114年成大電機碩士班-工程數學詳解

 國立成功大學114學年度碩士班招生考試

系所:電機工程學系

科目:工程數學

解答:$$\textbf{(a) }y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \quad \Rightarrow x^2y''-5xy'+8y=m(m-1)x^m-5mx^m+8x^m=(m^2-6m+8)x^m =0\\ \quad \Rightarrow m^2-6m+8= (m-4)(m-2)=0 \Rightarrow m=2,4 \Rightarrow \bbox[red, 2pt]{y=c_1x^2 +c_2x^4} \\ \textbf{(b) }W(x)= \begin{vmatrix} y_1& y_2\\ y_1' & y_2'\end{vmatrix} =  \begin{vmatrix} x^2& x^4\\ 2x & 4x^3\end{vmatrix}  =2x^5 \Rightarrow W(0)=0$$

解答:$$y_1=t代入原式\Rightarrow 0+t-t=0 \Rightarrow y_1=t為其中一解\\ y_2=e^t代入原式\Rightarrow (1-t)e^t+te^t-e^t=0 \Rightarrow y_2=e^t為其中一解\\ 因此通解為y(t)=c_1t+c_2e^t \Rightarrow  y'(t)=c_1+c_2e^t \Rightarrow \cases{y(0)=c_2=3\\ y'(0)=c_1+c_2=-1} \\  \Rightarrow \cases{c_1=-4\\ c_2=3}\Rightarrow \bbox[red, 2pt]{y(t)= -4t+3e^t}$$

解答:$$(e^{2x} y')'+\lambda e^{2x}y=0 \Rightarrow e^{2x}y''+ 2e^{2x}y'+ \lambda e^{2x}y=0 \Rightarrow y''+2y'+ \lambda y=0 \Rightarrow \alpha^2+2\alpha+ \lambda=0\\ \Rightarrow \alpha=   -1\pm \sqrt{1-\lambda}\\ \textbf{Case I }\lambda =1 \Rightarrow y=c_1e^{-x}+ c_2xe^{-x} \Rightarrow y(0)=c_1=0 \Rightarrow y(\pi) =c_2\pi e^{-\pi} =0 \Rightarrow c_2=0 \\\qquad \Rightarrow y=0\Rightarrow \text{ 1 is not the eigenvalue} \\ \textbf{Case II }1-\lambda \gt 0 \Rightarrow 1\gt \lambda \Rightarrow  y=c_1 e^{(-1+\sqrt{1-\lambda})x} + c_2 e^{(-1-\sqrt{1-\lambda}) x} \\\qquad \cases{y(0) =c_1+c_2=0 \\ y(\pi)=c_1 e^{(-1+\sqrt {1-\lambda}) \pi} +c_2 e^{(-1-\sqrt{1-\lambda})\pi}=0} \Rightarrow c_1=c_2=0 \Rightarrow y=0 \Rightarrow \text{ no eigenvalue for }1\gt \lambda \\\textbf{Case III }1- \lambda\lt 0 \Rightarrow \lambda \gt 1 \Rightarrow \alpha=-1\pm i \sqrt{\lambda-1} \\\qquad \Rightarrow y=e^{-x} (c_1 \cos(\sqrt{\lambda-1}x) +c_2 \sin (\sqrt{\lambda-1})x) \Rightarrow y(0)=c_1=0 \Rightarrow y(\pi) =c_2e^{-\pi} \sin(\sqrt{\lambda-1}\pi) =0 \\\qquad \Rightarrow \sin(\sqrt{\lambda-1}\pi) =0 \Rightarrow \sqrt{\lambda-1}=n \Rightarrow \lambda_n=n^2+1,n=1,2, \dots \Rightarrow y_n = e^{-x} \sin (nx) \\ \Rightarrow \text{eigenvalues } \bbox[red, 2pt]{\lambda_n=n^2+1,n=1,2,\dots} \\\text{ and the corresponding eigenfunctions } \bbox[red, 2pt]{y_n =e^{-x} \sin(nx),n=1,2,\dots}$$

解答:$$\text{By d'Alembert's Formula, }y(x,t)={1\over 2}[f(x-ct)+ f(x+ct)]+{1\over 2}\int_{x-ct}^{x+ct}g(s)\,ds\\ 將\cases{f(x)=8e^{-5|x|} \\ g(x)=0\\ c=4} 代入上式\Rightarrow y(x,t)={1\over 2}[ f(x-4t)+f(x+4t)]+0  \\={1\over 2} \left( 8e^{-5|x-4t|} + 8e^{-5|x+4t|} \right)\Rightarrow \bbox[red, 2pt]{y(x,t)=4\left( e^{-5|x-4t|} + e^{-5|x+4t|} \right)}$$

解答:$$\mathcal F(u(x,t)) =\hat u(\omega ,t) = \int_{-\infty}^\infty u(x,t)e^{-i \omega x}\,dx \\ \Rightarrow \mathcal F({\partial u\over \partial t}) = \mathcal F(k{\partial^2 u\over \partial x^2}) \Rightarrow {\partial \hat u\over \partial t}=k (i\omega)^2 \hat u(\omega,t) =-k\omega^2 \hat u \Rightarrow  \hat u(\omega, t) =A(\omega)e^{-k\omega^2 t} \\ 初始值u(x,0) =f(x) \Rightarrow \mathcal F(u(x,t)) =\mathcal F(f(x)) \Rightarrow \hat u(\omega,0)= \hat f(\omega) =A(\omega)e^{-k\omega^2 \cdot 0} \\ \Rightarrow  \hat u(\omega,t) =\hat f(\omega)e^{-k\omega^2 t} \Rightarrow u(x,t)= \mathcal F^{-1} \left( \hat f(\omega)e^{-k\omega^2 t} \right) =f(x) * g(x),\\ 其中g(x)= {1\over 2\pi} \int_{-\infty}^\infty e^{-k\omega^2 t}e^{i\omega x}\,d\omega = {1\over 2\pi} \int_{-\infty}^\infty e^{-k\omega^2 t}\cos(\omega x)\,d\omega = {1\over 2\pi } \sqrt{\pi\over kt} e^{-x^2/4kt} ={1\over \sqrt{4\pi kt}} e^{-x^2/4kt} \\ \Rightarrow u(x,t) =f(x)*g(x) \Rightarrow  \bbox[red, 2pt] {u(x,t) ={1\over \sqrt{4\pi k t}} \int_{-\infty}^\infty f(s)e^{-(x-s)^2/4kt}\,ds}$$

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

解題僅供參考，碩士班歷年試題及詳解