臺灣綜合大學系統107學年度學士班轉學生聯合招生考試
科目名稱：工程數學
類組代碼：D36
解答：$$f(x)=e^x \Rightarrow f^{[n]}(x)=e^x \Rightarrow f^{[n]}(0)=1,n\ge 0 \\\Rightarrow 泰勒級數＝\sum_{n=0}^\infty {1\over n!}x^nf^{[n]}(0) ＝ \bbox[red,2pt]{\sum_{n=0}^\infty {1\over n!}x^n= 1+x +{1\over 2}x^2 +{1\over 3!}x^3+\cdots +{1\over n!} x^n+\cdots}$$
解答：$$e^x \approx \sum_{n=0}^\infty {1\over n!}x^n= 1+x +{1\over 2}x^2 +{1\over 3!}x^3+\cdots +{1\over n!} x^n+\cdots\\ \Rightarrow e^{-x} \approx 1-x+{1\over 2}x^2-{1\over 3!}x^3+\cdots +{1\over n!}(-1)^{n}x^n +\cdots \\ \Rightarrow \int_0^1 {1-e^x \over x}\,dx \approx \int_0^1 1-{1\over 2}x+ {1\over 3!}x^2-\cdots+{1\over n!}(-1)^{n-1}x^{n-1}+\cdots\,dx \\=\left.\left[ x-{1\over 4}x^2+ {1\over 18}x^3-{1\over 96}x^4 +{1\over 600}x^5 -\cdots \right]\right|_0^1 \approx 1-{1\over 4}+{1\over 18}-{1\over 96}+{1\over 600} \approx \bbox[red,2pt]{0.797}$$解答：$$y'+4y=20 \Rightarrow \int {1\over 20-4y}dy=\int 1 dx \Rightarrow -{1\over 4}\ln(20-4y) =x+C_1 \Rightarrow \ln(20-4y) = -4x-4C_1\\ 將y(0)＝2代入\Rightarrow \ln 12=-4C_1 \Rightarrow \ln(20-4y)= -4x+\ln 12 \Rightarrow 20-4y = 12e^{-4x}\\ \Rightarrow \bbox[red, 2pt]{y=5-3e^{-4x}}$$
解答：$$取u={1\over y} \Rightarrow u'=-{1\over y^2}y' =-u^2y' \Rightarrow y'=-{1\over u^2} u'代回原式 \Rightarrow -{x\over u^2}u'-{3\over u} +{x^4\over u^2}=0 \\ \Rightarrow xu'+3u=x^4 \Rightarrow u'+{3\over x}u=x^3 \Rightarrow 積分因子I(x)=e^{\int (3/x)\,dx} =x^3 \Rightarrow x^3u'+3x^2u = x^6\\ \Rightarrow (x^3u)'=x^6 \Rightarrow x^3u = \int x^6\,dx = {1\over 7}x^7+ C \Rightarrow {x^3\over y}={1\over 7}x^7+ C \\ 將y(1)=1代入上式\Rightarrow 1={1\over 7}+C \Rightarrow C={6\over 7} \Rightarrow \bbox[red,2pt]{y={7x^3\over x^7+6}}$$
解答：$$y'''-5y''+8y'-4y=0 \Rightarrow \lambda^3-5\lambda^2+8\lambda -4=0 \Rightarrow (\lambda-1)(\lambda-2)^2=0 \\ \Rightarrow y=C_1e^x +C_2e^{2x}+ C_3xe^{2x} \Rightarrow y'= C_1e^x +(2C_2 +C_3)e^{2x} + 2C_3xe^{2x}\\ \Rightarrow y'' = C_1e^x+ (4C_2+4C_3)e^{2x} + 4C_3xe^{2x} \Rightarrow \cases{y(0)=0 = C_1+ C_2\\y'(0)=1 = C_1+ 2C_2+C_3\\ y''(0)=-1 = C_1+ 4C_2+ 4C_3} \Rightarrow \cases{C_1= -5\\ C_2=5\\ C_3=-4} \\ \Rightarrow \bbox[red,2pt]{ y=-5e^x +5e^{2x}-4xe^{2x}}$$
解答：$${d\omega\over dt} +{3C_D\over 2(2G+1)d}\omega^2 = {2(G-1)g\over 2G+1} \equiv \omega'+A\omega^2 = B\Rightarrow \int {1\over B-A\omega^2} \,d\omega = \int 1\,dt \\\Rightarrow \cfrac{\tanh^{-1}\left(\sqrt{A\over B}\omega \right)}{ \sqrt{AB}} =t+C,由初始值\omega(0)=0 \Rightarrow C=0 \Rightarrow \sqrt{A\over B}\omega=\tanh(t\sqrt{AB})\\ \Rightarrow \omega = \sqrt{B\over A}\tanh(t\sqrt{AB}) =\sqrt{4(G-1)gd\over 3C_D} \tanh(t\sqrt{ 3C_D(G-1)g \over (2G+1)^2d}) \\ \Rightarrow \bbox[red,2pt]{\omega(t)=\sqrt{4(G-1)gd/(3C_D)}\cdot \tanh \left(t\sqrt{3C_D(G-1)g/(2G+1)^2d} \right)}$$
解答：$$u(x,t)=X(x)T(t) \Rightarrow u_{xx}-3u_t= X''T-3XT'=0 \Rightarrow {X''\over X}=3{T'\over T} =k,k為常數 \\ \Rightarrow X''－kX=0,以下就k值分別討論：\\\text{Case I: } k=0\Rightarrow X''=0 \Rightarrow X=C_1x+ C_2，由於u(0,t)= u(2,t)=0,即X(0)=X(2)=0\\ \qquad \Rightarrow \cases{C_2=0\\ 2C_1+C_2=0} \Rightarrow X=0為明顯解，不討論\\ \text{Cases II: }k\gt 0 \Rightarrow X''-kX=0 \Rightarrow X=C_1e^{\sqrt k x} +C_2e^{-\sqrt kx} \Rightarrow \cases{X(0)= C_1+C_2 = 0\\ X(2)= C_1e^{2\sqrt k} +C_2e^{-2\sqrt k}=0} \\ \qquad \Rightarrow \cases{C_2=-C_1 \\ C_1e^ {4\sqrt k}+C_2=0} \Rightarrow C_1e^{4\sqrt k}-C_1=0 \Rightarrow C_1(e^{4\sqrt k}-1)=0 \Rightarrow C_1=0(k\gt 0 \Rightarrow e^{4\sqrt k}\ne 1) \\\qquad \Rightarrow C_1=C_2=0 \Rightarrow X=0為明顯解，不討論\\ \text{Cases III:}k\lt 0 \Rightarrow k=-t^2,t\ne 0,t\in\mathbb{R} \Rightarrow X''+t^2X=0 \Rightarrow X= C_1\cos tx +C_2\sin tx \\ \qquad \Rightarrow \cases{X(0)= 0 =C_1\\ X(2)=0 = C_1\cos 2t+ C_2\sin 2t} \\\qquad \Rightarrow C_2\sin 2t=0 \Rightarrow \cases{C_2=0 \Rightarrow C_1=C_2=0 \Rightarrow X=0為明顯解\\ 2t=m\pi,m\in \mathbb{Z}\Rightarrow t=m\pi/2} \\ \qquad \Rightarrow X=C_2\sin{m\pi x\over 2} ;又t=m\pi/2 \Rightarrow k=-m^2\pi^2/4 \Rightarrow T'-{k\over 3}T=0 \\ \qquad\Rightarrow T=C_3e^{kt/3} =C_3e^{-m^2\pi^2t/12} \\ 因此u(x,t)= X(x)T(t)= \sum_{n=1}^\infty A_ne^{-n^2\pi^2 t/12} \sin (n\pi x/2),再由u(x,0)=2\sin(3\pi x) \\ \Rightarrow \cases{A_6= 2\\ A_n=0,n\ne 6} \Rightarrow \bbox[red, 2pt]{u(x,t)=2e^{3\pi^2 t} \sin(3\pi x)}$$
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