115年成功大學電機碩士班-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:電機工程學系、通訊工程研究所
科目:工程數學

解答:$$\textbf{Existence: }\text{Yes. }\sqrt{|y-x|} \text{ is continuous everywhere, guaranteeing at least one solution.} \\ u=y-x \Rightarrow y'= \sqrt{|u|}-1 \Rightarrow \text{ On the line }y=x, \text{ the slope evaluates to }u'=-1 \\ \Rightarrow \text{the solution curve slices right through the line }y=x. \text{ It doesn't get stuck or split into }\\\text{ multiple paths, meaning the solution is strictly unique}\\ \text{That is, the solution exists and is unique on the line }y=x$$

解答:$$y''+y=0 \Rightarrow r^2+1=0 \Rightarrow r=\pm i \Rightarrow y_h= c_1\cos x+ c_2\sin x \\ \cases{y_1=\cos x\\ y_2=\sin x} \Rightarrow W(x,y)= \begin{vmatrix} \cos x& \sin x\\-\sin x& \cos x \end{vmatrix}=1 \Rightarrow \text{Using variation of parameters,} \\\text{ we have }y_p=-\cos x\int \sin x\sec ^3x\,dx +\sin x\int \cos x \sec^3x\,dx =-{1\over 2}\sec x+\sec x-\cos x \\ ={1\over 2}\sec x-\cos x \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_3\cos x+c_2\sin x+{1\over 2}\sec x}$$

解答:$$\cases{L\{tf(t)\} =-{d\over ds}F(s) \\ L\{\sin t\} ={1\over s^2+1}} \Rightarrow -{d\over ds} \left( {1\over s^2+1} \right) ={2s\over (s^2+1)^2} \Rightarrow L\{t\sin t\}={2s\over (s^2+1)^2} \\ \Rightarrow L\left\{{1\over 2}t\sin t \right\}={s\over (s^2+1)^2} \Rightarrow L^{-1} \left\{{s\over (s^2+1)^2} \right\} = \bbox[red, 2pt]{{1\over 2}t\sin t}$$

解答:$$f(x)=x^2 \text{ is an even function on }[-\pi, \pi] \Rightarrow f(x)={a_0\over 2}+ \sum_{n=1}^\infty a_n \cos(nx) \\ a_0= {1\over \pi} \int_{-\pi}^\pi x^2\,dx ={2\over \pi} \int_0^\pi x^2\,dx  ={2\over 3}\pi^2 \\a_n={1\over \pi} \int_{-\pi}^\pi x^2 \cos(nx)\,dx = {2\over \pi}\int_{0}^\pi x^2 \cos(nx)\,dx ={4\over n^2}(-1)^n \\ \Rightarrow x^2 ={\pi^2\over 3}+ \sum_{n=1}^\infty {4\over n^2}(-1)^n \cos(nx) \\ \text{Applying Parseval's Identity, we have } {1\over \pi} \int_{-\pi}^\pi [f(x)]^2\,dx ={a_0^2\over 2}+ \sum_{n=1}^\infty(a_n^2+b_n^2) \\ \Rightarrow {1\over \pi}\int_{-\pi}^\pi x^4\,dx ={2\over 5}\pi^4 ={{2\over 9}\pi^4} + \sum_{n=1}^\infty  \left( {4\over n^2}(-1)^n \right)^2 ={2\over 9}\pi^4+ \sum_{n=1}^\infty {16\over n^4} \\ \Rightarrow {8\over 45}\pi^4= 16\sum n^{-4} \Rightarrow \sum n^{-4} ={\pi^4\over 90} \quad \bbox[red, 2pt]{QED.}$$

解答:$$f(x) = \sum_{n=1}^{\infty} c_n J_0({\alpha_n\over a}x), \text{ where }\alpha_n \text{ are the positive roots of }J_0(x), \text{ such that }J_0(\alpha_n)=0 \\ c_n={2\over a^2 [J_1(\alpha_n)]^2} \int_0^a xf(x) J_0({\alpha_n\over a}x)\,dx = {2\over a^2 [J_1(\alpha_n)]^2} \int_0^a x  J_0({\alpha_n\over a}x)\,dx \\ \cases{u= {\alpha_n x\over a}\Rightarrow dx={a\over \alpha_n}du \\ {d\over dx}[xJ_1(x)] =xJ_0(x)} \Rightarrow \int_0^a x  J_0({\alpha_n\over a}x)\,dx = \left( {a\over \alpha_n} \right)^2 \int_0^{\alpha_n} uJ_0(u)\,du ={a^2\over \alpha_n} J_1(\alpha_n) \\ \Rightarrow c_n= {2\over a^2 [J_1(\alpha_n)]^2}\cdot {a^2\over \alpha_n} J_1(\alpha_n) ={2\over \alpha_nJ_1(\alpha_n)} \\ \Rightarrow \bbox[red, 2pt]{f(x) = \sum_{n=1}^\infty {2\over \alpha_nJ_1(\alpha_n)} J_0 \left( {\alpha_n\over a}x \right)}$$

解答:$$\text{Using d'Alembert's formula: }u(x,t)={1\over 2} \left[ f(x-ct)+f(x+ct) \right]+{1\over 2c}\int_{x-ct}^{x+ct} g(s)ds \\ \cases{f(x)=u(x,0)=\sin x \\ g(x)=u_t(x,0) =\cos x} \Rightarrow u(x,t)={1\over 2}[\sin(x-ct)+ \sin(x+ct)]+{1\over 2c}\int_{x-ct}^{x+ct} \cos s\,ds \\=\sin x\cos ct +{1\over 2c} \left. \left[  \sin s \right] \right|_{x-ct}^{x+ct} \Rightarrow \bbox[red, 2pt]{u(x,t)=\sin x\cos ct+{1\over c} \cos x\sin ct}$$

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解題僅供參考，碩士班歷年試題及詳解