, (3,6),\dots, (6,6)\}\\ \Rightarrow \begin{array}{}X & P(X)\\ \hline 2 & 1/36\\ 3& 2/36\\ 4 & 3/36\\ 5& 4/36 \\ 6 & 5/36\\ 7&6/36 \\ 8& 5/36 \\ 9 & 4/36 \\ 10 & 3/36\\ 11& 2/36\\ 12 & 1/36 \end{array} \Rightarrow \cases{EX=\sum x_ip(x_i)=7 \\EX^2= \sum x_i^2p(x_i)= 329/6} \\\Rightarrow \cases{\mu(X)=7\\ \sigma(X)= \sqrt{EX^2-(EX)^2}=\sqrt{210}/6=2.415} \Rightarrow \bbox[red,2pt]{\cases{\mu(X)=7\\ \sigma(X)=2.415}}$$
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