國立臺北科技大學113學年度碩士班招生考試
系所組別: 製造科技研究所
第一節 微分方程
解答: $$y'=y^2e^{-x} \Rightarrow {1\over y^2}dy = e^{-x} dx \Rightarrow -{1\over y}=-e^{-x}+ c_1 \Rightarrow \bbox[red, 2pt]{y={1\over e^{-x}-c_1}}$$
解答: $$\cases{P(x,y)= \cos x-2xy\\ Q(x,y)=e^y-x^2} \Rightarrow P_y=-2x=Q_x \Rightarrow \text{Exact} \\ \Rightarrow \Phi(x,y)=\int (\cos x-2xy)\,dx = \int (e^y-x^2)\,dy \Rightarrow \sin x-x^2y +\phi(y) =e^y-x^2y+\rho(x) \\ \Rightarrow \Phi(x,y)= \bbox[red, 2pt]{\sin x-x^2y+e^y +c_1 =0}$$
解答: $$y'+{1\over x}y ={1\over x^3}y^{-4/3} \Rightarrow x^3y^{4/3} y'+x^2y^{7/3}=2 \Rightarrow (x^2y^{7/3}-2)dx+ x^3y^{4/3} \,dy=0 \\ \text{Let }\cases{P(x,y)=x^2y^{7/3} -2\\ Q(x,y)= x^3y^{4/3}} \Rightarrow \cases{P_y={7\over 3}x^2y^{4/3} \\ Q_x=3x^2y^{4/3}} \Rightarrow {Q_x-P_y\over Q} ={2\over 3x} \text{ (depend on }x) \\ \Rightarrow u'={2\over 3x} u \Rightarrow \text{integrating factor }u(x)=x^{-2/3} \Rightarrow \cases{uP= x^{4/3}y^{7/3}-2x^{-2/3} \\ uQ=x^{7/3}y^{4/3}} \\ \Rightarrow (uP)_y= {7\over 3}x^{4/3}y^{4/3} =(uQ)_x \Rightarrow \text{ exact} \Rightarrow \Phi(x,y)=\int uP\, dx =\int uQ\,dy \\ \Rightarrow \Phi = \int \left(x^{4/3}y^{7/3}-2x^{-2/3} \right)\,dx =\int x^{7/3}y^{4/3} \, dy \\\Rightarrow \Phi = {3\over 7}x^{7/3} y^{7/3} -6x^{1/3} +\phi(y)={3\over 7}x^{7/3} y^{7/3} + \rho(x) \Rightarrow \Phi= {3\over 7}x^{7/3} y^{7/3} -6x^{1/3}=c_1 \\ \Rightarrow y= \left( {(7/3)(6x^{1/3}+c_1)\over x^{7/3}} \right)^{3/7} \Rightarrow \bbox[red, 2pt]{y= \left( {14x^{1/3}+c_2 \over x^{7/3}} \right)^{3/7} }$$
解答: $$y''+2y'+2y=0 \Rightarrow \lambda^2+2\lambda +2=0 \Rightarrow \lambda =-1\pm i \Rightarrow y_h= e^{-x}(c_1 \cos x+ c_2\sin x)\\ y_p=A\cos(3x)+B\sin(3x) \Rightarrow y_p'=-3A\sin(3x)+3B\cos(3x) \\ \Rightarrow y_p''=-9A\cos(3x)-9B\sin(3x)\\ \Rightarrow y_p''+2y_p'+2y_p =(-7A+6B) \cos(3x)-(6A+7B) \sin (3x)=3.5 \sin (3x)-3\cos(3x) \\ \Rightarrow \cases{-7A+6B=-3\\ 6A+7B=-3.5} \Rightarrow \cases{A＝0\\B=-1/2} \Rightarrow y_p=-{1\over 2}\sin(3x) \\ \Rightarrow y=y_h+y_p = e^{-x}(c_1 \cos x+ c_2\sin x)-{1\over 2}\sin(3x) \\\Rightarrow y'=e^{-x}\left( (c_2-c_1) \cos x-(c_1+c_2) \sin x\right) -{3\over 2}\cos(3x) \Rightarrow \cases{y(0)=c_1=0 \\ y'(0)=c_2-3/2=1/2} \Rightarrow c_2= 2\\ \Rightarrow \bbox[red, 2pt]{y= 2e^{-x} \sin x-{1\over 2}\sin(3x)}$$
解答: $$L\{y''\} +4L\{ y\}=L\{f(t)\} =L\{ t u(t-3)\} \Rightarrow s^2Y(s)+4Y(s)=e^{-3s}({3\over s}+{1\over s^2}) \\ \Rightarrow Y(s)= e^{-3s} \left({3 \over s(s^2+4)} +{1\over s^2(s^2+4)} \right) \\ \Rightarrow y(t)=L^{-1}\{Y(s) \} =L^{-1}\left\{ e^{-3s} \left({3 \over s(s^2+4)} +{1\over s^2(s^2+4)} \right) \right\} \\=u(t-3)L^{-1}\left\{ {3 \over s(s^2+4)} +{1\over s^2(s^2+4)} \right\} (t-3) \\\Rightarrow \bbox[red, 2pt]{y(t)=u(t-3)\left( {3\over 4}-{3\over 4}\cos(2(t-3))+{t-3\over 4}-{1\over 8}\sin(2(t-3))\right)}$$
解答: $$\text{Let }u(x,t)= X(x)T(t), \text{then } \frac{\partial u}{\partial t} =k \frac{ \partial^2 u}{\partial x^2} \Rightarrow XT'=kX''T \Rightarrow {T'\over T}=k{X''\over X} \\ \textbf{BC.} \cases{u(0,t)=X(0)T(t)=0 \\ u(L,t)=X(L)T(t) =0} \Rightarrow \cases{X(0)=0\\ X(L)=0}\\ \text{Suppose that } {T'\over kT}={X''\over X}=\lambda\\ \textbf{Cases I. }\lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow \textbf{BC.} \cases{X(0)=c_2=0\\ X(L)=c_1L+ c_2=0} \\\qquad \Rightarrow c_1=c_2=0 \Rightarrow X=0 \Rightarrow u=0\\ \textbf{Case II. }\lambda \gt 0 \Rightarrow \lambda = \rho^2 (\rho \ne 0) \Rightarrow X''-\rho^2 X=0 \Rightarrow X=c_1e^{\rho x }+c_2 e^{-\rho x} \\\qquad \textbf{BC.} \cases{X(0)=c_1+c_2=0\\ X(L)= c_1e^{\rho L}+ c_2 e^{-\rho L}=0 } \Rightarrow c_2=-c_1 \Rightarrow c_1e^{\rho L}-c_1e^{-\rho L}=0 \\ \qquad \Rightarrow c_1(e^{2\rho L}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0 \Rightarrow u=0\\ \textbf{Case III. }\lambda \lt 0 \Rightarrow \lambda=-\rho^2 (\rho \gt 0) \Rightarrow X''+\rho^2 X=0 \Rightarrow X=c_1\cos(\rho x)+ c_2\sin (\rho x) \\ \qquad \Rightarrow \text{BC.} \cases{X(0)=c_1=0 \\ X(L)=c_2 \sin(\rho L)=0} \Rightarrow \sin(\rho L)=0 \Rightarrow \rho ={n\pi \over L} \Rightarrow X=\sin {n\pi x\over L},n \in \mathbb N\\ T'=\lambda kT \Rightarrow T=e^{\lambda k t} =e^{-n^2 \pi^2 k t/L^2} \Rightarrow u_n(x,t)= e^{-n^2 \pi^2 k t/L^2} \sin {n\pi x\over L} \\ \Rightarrow u(x,t)= \sum_{n=1}^\infty a_n e^{-n^2 \pi^2 k t/L^2} \sin {n\pi x\over L} \Rightarrow u(x,0)= \sum_{n=1}^\infty a_n \sin {n\pi x\over L} =A \\ \Rightarrow a_n={2\over L} \int_0^L A\sin{n\pi x\over L}\,dx \\ \Rightarrow \bbox[red, 2pt]{u(x,t)= \sum_{n=1}^\infty a_n e^{-n^2 \pi^2 k t/L^2} \sin {n\pi x\over L}, a_n={2\over L} \int_0^L A\sin{n\pi x\over L}\,dx}$$
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