115年成功大學環境碩士班-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:環境工程學系
科目:工程數學

解答:$$\textbf{(a) }2y''-5y'+4x=\cos^2x \Rightarrow 2y''-5y'=\cos^2 x-4x \\ \quad \Rightarrow 2r^2-5r=0 \Rightarrow r(2r-5)=0 \Rightarrow r=0,5/2 \Rightarrow y_h= c_1+c_2 e^{5x/2} \\ \cos^2x-4x=  {1\over 2}(\cos 2x+1)-4x \Rightarrow y_p=A\cos 2x+B\sin 2x+ Cx^2+Dx \\ \Rightarrow y_p'=-2A\sin 2x+2B\cos 2x+ 2Cx+D \Rightarrow y_p''=-4A\cos 2x-4B\sin 2x +2C\\ \Rightarrow 2y_p''-5y_p'=(-8A-10B)\cos 2x +(10A-8B)\sin 2x  -10Cx+4C-5D ={1\over 2}\cos 2x-4x+{1\over 2} \\ \Rightarrow \cases{-8A-10B=1/2\\ 10A-8B=0 \\-10C=-4\\ 4C-5D=1/2} \Rightarrow \cases{A=-1/41\\ B=-5/164\\ C=2/5\\ D=11/50} \Rightarrow y_p=-{1\over 41}\cos 2x-{5\over 164} \sin 2x +{2\over 5}x^2+{11\over 50}x \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt] {y= c_1+c_2 e^{5x/2}- {1\over 41}\cos 2x-{5\over 164} \sin 2x +{2\over 5}x^2+{11\over 50}x} \\\textbf{(b) } 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= \begin{vmatrix} y_1&y_2\\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix} \cos x& \sin x\\ -\sin x& \cos x \end{vmatrix} =1 \\ \text{Using variations of parameters: }y_p=-\cos x \int  \sin x\sec x\,dx + \sin x\int \cos x\sec x\,dx \\ \Rightarrow y_p=-\cos x \int \tan x\,dx +\sin x \int 1\,dx = \cos x \ln \cos x+ x\sin x \Rightarrow y =y_h+ y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1 \cos x+c_2 \sin x+\cos x \ln \cos x+ x\sin x} \\\textbf{(c) } y''-y=0 \Rightarrow r^2-1=0 \Rightarrow r=\pm 1 \Rightarrow y_h= c_1e^x+c_2e^{-x} \\ y_p=A\cos x+B\sin x+ Cx+D \Rightarrow y_p'=-A\sin x+B\cos x+C \Rightarrow y_p''=-A\cos x-B\sin x \\ \Rightarrow y_p''-y_p=-2A\cos x-2B\sin x-Cx-D=x+\sin x \Rightarrow \cases{A=0\\ B=-1/2\\ C=-1\\ D=0} \\ \Rightarrow y_p=-{1\over 2}\sin x -x \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^x+c_2 e^{-x}-{1\over 2}\sin x-x} \\\textbf{(d) } L\{y''\}+ 4L\{y'\} +13L\{y\} = L\{\delta(t-\pi)\} +L\{\delta(t-3\pi)\} \\ \Rightarrow s^2Y(s)-s-1+4(sY(s)-1)+13Y(s) =e^{-\pi s}+e^{-3\pi s} \\ \Rightarrow   Y(s)={1\over s^2+4s+13} \left( e^{-\pi s}+e^{-3\pi s}+s+5  \right) \Rightarrow y(t)=L^{-1}\{Y(s)\} \\ \Rightarrow y(t)=L^{-1}\left \{ {e^{-\pi s}\over s^2+4s+13} \right\}+ L^{-1}\left \{ {e^{-3\pi s}\over s^2+4s+13}\right\} +L^{-1}\left \{ {s+5\over s^2+4s+13}\right\}  \\ \Rightarrow y(t)= {1\over 3}u(t-\pi)e^{-2(t-\pi)} \sin(3(t-\pi)) + {1\over 3} u(t-3\pi) e^{-2(t-3\pi)} \sin(3(t-3\pi)) +e^{-2t}(\cos(3t)+ \sin(3t)) \\ \Rightarrow \bbox[red, 2pt]{y(t) =-{1\over 3}u(t-\pi)e^{-2(t-\pi)} \sin(3t) - {1\over 3} u(t-3\pi) e^{-2(t-3\pi)} \sin(3t ) +e^{-2t}(\cos(3t)+ \sin(3t))}$$

解答:$$\cases{\text{tank diameter: }D \\ \text{Hole diameter: }d} \Rightarrow \cases{\text{tank corss-sectional area: }A=\pi D^2/4 \\ \text{hole area: }a=\pi d^2/4} \\\text{From Torricelli's law, the velocity of water leaving the hole is }v=\sqrt{2gh}\\ \Rightarrow \text{the outflow rate is }Q_{out} =a\sqrt{2gh}\\ \text{The rate of change of water volume is } {dV\over dt}= {d\over dt}(Ah)= A{dh\over dt}=Q-Q_{out} \\ \Rightarrow A{dh\over dt}=Q-a\sqrt{2gh} \Rightarrow {dh\over dt}=k_1-k_2 \sqrt h, \text{ where }\bbox[red, 2pt]{\cases{k_1=Q/A\\ k_2=a\sqrt{2g}}/A} \\ \Rightarrow \int {dh\over k_1-k_2\sqrt h} =\int 1\,dt \Rightarrow \bbox[red, 2pt]{t=-{2\sqrt h\over k_2}-{2k_1\over k_2^2} \ln |k_1-k_2\sqrt h|+C}$$

解答:$$$$

解答:$$u(x,y) =X(x)Y(y) \Rightarrow X''Y+XY''=0 \Rightarrow {X''\over X}=-{Y''\over Y}=\lambda \Rightarrow \cases{X''-\lambda X=0\\ Y''+\lambda Y=0} \\ \textbf{Solving for }Y(y):\\ \left. {\partial u\over \partial y}\right|_{y=0} =0 \Rightarrow XY'(0)=0 \Rightarrow Y'(0)=0 \\ \left. {\partial u\over \partial y}\right|_{y=1} =-u(x,1) \Rightarrow XY'(1)=-X(x)Y(1) \Rightarrow X(Y(1)+Y'(1)) =0 \Rightarrow Y(1)+Y'(1)=0\\ \text{Trivail solutions occur for }\lambda\le 0, \text{ we ocus on }\lambda \gt 0\\ \text{Let }\lambda=\alpha^2 \Rightarrow Y''+\alpha^2 Y =0 \Rightarrow Y=A\cos(\alpha y)+B\sin(\alpha y) \Rightarrow Y'=-A\alpha \sin(\alpha y)+ \alpha B\cos(\alpha y) \\ \Rightarrow Y'(0) =\alpha B=0 \Rightarrow B=0 \Rightarrow Y=A\cos(\alpha y) \\ \Rightarrow Y'(1)+Y(1)= -A\alpha \sin(\alpha)+ A \cos(\alpha) =0 \Rightarrow \cot (\alpha)=\alpha \Rightarrow Y_n= \cos(\alpha_n y) \\\textbf{Solving for }X(x): \\ X_n''-\alpha_n^2X_n=0 \Rightarrow X_n= C_ne^{\alpha_n x} +D_n e^{-\alpha_n x} \\ \lim_{x\to \infty} u(x,y)=0 \Rightarrow C_n=0 \Rightarrow X_n=D_ne^{-\alpha_n x} \\ \Rightarrow u(x,y)= \sum_{n=1}^\infty D_n e^{-\alpha_n x} \cos(\alpha_n y) \Rightarrow u(0,y)= \sum_{n=1}^\infty D_n\cos(\alpha_n y)=1 \\ \Rightarrow D_n= {\int_0^1 \cos(\alpha_n y)\,dy\over \int_0^1 \cos^2(\alpha_n y)\,dy} = {\sin(\alpha_n)/\alpha_n \over (1+\sin^2(\alpha_n))/2} ={2\sin(\alpha_n) \over \alpha_n(1+ \sin^2(\alpha_n))} \\ \Rightarrow \bbox[red, 2pt]{u(x,y) =\sum_{n=1}^\infty {2\sin(\alpha_n) \over \alpha_n(1+\sin^2(\alpha_n))}e^{-\alpha_n x} \cos(\alpha_n y), \alpha_n \text{ are the positive roots of }\cot(\alpha)=\alpha}$$

解答:$$U(x,s) =L\{u(x,t)\} =\int_0^\infty u(x,t)e^{-st}\,dt \Rightarrow L\{u_t\} =L\{ku_{xx}\} \Rightarrow sU(x,s)-u(x,0)=k{\partial^2 U\over \partial x^2} \\ \Rightarrow {\partial^2 U\over \partial x^2} -{s\over k}U=0 \Rightarrow U(x,s)=A(s)e^{x\sqrt{s/k}} +B(s)e^{-x\sqrt{s/k}} \\ u(\infty,t)=0 \Rightarrow U(\infty,s)=0 \Rightarrow A(s)=0 \Rightarrow U(x,s)=B(s)e^{-x\sqrt{s/k}} \\ u(0,t)=1 \Rightarrow U(0,s)=L\{ 1\} ={1\over s} \Rightarrow U(0,s)=B(s)={1\over s} \Rightarrow U(x,s)={1\over s}e^{-x\sqrt{s/k}} \\ \Rightarrow u(x,t) =L^{-1} \left\{ {1\over s}e^{-x\sqrt{s/k}} \right\} = \bbox[red, 2pt]{\text{erfc} \left( {x\over 2\sqrt{kt}} \right)}$$

解答:$$\hat y_i =ax_i+b \Rightarrow \text{ error }E(a,b)=\sum_{i=1}^n (y_i-\hat y_i)^2 =\sum_{i=1}^n (y_i-ax_i-b)^2 \\ \Rightarrow {\partial E\over\partial b}= -2\sum_{i=1}^n (y_i-ax_i-b)=0 \Rightarrow \sum_{i=1}^n (y_i-ax_i-b)=0 \Rightarrow \sum_{i=1}^n y_i-a\sum_{i=1}^n x_i-b\sum_{i=1}^n1=0 \\ \Rightarrow \sum_{i=1}^ny_i=a\sum_{i=1}^nx_i+ nb \Rightarrow b={1\over n}\sum_{i=1}^n y_i-a\cdot {1\over n} \sum_{i=1}^nx_i\Rightarrow \bbox[red, 2pt]{b=\bar y-a\bar x}\\ \Rightarrow {\partial E\over \partial a} =-2\sum_{i=1}^n (x_iy_i-ax_i^2-bx_i)=0 \Rightarrow \sum_{i=1}^n (x_iy_i-ax_i^2-bx_i)=0 \Rightarrow \sum_{i=1}^n x_iy_i-a\sum_{i=1}^nx_i^2-b\sum_{i=1}^nx_i=0 \\ \Rightarrow \sum_{i=1}^nx_iy_i= a\sum_{i=1}^n x_i^2+ b\sum_{i=1}^n x_i =a \sum_{i=1}^n x_i^2+ \left( {1\over n}\sum_{i=1}^n y_i-a\cdot {1\over n} \sum_{i=1}^nx_i \right)\sum_{i=1}^nx_i \\ \Rightarrow n\sum_{i=1}^nx_iy_i= na\sum_{i=1}^nx_i^2+ \left(\sum_{i=1}^ny_i-a\sum_{i=1}^nx_i  \right) \sum_{i=1}^nx_i= na\sum_{i=1}^nx_i^2 + \left( \sum_{i=1}^nx_i \right) \left( \sum_{i=1}^ny_i \right)-a \left( \sum_{i=1}^nx_i \right)^2 \\ \Rightarrow \bbox[red, 2pt]{a= {{n\sum_{i=1}^n x_iy_i}- \left( \sum_{i=1}^n x_i \right) \left( \sum_{i=1}^ny_i \right) \over n\sum_{i=1}^n x_i^2 - \left( \sum_{i=1}^nx_i \right)^2}}$$

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解題僅供參考，碩士班歷年試題及詳解