112年成大地球科學碩士班-應用數學詳解

 國立成功大學112學年度碩士班招生考試

編號: 50
系所: 地球科學
科目:應用數學

解答: $$\textbf{1.}\;y'=1+y^2 \Rightarrow {1\over 1+y^2}dy = dx \Rightarrow \tan^{-1} y= x+c_1 \Rightarrow \bbox[red, 2pt]{y= \tan(x+c_1)} \\ \textbf{2.}\; y'= \sec^2 y ={1\over \cos^2 y} \Rightarrow \cos^2y dy=dx \Rightarrow \bbox[red, 2pt]{{1\over 4}\sin(2y)+ {1\over 2}y=x+c_1} \\\textbf{3.}\; 2xydx+x^2dy =0   \Rightarrow {2\over x}dx =-{1\over y}dy \Rightarrow 2\ln x+c_1=-\ln y \Rightarrow \ln {1\over y}= \ln c_2x^2 \\ \quad \Rightarrow y= {1\over c_2 x^2} \Rightarrow \bbox[red, 2pt]{y={c_3\over x^2}} \\\textbf{4.}\; y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda= \pm i \Rightarrow \bbox[red, 2pt]{y=c_1\cos x+c_2\sin x} \\ \textbf{5.}\; y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''= m(m-1)x^{m-2} \\\quad \Rightarrow x^2y''+1.5xy'-0.5y= (m(m-1)+1.5m -0.5)y^m=0 \Rightarrow m^2+0.5m-0.5=0\\\quad \Rightarrow (m-0.5)(m+1)=0 \Rightarrow m=0.5,-1 \Rightarrow \bbox[red, 2pt]{ y=c_1\sqrt x+ {c_2 \over x}} \\\textbf{6.}\; y''+y=0 \Rightarrow y_h= c_1\cos x+c_2\sin x \\\quad y_p= Ax^2+ Bx+C \Rightarrow y_p'=2Ax+B \Rightarrow y_p''=2A \Rightarrow y_p''+y_p =Ax^2+Bx +C+2A=0.001x^2 \\\quad \Rightarrow \cases{A=0.001 \\B=0\\ C=-0.002} \Rightarrow y_p= 0.001x^2-0.002 \Rightarrow y=y_h+y_p \\\quad\Rightarrow \bbox[red, 2pt]{y=c_1\cos x+c_2\sin x+0.001x^2-0.002 } \\\textbf{7.}\; y''+5y'+4y=0 \Rightarrow \lambda^2 +5\lambda+4=0 \Rightarrow (\lambda+4)(\lambda +1)=0 \Rightarrow \lambda=-4,-1\\ \quad \Rightarrow y_h=c_1e^{-4x}+ c_2e^{-x} \\ \quad y_p=Ae^{-3x} \Rightarrow y_p'=-3Ae^{-3x} \Rightarrow y_p''=9Ae^{-3x} \Rightarrow y_p''+5y_p'+ 4y_p =-2Ae^{-3x} =10e^{-3x} \\\quad \Rightarrow -2A=10 \Rightarrow A=-5 \Rightarrow y_p =-5e^{-3x}  \Rightarrow y=y_h+y_p \\ \quad \Rightarrow \bbox[red, 2pt]{y=c_1e^{-4x}+ c_2e^{-x}-5e^{-3x}} \\\textbf{8.}\; y''+ y=0 \Rightarrow y_h=c_1\cos x+c_2\sin x \\ \quad y_p =Ax\cos x +Bx\sin x \Rightarrow y_p'=A\cos x-Ax\sin x+B\sin x+ Bx\cos x \\ \quad \Rightarrow y_p''=-2A\sin x+2B\cos x-Ax\cos x-Bx\sin x \\ \quad \Rightarrow y_p''+y_p =-2A\sin x+2B\cos x=\cos x-\sin x \Rightarrow A=B={1\over 2} \Rightarrow y_p={1\over 2}x(\cos x+\sin x) \\ \quad \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos x+c_2\sin x+ {1\over 2}x(\cos x+\sin x)}$$

解答: $$A=\begin{bmatrix}4 & 6 & 6 \\1 & 3 & 2 \\-1 & -5 & -2 \end{bmatrix} \Rightarrow \det(A-\lambda I)= -(\lambda-1)(\lambda-2)^2 =0 \Rightarrow \lambda=1,2\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}3 & 6 & 6 \\1 & 2 & 2 \\-1 & -5 & -3 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1  +\frac{4}{3}x_3  =  0 \\x_2  +\frac{1}{3}x_3  =  0} \\\qquad \Rightarrow v=x_3 \begin{pmatrix}\frac{-4}{3} \\\frac{-1}{3} \\1 \end{pmatrix} , \text{choose }v_1= \begin{pmatrix}\frac{-4}{3} \\\frac{-1}{3} \\1 \end{pmatrix}\\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}2 & 6 & 6 \\1 & 1 & 2 \\-1 & -5 & -4 \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1 +\frac{3}{2}x_3  =  0 \\x_2  +\frac{1}{2}x_3  =  0} \\ \qquad \Rightarrow v= x_3 \begin{pmatrix} \frac{-3}{2} \\\frac{-1}{2} \\1 \end{pmatrix}, \text{choose }v_2= \begin{pmatrix}\frac{-3}{2} \\\frac{-1}{2} \\1 \end{pmatrix}\\ \Rightarrow \text{eighenvalues: }\bbox[red, 2pt]{1,2}, \text{ eigenvectros: } \bbox[red, 2pt]{\begin{pmatrix}\frac{-4}{3} \\\frac{-1}{3} \\1 \end{pmatrix}, \begin{pmatrix}\frac{-3}{2} \\\frac{-1}{2} \\1 \end{pmatrix}}$$

解答: $$A=\left[\begin{matrix}1 & -1 & -2\\3 & -1 & 1\\-1 & 3 & 4\end{matrix}\right] \Rightarrow [A\mid I]= \left[\begin{matrix}1 & -1 & -2 & 1 & 0 & 0\\3 & -1 & 1 & 0 & 1 & 0\\-1 & 3 & 4 & 0 & 0 & 1\end{matrix}\right] \xrightarrow{R_2-3R_1 \to R_2, R_1+R_3\to R_3} \\ \left[\begin{matrix}1 & -1 & -2 & 1 & 0 & 0\\0 & 2 & 7 & -3 & 1 & 0\\0 & 2 & 2 & 1 & 0 & 1\end{matrix}\right] \xrightarrow{R_1+0.5R_2\to R_1,R_3-R_2\to R_3} \left[\begin{matrix}1 & 0 & \frac{3}{2} & - \frac{1}{2} & \frac{1}{2} & 0\\0 & 2 & 7 & -3 & 1 & 0\\0 & 0 & -5 & 4 & -1 & 1\end{matrix}\right] \xrightarrow{R_2/2\to R_2, -R_3/5 \to R_3} \\ \left[\begin{matrix}1 & 0 & \frac{3}{2} & - \frac{1}{2} & \frac{1}{2} & 0\\0 & 1 & \frac{7}{2} & - \frac{3}{2} & \frac{1}{2} & 0\\0 & 0 & 1 & - \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{matrix}\right] \xrightarrow{R_1-(3/2)R_3\to R_1, R_2-(7/2) R_3 \to R_2} \left[\begin{matrix}1 & 0 & 0 & \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\0 & 1 & 0 & \frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\0 & 0 & 1 & - \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{matrix}\right] \\ \Rightarrow A^{-1}= \bbox[red, 2pt]{\left[ \begin{matrix} \frac{7}{10} & \frac{1}{5} & \frac{3}{10}\\\frac{13}{10} & - \frac{1}{5} & \frac{7}{10}\\- \frac{4}{5} & \frac{1}{5} & - \frac{1}{5}\end{matrix}\right]}$$

解答: $$a_0= {1\over 2\pi} \int_{-\pi}^\pi (x+\pi)\,dx ={1\over 2\pi} \left. \left[ {1\over 2}x^2+\pi x \right] \right|_{-\pi}^\pi =\pi \\a_n={1\over \pi} \int_{-\pi}^\pi (x+\pi)\cos(nx)\,dx ={1\over \pi}\left. \left[ {x\over n}\sin(nx)+ {1\over n^2}\cos(nx)+ {\pi\over n}\sin (nx)\ \right] \right|_{-\pi}^\pi=0\\ b_n={1\over \pi} \int_{-\pi}^\pi (x+\pi)\sin(nx)\,dx =-{2\over n}(-1)^n \\ \Rightarrow f(x)=a_0+ \sum_{n=1}^\infty b_n\sin(nx)\Rightarrow \bbox[red, 2pt]{f(x) =\pi-\sum_{n=1}^\infty {2\over n}(-1)^n\sin(nx)}$$

解答: $$f(x)=x^2 \Rightarrow f(-x)=f(x) \Rightarrow f \text{ is even } \Rightarrow b_n=0, n \in \mathbb N\\ a_0={1\over 2} \int_{-1}^1 x^2\,dx = {1\over 2}\left. \left[ {1\over 3}x^3 \right] \right|_{-1}^1 ={1\over 3} \\ a_n= \int_{-1}^1 x^2 \cos(n\pi x)\,dx ={4\over n^2\pi^2}(-1)^n \\ \Rightarrow f(x)=a_0+ \sum_{n=1}^\infty a_n\cos(n\pi x) \Rightarrow  \bbox[red, 2pt]{f(x)={1\over 3}+ {4\over \pi^2}  \sum_{n=1}^\infty{1\over n^2}(-1)^n \cos(n\pi x)}$$

解答: $$u(x,t)=X(x)T(t) \Rightarrow \frac{\partial u}{\partial t}=c^2 \frac{\partial^2 u}{\partial x^2} \Rightarrow XT'=c^2X''T \Rightarrow {X'' \over X}= {T'\over c^2T} =k\\ \Rightarrow \cases{X''-kX=0\\ T'-kc^2T=0} ,\text{ then B.C: } \cases{u(0,t)=0\\ u(L,t)=0} \Rightarrow \cases{X(0)T(t)=0\\ X(L)T(t)=0} \Rightarrow \cases{X(0)=0\\ X(L)=0} \\ \textbf{Case 1: }k= \mu^2 \gt 0(\mu\gt 0) \Rightarrow X''-\mu^2X=0 \Rightarrow X=c_1e^{\mu x} +c_2 e^{-\mu x} \\ \qquad \text{B.C: }\cases{X(0)=c_1+c_2=0 \\ X(L)=c_1e^{\mu L} +c_2e^{-\mu L} =0} \Rightarrow c_1(e^{2\mu L}-1)=0 \Rightarrow c_1=0 \Rightarrow c_2=0 \Rightarrow X=0\\ \textbf{Case 2: }k=0 \Rightarrow X''=0 \Rightarrow X=ax+b \Rightarrow \text{ B.C: }\cases{X(0)=b=0\\ X(L)=aL+b=0} \Rightarrow X=0\\ \textbf{Case 3: }k=-\mu^2 \lt 0\;(\mu \gt 0) \Rightarrow X^2+\mu^2X=0 \Rightarrow X=c_1\cos (\mu x)+ c_2\sin (\mu x) \\ \qquad \Rightarrow \text{B.C: }\cases{X(0)=c_1=0 \\ X(L)=c_2 \sin(\mu L)=0} \Rightarrow \sin(\mu L)=0 \Rightarrow \mu L=n\pi \Rightarrow \mu=\mu_n = {n\pi \over L} \\ \qquad \Rightarrow X_n=\sin({n\pi x\over L}) \Rightarrow T'=c^2kT= -c^2\mu^2 T \Rightarrow T_n=e^{-c^2\mu^2 t} =e^{-c^2 n^2 \pi^2/L^2} \\ \Rightarrow u_n(x,t)=X_n(x)T_n(t) = \sin({n\pi x\over L})e^{-c^2 n^2 \pi^2t/L}, n\in \mathbb N \\ \Rightarrow u(x,t)= a_0+ \sum_{n=1}^\infty a_n  \sin({n\pi x\over L})e^{-c^2 n^2 \pi^2t/L} \\ \Rightarrow \text{I.C: } u(x,0)=f(x)=  a_0+ \sum_{n=1}^\infty a_n  \sin({n\pi x\over L}) \\ \Rightarrow \bbox[red, 2pt]{ u(x,t)=a_0+ \sum_{n=1}^\infty a_n  \sin({n\pi x\over L})e^{-c^2 n^2 \pi^2t/L}, \text{ where } \cases{a_0={1\over L}\int_0^L f(x)\,dx \\ a_n= {2\over L}\int_0^L f(x)\sin(n\pi x)\,dx}}$$

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