110年北科大製造碩士班-微分方程詳解

 國立臺北科技大學110學年度碩士班招生考試

系所組別:製造科技研究所
科目:微分方程

解答: $$x-xy-y'=0 \Rightarrow y'=x(1-y) \Rightarrow {1\over 1-y}dy=x\,dx \Rightarrow -\ln(1-y)={1\over 2}x^2+c_1 \\ \Rightarrow 1-y=e^{-x^2/2-c_1} \Rightarrow \bbox[red, 2pt]{y=1+c_2e^{-x^2/2}}$$

解答: $$\textbf{(1)}\; 2y+3xy'=0 \Rightarrow (2y)dx+(3x)dy=0 \Rightarrow \cases{P(x,y)=2y\\ Q(x,y)=3x} \Rightarrow \cases{P_y=2\\ Q_x=3} \\\qquad \Rightarrow P_y\ne Q_x \Rightarrow \text{Not exact}, \bbox[red, 2pt]{Q.E.D.} \\ \textbf{(2)}\;{P_y-Q_x\over Q}=-{1\over 3x} \text{ dependent on }x \Rightarrow u'=-{1\over 3x}u \Rightarrow \text{integrating factor }u=\bbox[red, 2pt]{x^{-1/3}} \\ \textbf{(3)}\; \cases{uP=2x^{-1/3}y\\ uQ=3x^{2/3}} \Rightarrow (uP)_y= 2x^{-1/3}=(uQ)_x \Rightarrow \Phi(x,y)=\int uP\, dx=\int uQ\,dy\\ \qquad \Rightarrow \int 2x^{-1/3}y\,dx = \int 3x^{2/3}\,dy \Rightarrow 3x^{2/3}y+ \phi(y)=3x^{2/3}y +\rho(x) \\\qquad \Rightarrow 3x^{2/3}y+c_1=0\Rightarrow \bbox[red, 2pt]{y=c_2x^{-2/3}}$$

解答: $$xy'+y-e^x=0 \Rightarrow (xy)'=e^x \Rightarrow xy=e^x+c_1\\ y(1)=e \Rightarrow e=e+c_1 \Rightarrow c_1=0 \Rightarrow \bbox[red, 2pt] {y={e^x\over x}}$$

解答: $$\bbox[cyan,2pt]{題目有誤},應該是x^2y''+xy'-y={1 \over x+1} \Rightarrow y''+{y'\over x}-{y\over x^2}={1\over x^2(x+1)} \\ \Rightarrow y''+({y\over x})'={1\over x^2(x+1)} \Rightarrow (y'+{y\over x})'={1\over x^2}-{1\over x}+{1\over x+1} \\ \Rightarrow y'+{y\over x}=\int \left({1\over x^2}-{1\over x}+{1\over x+1} \right)\,dx =-{1\over x}-\ln x+\ln(x+1)+ c_1 \\ \Rightarrow xy'+y=(xy)'=-1-x\ln x+x\ln (x+1)+c_1x \\\Rightarrow xy=\int \left( -1-x\ln x+x\ln (x+1)+c_1x\right)\,dx \\=-x-({1\over 2}x^2\ln x-{1\over 4}x^2)+{1\over 2}(x^2-1)\ln(x+1)-{1\over 4}x^2+{1\over 2}x+{1\over 2}c_1x^2+c_2 \\ =-{1\over 2}x+c_3x^2+c_2-{1\over 2}x^2\ln x+{1\over 2}(x^2-1)\ln(x+1)\\\Rightarrow \bbox[red, 2pt]{y= -{1\over 2}+c_3x+{c_2\over x}-{1\over 2}x\ln x+{1\over 2x}(x^2-1)\ln(x+1)}$$

解答: $$u(x,t)=X(x)T(t) \Rightarrow \cases{\text{PDE: }X''T=XT''+XT \Rightarrow {X''\over X}={T''\over T}+1 \\ BC \cases{u(0,t)=X(0)T(t)=0\\ u(\pi,t)=X(\pi)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(\pi)=0} } \\{X''\over X}={T''\over T}+1 =\lambda\\ \textbf{Case 1: } \lambda=0 \Rightarrow X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow BC\cases{X(0)=c_2=0\\ X(\pi)=c_1\pi+ c_2=0} \\\qquad \Rightarrow c_1=c_2=0 \Rightarrow X=0 \Rightarrow u=0\\ \textbf{Case 2: }\lambda \gt 0 \Rightarrow \lambda=k^2(k\gt 0) \Rightarrow X''-k^2X=0 \Rightarrow X=c_1e^{kx} +c_2e^{-kx} \\ \qquad \Rightarrow BC\cases{X(0)=c_1+c_2=0\\ X(\pi)=c_1e^{k\pi}+ c_2e^{-k\pi}=0} \Rightarrow c_1e^{k\pi}-c_1e^{-k\pi}=0 \Rightarrow c_1(e^{2k\pi}-1)=0 \\\qquad \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0 \Rightarrow u=0 \\ \textbf{Case 3: }\lambda \lt 0 \Rightarrow \lambda=-k^2(k\gt 0) \Rightarrow X''+k^2X=0 \Rightarrow X= c_1\cos(kx)+ c_2\sin(kx) \\\qquad BC \cases{X(0)= c_1 =0\\ X(\pi)=c_2\sin(k\pi)=0} \Rightarrow \sin(k\pi)=0 \Rightarrow k=n \Rightarrow X_n= \sin(nx),n\in \mathbb N  \\ \Rightarrow {T''\over T}+1=-k^2 \Rightarrow T''+(1+k^2)T=0 \Rightarrow T=c_3\cos(\sqrt{k^2+1}t)+ c_4\sin(\sqrt{k^2+1}t) \\ \Rightarrow T'=-c_3\sqrt{k^2+1} \sin(\sqrt{k^2+1}t) +c_4\sqrt{k^2+1} \cos(\sqrt{k^2+1}t)\\ IC: X(x)T'(0)=0 \Rightarrow T'(0)=0 \Rightarrow c_4=0 \Rightarrow T=c_3\cos(\sqrt{k^2+1}t) \\ \Rightarrow T_n=c_n \cos(\sqrt{n^2+1}t) \Rightarrow u(x,t)=\sum_{n=1}^\infty a_n \sin(nx) \cos(\sqrt{n^2+1}t) \\ \Rightarrow u(x,0) =\sum_{n=1}^\infty a_n \sin(nx) =f(x)= \begin{cases} x& 0\lt x\lt \pi/2\\ \pi-x& \pi/2\lt x\lt \pi\end{cases}\\ \Rightarrow a_n={2\over \pi} \int_0^\pi f(x)\sin(nx)\,dx ={4\over n^2\pi} \sin(n\pi/2)\\ \Rightarrow \bbox[red, 2pt]{ u(x,t)=\sum_{n=1}^\infty {4\over n^2\pi} \sin(n\pi/2) \sin(nx) \cos(\sqrt{n^2+1}t) }$$

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