115年成功大學工程科學碩士班-工程數學詳解

國立成功大學115學年度碩士班招生考試試題

系所:工程科學系
科目:工程數學

 

解答:$$\textbf{(a) }\cases{P(x,y)=6x^2+4axy\\ Q(x,y)=4bx^2+3y^2} \Rightarrow \cases{P_y=4ax\\ Q_x= 8bx} \Rightarrow P_y=Q_x \Rightarrow a=2b \Rightarrow \bbox[red, 2pt]{(B)} \\\textbf{(b) } u(x,y)= \int (6x^2+4axy)\,dx = \int (4bx^2+3y^2)\,dy \\\Rightarrow u(x,y)=2x^3+2ax^2y+ \phi(y) =4bx^2y+y^3+ \rho(x) \Rightarrow u(x,y)=2x^3+2ax^2y+y^3=c \Rightarrow \bbox[red, 2pt]{(E)}$$

解答:$$\text{Solving for homogeneous solution }y_h: \\t=x+1 \Rightarrow t^2 y''-4ty'+6y=0 \\ y=t^m \Rightarrow y'=mt^{m-1} \Rightarrow y''=m(m-1)t^{m-2} \Rightarrow m(m-1)t^m-4mt^m+6t^m=0 \\ \Rightarrow (m^2-5m+6)t^m=0 \Rightarrow m^2-5m+6=(m-3)(m-2)=0 \Rightarrow m=3,2 \\ \Rightarrow y_h=c_1t^2+c_2t^3 =c_1(x+1)^2+ c_2(x+1)^3 \\ y_p=A(x+1)+B \Rightarrow y_p'=A \Rightarrow y_p''=0 \Rightarrow 0-4(x+1)A+6A(x+1)+ 6B=x+1 \\ \Rightarrow \cases{A=1/2\\ B=0} \Rightarrow y_p={1\over 2}(x+1) \Rightarrow y= y_h+ y_p \Rightarrow y= c_1(x+1)^2+ c_2(x+1)^3+{1\over 2}(x+1)  \Rightarrow \bbox[red, 2pt]{(C)}$$

解答:$$\cases{L=1\\ R=1} \Rightarrow i'(t)+i(t)=E(t) =u(t)-u(t-1)+u(t-2)-u(t-3) \\ \Rightarrow L\{ i'(t)\}+ L\{ i(t)\}=L\{ u(t)-u(t-1)+u(t-2)-u(t-3) \}\\ \Rightarrow sI(s)+I(s)={1\over s}-{e^{-s} \over s}+{e^{-2s} \over s}-{e^{-3s} \over s}\\ \Rightarrow I(s) ={1\over s(s+1)}-{e^{-s} \over s(s+1)}+{e^{-2s} \over s(s+1)}-{e^{-3s} \over s(s+1)}\\= \left( {1\over s}-{1\over s+1} \right) \left( 1-e^{-s} +e^{-2s}-e^{-3s}\right) \\ \Rightarrow i(t) =L^{-1}\{I(s)\} = (1-e^{-t}) - \left( 1-e^{-(t-1)} \right)u(t-1)+ \left( 1-e^{-(t-2)} \right)u(t-2) \\\qquad -\left( 1-e^{-(t-3)} \right)u(t-3) \Rightarrow \bbox[red, 2pt]{(D)}$$

解答:$$f(x) =u(x)-u(x-\pi) =\begin{cases}1, & 0\lt x\lt \pi\\ 0,& \pi \lt x\lt 2\pi \end{cases} \\ a_0={1\over \pi} \int_0^{2\pi} f(x)\,dx ={1\over \pi} \int_0^\pi 1\,dx =1\\ a_n= {1\over \pi}\int_0^{2\pi} f(x) \cos(nx)\,dx = {1\over \pi}\int_0^{\pi}  \cos(nx)\,dx =0\\ b_n= {1\over \pi}\int_0^{2\pi} f(x) \sin(nx)\,dx = {1\over \pi}\int_0^{\pi}  \sin(nx)\,dx  ={1\over n\pi} \left( 1-(-1)^n \right) \\ \Rightarrow f(x)\sim{a_0\over 2}+ \sum_{n=1}^\infty [a_n\cos(nx) +b_n\sin(nx)] ={1\over 2} + \sum_{n=1}^\infty {1\over n\pi} \left( 1-(-1)^n \right)\sin(nx) \\\Rightarrow \bbox[red, 2pt]{(B)}$$

解答:$$r(t) =\begin{cases}-1,& -\pi\lt t\lt 0\\ 1,& 0\lt t\lt \pi \end{cases} \Rightarrow r(-t)=-r(t) \Rightarrow r(t)\text{ is odd} \Rightarrow a_n=0, n=0,1,2,\dots \\ b_n={1\over \pi} \int_{-\pi}^\pi r(t)\sin(nt)\,dt = {2\over \pi} \int_{0}^\pi r(t)\sin(nt)\,dt ={2\over \pi} \int_{0}^\pi  \sin(nt)\,dt = {2\over n\pi} (1-(-1)^n) \\ \Rightarrow b_n= \begin{cases} 0,& n=2,4,6,\dots\\ {4\over n\pi},& n=1,3,5,\dots\end{cases} \Rightarrow \bbox[red, 2pt]{(C)}$$

解答:$$y''+\omega^2 y=0 \Rightarrow y_h=c_1 \cos(\omega t)+ c_2 \sin(\omega t) \\ y_p= \sum_{n=1,3,5,\dots} A_n \sin(nt) \Rightarrow y_p'' = \sum_{n=1,3,5,\dots}-n^2 A_n \sin(nt)  \\ \Rightarrow y_p''+\omega^2y_p=  \sum_{n=1,3,5,\dots} (\omega^2-n^2)A_n \sin(nt)  =r(t) = \sum_{n=1,3,5,\dots} {4\over n\pi} \sin(nt)  \\ \Rightarrow A_n={4\over n\pi(\omega^2-n^2)} \Rightarrow y=y_h+y_p =c_1 \cos(\omega t)+ c_2 \sin(\omega t)+  \sum_{n=1,3,5,\dots} {4\over n\pi(\omega^2-n^2)} \sin(nt) \\ \text{Choosing }c_1=c_2=0, \text{we get one solution: } y= \sum_{n=1,3,5,\dots} {4\over n\pi(\omega^2-n^2)} \sin(nt) \Rightarrow \bbox[red, 2pt]{(C)}\\ \bbox[cyan, 2pt]{題目有誤}, \text{solved in 1}應該是\text{solved in 5}$$

解答:$$\oint_C {e^z\over z^n} \,dz = {2\pi i\over (n-1)!} \left.  {d^{n-1}\over dz^{n-1}} e^z\right|_{z=0}  = {2\pi i\over (n-1)!} \left.    e^z\right|_{z=0} ={2\pi i\over (n-1)!} \Rightarrow \bbox[red, 2pt]{(C)}$$

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