113年成大電機碩士班-工程數學詳解

 國立成功大學113學年度碩士班招生考試

系所:電機工程、電腦與通訊工程研究所、電機資訊學院-微電、奈米聯招
科目:工程數學

解答: $$(y-1)y''+(y')^2-y'=0 \Rightarrow y'=0,即y=c_1為一明顯解\\此外, 原式\;yy''-y''+(y')^2-y'=0 \Rightarrow yy''+(y')^2=y''+y' \Rightarrow (yy')'=y''+y' \\ \Rightarrow yy'=\int (y''+y')dy = y'+y+c_2 \Rightarrow (y-1)y'=y+c_2 \Rightarrow {y-1\over y+c_2}dy= 1\,dx \\ \Rightarrow \int {y-1\over y+c_2}\,dy= \int\left(1-{1+c_2\over y+c_2} \right)dy = \int 1\,dx \Rightarrow y-(1+c_2)\ln(y+c_2)=x+c_3 \\ \Rightarrow \ln {e^y \over(y+c_2)^{(1+c_2)}} =x+c_3 \Rightarrow {e^y \over(y+c_2)^{(1+c_2)}}=e^{(x+c_3)} \\ \Rightarrow 其解為\bbox[red, 2pt]{y=c_1或{e^y \over(y+ c_2)^{(1+c_2)}}= e^{(x+c_3)} ,其中c_1,c_2,c_3為常數}$$

解答: $$\cases{x(t) =\cos(at) ={1\over 2}(e^{jat}+e^{-jat})\\ y(t) = \sin (at) =-{1\over 2}(e^{jat}-e^{-jat})} \Rightarrow \cases{X(\omega) =\mathcal F(x(t)) = \pi[\delta(\omega-a)+ \delta(\omega+a)] \\ Y(\omega) =\mathcal F(y(t)) ={\pi\over j} [\delta(\omega-a)- \delta(\omega+a)]} \\ z(t)= x(t)*y(t) \Rightarrow Z(\omega) =\mathcal F(z(t)) ={\pi\over j} [\delta(\omega-a)- \delta(\omega+a)] \cdot \pi[\delta(\omega-a)+ \delta(\omega+a)] \\={\pi^2\over j}\left(\delta(\omega-a) \delta(\omega-a)+ \delta(\omega-a)\delta(\omega+a) -\delta( \omega+a) \delta(\omega +a) -\delta( \omega-a) \delta( \omega+a)\right) \\={\pi^2\over j}\left( \delta(0) \delta( \omega-a)+ \delta(-2a)\delta(\omega+a) -\delta(2a) \delta(\omega-a) -\delta( 0) \delta( \omega-a) \right) \\= {\pi^2\over j}\left( \delta(0) \delta( \omega-a)  -\delta( 0) \delta( \omega-a) \right)= \pi \delta(0){\pi\over j}\left( \delta(0)\delta(\omega-a)  -\delta( 0) \delta( \omega-a) \right) \\ \Rightarrow z(t)= \pi \delta(0)\sin(at) \Rightarrow 2\sin(at)*\cos(at) = \bbox[red, 2pt]{2\pi \delta(0)\sin(at)}$$

解答: $$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow u_{xx}+u_{yy}=0 \Rightarrow u_{rr}+{1\over r}u_r+{1\over r^2} u_{\theta \theta}=0 \\ \Rightarrow u(r,\theta)=A_0+\sum_{n=1}^\infty r^n(A_n \cos(n\theta) +B_n \sin(n\theta)) \\ \Rightarrow u_r= \sum_{n=1}^\infty nr^{n-1}(A_n \cos(n\theta) +B_n \sin(n\theta))  \\ \Rightarrow f(\theta)=u_r(2,\theta)=8\cos \theta \sin^2\theta =\sum_{n=1}^\infty (nA_n2^{n-1} \cos(n\theta) +nB_n2^{n-1} \sin(n\theta)) \\ \Rightarrow \cases{nA_n 2^{n-1}= {1\over \pi} \int_{-\pi}^\pi f(\theta) \cos (n\theta)\,d\theta=0 \\ nB_n2^{n-1} = {1\over \pi}\int_{-\pi}^\pi f(\theta) \sin(n\theta) \,d\theta=0} \Rightarrow \bbox[red, 2pt]{u(r,\theta)=0},??$$

解答: $$u(x,t)=T(t)X(x) \Rightarrow \cases{u(0,t)= X(0)T(t) =0 \\ u(8,t)= X(8)T(t)=0} \Rightarrow BC:\cases{X(0)=0\\ X(8)=0} \\又 u_t=u_{xx} \Rightarrow T'X=TX'' \Rightarrow {T'\over T}={X'' \over X} =\lambda \\ \textbf{Case I }\lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow BC: \cases{X(0) =c_2=0\\ X(8)=8c_1+c_2=0} \Rightarrow c_1=c_2=0 \Rightarrow X=0 \\ \textbf{Case II }\lambda \gt 0 \Rightarrow \lambda=\rho^2 (\rho\gt 0) \Rightarrow X''-\rho^2 X=0 \Rightarrow X=c_1e^{\rho x} +c_2 e^{-\rho x} \\ \qquad \Rightarrow BC: \cases{X(0)= c_1+c_2=0\\ X(8) =c_1e^{8\rho} +c_2e^{-8 \rho} =0} \Rightarrow c_1e^{8\rho} -c_1e^{-8 \rho} =0 \Rightarrow c_1(e^{16\rho}-1)=0 \Rightarrow c_1=0 \\ \qquad \Rightarrow c_2=0 \Rightarrow X=0\\ \textbf{Case III } \lambda\lt 0 \Rightarrow \lambda=-\rho^2(\rho\gt 0) \Rightarrow X''+\rho^2X=0 \Rightarrow X=c_1 \cos(\rho x)+c_2 \sin(\rho x)\\ \qquad \Rightarrow BC: \cases{X(0)=c_1=0\\ X(8)=c_2 \sin(8\rho)=0} \Rightarrow \sin (8\rho) =0 \Rightarrow \rho ={n \pi\over 8} \Rightarrow X=\sin{n\pi x\over 8},n\in \mathbb N\\ \Rightarrow T'+\rho^2 T=0  \Rightarrow T=e^{-\rho^2 t} =e^{-n^2\pi^2t/64} \Rightarrow u(x,t) =\sum_{n= 1}^\infty c_ne^{-n^2\pi^2t/64} \sin{n\pi x\over 8}\\ \Rightarrow u(x,0)= \sum_{n=1}^\infty c_n\sin{n\pi x\over 8} =\sin {\pi x\over 8}+ \sin(8\pi x) \Rightarrow \cases{c_8=c_{64}=1\\ c_k=0, k\ne8, k\ne 64} \\ \Rightarrow \bbox[red, 2pt]{u(x,t)= e^{-\pi^2 t} \sin(\pi x)+ e^{-64\pi^2 t} \sin(8\pi x)}$$

解答: $$依\text{d'Alembert's solution }u_{tt}=c^2u_{xx}的通解為 u(x,t)=F(x+ct) + G(x-ct)\\ 依題意 \cases{u(x,0)=f(x) \Rightarrow F(x)+G(x)=f(x) \cdots(1)\\ u_t(x,0) =g(x) \Rightarrow cF'(x )-cG'(x )=g(x) \cdots(2)} \\ \Rightarrow \int_{x_0}^x g(s)\,ds =c(F(x)-G(x))-c(F(x_0)-G(x_0)) \cdots(3)\\ 由(1)及(3)可得\cases{F(x)= {1\over 2}\left(f(x)+{1\over c}\int_{x_0}^x g(s)\,ds +F(x_0)-G(x_0) \right) \\ G(x) ={1\over 2}\left( f(x)-{1\over c}\int_{x_0}^x g(s)\,ds-F(x_0)+G(x_0)\right)} \\ \Rightarrow \cases{F(x+ ct)= {1\over 2}\left(f(x+ ct)+{1\over c}\int_{x_0}^{x+ct} g(s)\,ds +F(x_0)-G(x_0) \right) \\ G(x-ct) ={1\over 2}\left( f(x-ct)-{1\over c}\int_{x_0}^{x-ct} g(s)\,ds-F(x_0)+G(x_0)\right)} \\ \Rightarrow F(x+ct)+ G(x-ct) ={1\over 2}(f(x+ct)+ f(x-ct)) +{1\over 2c}\int_{x-ct}^{x+ct} g(s)\,ds \\ \Rightarrow \bbox[red, 2pt]{u(x,t)={1\over 2}(f(x+ct)+ f(x-ct)) +{1\over 2c}\int_{x-ct}^{x+ct} g(s)\,ds}$$

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解題僅供參考，碩士班歷年試題及詳解