114年成大生醫碩士班-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:生物醫學工程科學系
科目:工程數學

解答: $$\textbf{(1) }\lambda^3+ \lambda^2+2\lambda-4=0 \Rightarrow (\lambda-1)(\lambda^2+2\lambda+4)=0 \Rightarrow \lambda=1,-1\pm \sqrt 3 i\\\quad  \Rightarrow y_h=c_1e^x +e^{-x}(c_2 \cos \sqrt 3 x +c_3\sin \sqrt 3 x) \\ \quad y_p = a_1\cos(2x)+ a_2\sin(2x)+ a_3x\cos(2x)+ a_4 x\sin(2x)+ e^x(a_5x^2+ a_6x+ a_7) \\ \Rightarrow y_p'''+y_p''+2 y_p'-4y_p= x\cos(2x) +xe^x \\\quad  \Rightarrow y_p = {7\over 200}\cos(2x)+ {13\over 100} \sin(2x)- {1\over 10}x\cos(2x)- {1\over 20}x\sin(2x)+ e^x( {1\over 14}x^2-  {4\over 49}x+ a_7) \\ \Rightarrow y=y_h+ y_p\\ \Rightarrow \bbox[red, 2pt]{y={7\over 200}\cos(2x)+ {13\over 100}\sin(2x)-{1\over 10}x\cos(2x)- {1\over 20}x\sin(2x)+ e^x( {1\over 14}x^2-  {4\over 49}x+ c_1)} \\\qquad \qquad \bbox[red, 2pt]{+e^{-x}(c_2\cos \sqrt 3x+ c_3 \sin \sqrt 3 x)}$$

解答: $$\textbf{(1) }\text{div }(f\nabla g) =\text{div }(fg_x, fg_y,fg_z) = f_xg_x+fg_{xx} +f_yg_y+ fg_{yy}+ f_zg_z+ fg_{zz} \\=f(g_{xx}+ g_{yy}+ g{zz}) +(f_x,f_y,f_z) \cdot (g_x,g_y,g_z) =f\nabla^2 g+ \nabla f\cdot \nabla g \\ \quad \text{By Divergence Theorem of Gauss, }\iint_S (f\nabla g)d\vec \sigma =\iiint_D \text{div }(f\nabla g)\,dV \\\qquad \qquad =\iiint_D (f\nabla^2 g+ \nabla f\cdot \nabla g) \,dV \qquad \bbox[red, 2pt]{\text{QED}} \\\textbf{(2) }\text{div }(f\nabla g-g\nabla f) =\text{div }((fg_x,fg_y,fg_z)-(gf_x,gf_y,gf_z)) =\text{div }((fg_x-gf_x, fg_y-gf_y, fg_z-gf_z)) \\ =fg_{xx} +f_xg_x-g_xf_x-gf_{xx}+f_yg_y +fg_{yy}-g_yf_y-gf_{yy}+f_zg_z+ fg_{zz}-g_zf_z-gf_{zz} \\=f(g_{xx}+g_{yy} +g_{zz}) -g(f_{xx} + f_{yy}+f_{zz}) =f\nabla^2 g-g\nabla^2 f\\ \quad \text{By Divergence Theorem of Gauss, } \iint_S (f\nabla g-g\nabla f)\,d\vec \delta = \iiint_D \text{div }(f\nabla g-g\nabla f)\,dV \\=\iiint_D (f\nabla^2 g-g\nabla^2 f)\,dV \qquad \bbox[red, 2pt]{QED}$$

解答: $$\cases{U_{tt}=c^2 U_{xx},c=\sqrt{15} \\ U(x,t)=f(x) =\cos(\pi x)\\ U_t(x,0)= g(x) =e^{-2x}\sin x} \\\Rightarrow \text{d'Alembert's solution: }U(x,t)= {1\over 2}(f(x+ct)+ f(x-ct))+{1\over 2c} \int_{x-ct}^{x+ct} g(s)\,ds \\ ={1\over 2}(\cos\pi(x+\sqrt{15}t)+ \cos \pi(x-\sqrt{15}t)) +{1\over 2\sqrt{15}} \int_{x-\sqrt{15}t}^{x+\sqrt{15}t} e^{-2s}\sin s\,ds \\ ={1\over 2}(\cos\pi(x+\sqrt{15}t)+ \cos \pi(x-\sqrt{15}t)) +{1\over 2\sqrt{15}}\left. \left[ -{1\over 5}e^{-2s}(2\sin s+\cos s) \right] \right|_{x-\sqrt{15}t}^{x+\sqrt{15}t}\\ \Rightarrow \bbox[red, 2pt]{U(x,t)= {1\over 2}(\cos\pi(x+\sqrt{15}t)+ \cos \pi(x-\sqrt{15}t))} \\ \bbox[red, 2pt]{-{1\over 10\sqrt{15}} \left[ e^{-2(x+\sqrt{15}t)} \left(2\sin(x +\sqrt{15}t) + \cos(x+\sqrt{15}t) \right) -e^{-2(x-\sqrt{15}t)} \left(2\sin(x-\sqrt{15}t)+\cos(x-\sqrt{15}t) \right) \right]}$$

解答: $$\textbf{(1) }z(t)= \cos(2t)+5\delta(t) \Rightarrow \mathcal F[z] =\mathcal F[\cos(2t)] +5 \mathcal F[\delta(t)] \Rightarrow Z(\omega ) =\pi[\delta(\omega-2)+ \delta(\omega+2)] +5\\\mathcal F[y']+5 \mathcal F[y]= \mathcal F \left[\int_{-\infty}^\infty x(v)z(t-v)\,dv \right]- 3\mathcal F[x] \\ \Rightarrow i\omega Y(\omega) +5 Y(\omega) = X(\omega) Z(\omega)-3X(\omega) \Rightarrow (i\omega+5) Y(\omega) =X(\omega) (\pi[\delta(\omega-2)+ \delta(\omega+2)] +2) \\ \Rightarrow {Y(\omega) \over X(\omega)}=\bbox[red, 2pt]{H(\omega) ={\pi[\delta(\omega-2)+ \delta(\omega+2)] +2\over i\omega+5}}\\ \textbf{(2)} h(t)= \mathcal F^{-1}[H(\omega)] = ...$$

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解題僅供參考，碩士班歷年試題及詳解