112年中山海下碩士班-工程數學詳解

 國立中山大學112學年度碩士班暨碩士在職專班招生考試試題

科目名稱:工程數學【海下所碩士班】

解答:$$f(x)=e^x \Rightarrow f^{[n]}(x)=e^x \Rightarrow f^{[n]}(0)=1 \Rightarrow f(x)=f(0)+f'(0)x+ {f''(0) \over 2!}x^2 +{f'''(0) \over 3!}x^2 +\cdots \\ \Rightarrow \bbox[red, 2pt] {e^x=1+x+{x^2\over 2!}+{x^3\over 3!}+ {x^4\over 4!}+ \cdots} \\ g(x)=\sin (x) \Rightarrow g'(x)=\cos(x) \Rightarrow g''(x)=-\sin (x) \Rightarrow g'''(x)=-\cos (x) \Rightarrow g^{[4]}(x)= \sin (x) \\ \Rightarrow g(0)=0 \Rightarrow g'(0)=1 \Rightarrow g''(0)=0 \Rightarrow g'''(0)=-1 \Rightarrow g^{[4]}(0)=0 \\ \Rightarrow \bbox[red, 2pt]{\sin (x)=x-{x^3\over 3!} +{x^5\over 5!}-{x^7\over 7!}+{x^9\over 9!}-\cdots}$$

解答:$$[A|I] =\left[\begin{matrix}1 & 2 & 3 & 1 & 0 & 0\\3 & 2 & 1 & 0 & 1 & 0\\1 & 3 & 2 & 0 & 0 & 1\end{matrix}\right] \xrightarrow{R_2-3R_1 \to R_2, R_3-R_1\to R_3} \left[\begin{matrix}1 & 2 & 3 & 1 & 0 & 0\\0 & -4 & -8 & -3 & 1 & 0\\0 & 1 & -1 & -1 & 0 & 1\end{matrix}\right] \xrightarrow{R_1-2R_3\to R_1,R_2+4R_3\to R_2} \\\left[\begin{matrix}1 & 0 & 5 & 3 & 0 & -2\\0 & 0 & -12 & -7 & 1 & 4\\0 & 1 & -1 & -1 & 0 & 1\end{matrix}\right] \xrightarrow{R_2\leftrightarrow R_3} \left[ \begin{matrix} 1 & 0 & 5 & 3 & 0 & -2\\0 & 1 & -1 & -1 & 0 & 1\\0 & 0 & -12 & -7 & 1 & 4\end{matrix} \right] \xrightarrow{R_3/(-12) \to R_3} \\\left[\begin{matrix}1 & 0 & 5 & 3 & 0 & -2\\0 & 1 & -1 & -1 & 0 & 1\\0 & 0 & 1 & \frac{7}{12} & - \frac{1}{12} & - \frac{1}{3}\end{matrix} \right] \xrightarrow{R_1-5R_3\to R_1,R_2+R_3 \to R_2} \left[\begin{matrix}1 & 0 & 0 & \frac{1}{12} & \frac{5}{12} & - \frac{1}{3}\\0 & 1 & 0 & - \frac{5}{12} & - \frac{1}{12} & \frac{2}{3}\\0 & 0 & 1 & \frac{7}{12} & - \frac{1}{12} & - \frac{1}{3}\end{matrix}\right] \\ \Rightarrow A^{-1} =\left[\begin{matrix} \frac{1}{12} & \frac{5}{12} & - \frac{1}{3}\\  - \frac{5}{12} & - \frac{1}{12} & \frac{2}{3}\\ \frac{7}{12} & - \frac{1}{12} & - \frac{1}{3}\end{matrix}\right] \Rightarrow A^{-1}B= \bbox[red, 2pt] {\begin{bmatrix}{1\over 12} & {1\over 2}& -{1\over 3}\\ -{5\over 12}& -{1\over 2}& {2\over 3} \\ {7\over 12}& {1\over 2}& -{1\over 3} \end{bmatrix}}$$

解答:$${dy\over dx}+y=0 \Rightarrow \lambda+1=0 \Rightarrow \lambda=-1 \Rightarrow y_h=c_1e^{-x}\\ y_p=Ax+B \Rightarrow y_p'=A \Rightarrow y_p'+y_p=Ax+(A+B) =x \Rightarrow \cases{A=1\\ B=-1} \Rightarrow y_p=x-1\\ \Rightarrow y=y_h+ y_p = c_1e^{-x}+x-1 \Rightarrow y(0)=c_1-1=10 \Rightarrow c_1=11 \Rightarrow \bbox[red, 2pt]{y=11e^{-x}+x-1}$$

解答:$${dy\over dx}=-{2xy\over 1+x^2} \Rightarrow (2xy)dx +(1+x^2)dy=0 \Rightarrow \cases{P(x,y)=2xy\\ Q(x,y) =1+x^2} \\ \Rightarrow P_y=2x = Q_x \Rightarrow \text{Exact} \\\Rightarrow \Phi(x,y) = \int 2xy\, dx=\int (1+x^2)\,dy \Rightarrow \Phi=x^2y+ \rho(y) =y+x^2y+ \phi(x) \\ \Rightarrow x^2y+y=c_1 \Rightarrow \bbox[red, 2pt]{y={c_1\over x^2+1}}$$

解答:$$u_{xy} =4y \sin x+2y \Rightarrow u_x= \int (4y \sin x+2y)\,dy = 2y^2 \sin x+y^2 + \phi_1(x) \\ \Rightarrow u=\int (2y^2 \sin x+y^2 + \phi_1(x))\,dx = -2y^2 \cos x+xy^2 +\phi_2(x)+ \phi_3(y) \\ \Rightarrow \bbox[red, 2pt]{u(x,y)= -2y^2 \cos x+xy^2 +\phi(x)+ \rho(y)}$$

解答:$$a_0= {1\over 2} \int_0^2 f(x)\,dx = {1\over 2} \int_0^1 2x\,dx = {1\over 2} \left. \left[x^2 \right] \right|_0^1 ={1\over 2} \\ a_n= \int_0^2 f(x) \cos(n\pi x)\,dx =\int_0^1 2x \cos(n\pi x)\,dx ={2\over n^2\pi^2} \left((-1)^n-1 \right) \\b_n=\int_0^2 f(x)\sin(n\pi x)\,dx =\int_0^1 2x \sin (n\pi x)\,dx =-{2\over n\pi} (-1)^n \\ f(x)\sim a_0+ \sum_{n=1}^\infty \left[ a_n\cos(n\pi x)+ b_n \sin(n\pi x)\right]\\ ={1\over 2}+\sum_{n=1}^\infty \left[ {2\over n^2\pi^2} \left((-1)^n-1 \right)\cos(n\pi x) -{2\over n\pi} (-1)^n \sin(n\pi x)\right] \\ = \bbox[red, 2pt]{{1\over 2}+\sum_{n=1}^\infty {2\over n\pi}\left[ {1\over n\pi} \left((-1)^n-1 \right)\cos(n\pi x) -(-1)^n \sin(n\pi x)\right] }$$

解答:$$F(s)= L\{f(t)\} =\int_0^\infty f(t)e^{-st}\,dt \Rightarrow {1\over a}F({s\over a}) ={1\over a }\int_0^\infty f(t)e^{-st/a} \,dt \cdots(1)\\ u=at \Rightarrow du=a\,dt \Rightarrow L\{f(at)\} =\int_0^\infty e^{-st}f(at)\,dt =\int_0^\infty e^{-su/a}f(u)\cdot {1\over a}du \\={1\over a} \int_0^\infty e^{-su/a} f(u)\,du \cdots(2)\\ (1)=(2) \Rightarrow L\{f(at)\}= {1\over a}F({s\over a})\qquad \bbox[red, 2pt]{QED}$$

解答:$$L\{e^{at}\} =\int_0^\infty e^{at}e^{-st}\,dt= \int_0^\infty e^{(a-s)t}\,dt= \left. \left[ {1\over a-s} e^{(a-s)t} \right] \right|_0^\infty =0-{1\over a-s} =\bbox[red, 2pt]{1\over s-a}$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解