115年中央大學大氣碩士班-應用數學詳解

國立中央大學115學年度碩士班考試入學

系所:大氣科學碩士班
科目:應用數學

解答:$$ e^{2x}(2\cos y\,dx - \sin y\,dy) = 0 \Rightarrow  2e^{2x}\cos y \, dx - e^{2x}\sin y \, dy = 0  \Rightarrow \cases{ M(x, y) = 2e^{2x}\cos y\\ N(x, y) = -e^{2x}\sin y} \\ \Rightarrow M_y=-2e^{2x} \sin y= N_x \Rightarrow \text{ Exact} \Rightarrow \Psi(x,y) = \int 2e^{2x}\cos y\,dx = \int -e^{2x}\sin y\,dy\\ \Rightarrow \Psi = e^{2x} \cos y + \phi(y) =e^{2x}\cos y+ \rho(x) \Rightarrow e^{2x}\cos y=C \\ y(2)=\pi \Rightarrow e^{4} \cos \pi =-e^4=C \Rightarrow e^{2x} \cos y=-e^4 \Rightarrow \bbox[red, 2pt]{e^{2x} \cos y+e^4=0}$$

解答:$$ \mathcal{L}\left\{x \frac{\partial w}{\partial x} + \frac{\partial w}{\partial t}\right\} = \mathcal{L}\{xt\}  \Rightarrow  x \frac{d}{dx}W(x,s) + [sW(x,s) - w(x,0)] = x \cdot \frac{1}{s^2} \\ \Rightarrow x \frac{dW}{dx} + sW = \frac{x}{s^2} \Rightarrow \frac{dW}{dx} + \frac{s}{x}W = \frac{1}{s^2} \Rightarrow \text{ integrating factor }I=e^{\int(s/x)dx} =x^s \\ \Rightarrow  x^s \frac{dW}{dx} + s x^{s-1}W = \frac{x^s}{s^2} \Rightarrow \frac{d}{dx}(x^s W) = \frac{x^s}{s^2} \Rightarrow x^s W = \int \frac{x^s}{s^2} dx  = \frac{x^{s+1}}{s^2(s+1)} + C(s) \\ \Rightarrow W(x,s) = \frac{x}{s^2(s+1)} + C(s)x^{-s} \\ w(0,t)=0 \Rightarrow W(0,s)=0 \Rightarrow W(0,s)=0+C(s)0^{-s}  \Rightarrow C(s)=0 \Rightarrow  W(x,s) = \frac{x}{s^2(s+1)} \\ \Rightarrow  W(x,s) = x \left( -\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s+1} \right) \Rightarrow w(x,t) = x \cdot \mathcal{L}^{-1} \left\{ -\frac{1}{s} + \frac{1}{s^2} + \frac{1}{s+1} \right\}  = x (-1 + t + e^{-t})  \\ \Rightarrow \bbox[red, 2pt]{w(x,t)=x(t-1+e^{-t})}$$

解答:$$A = \begin{bmatrix} -1 & 4 & -2 \\ -3 & 4 & 0 \\ -3 & 1 & 3 \end{bmatrix} \Rightarrow \det(A-\lambda I) = -\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0 =0 \Rightarrow -(\lambda-1)(\lambda-2)(\lambda-3)=0 \\ \Rightarrow \lambda=1,2,3\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix} -2 & 4 & -2 \\ -3 & 3 & 0 \\ -3 & 1 & 2 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \\ x_3\end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2=x_3} \\ \qquad \Rightarrow v=x_1 \begin{bmatrix}1\\1\\1 \end{bmatrix} \Rightarrow \text{ choosing }v_1= \begin{bmatrix}1\\1\\1 \end{bmatrix} \\ \lambda_2=2 \Rightarrow(A-\lambda_2 I)v =0 \Rightarrow   \begin{bmatrix} -3 & 4 & -2 \\ -3 & 2 & 0 \\ -3 & 1 & 1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{3x_1=2x_3\\ x_2=x_3} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}2/3\\1\\1 \end{bmatrix} \Rightarrow \text{ choosing }v_2= \begin{bmatrix}2\\3\\3 \end{bmatrix} \\ \lambda_3= 3 \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow  \begin{bmatrix} -4 & 4 & -2 \\ -3 & 1 & 0 \\ -3 & 1 & 0 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{-x_1+3x_2-2x_3=0 \\ -8x_2+6x_3=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}1/4\\ 3/4\\ 1 \end{bmatrix}\Rightarrow \text{ choosing }v_3= \begin{bmatrix}1\\3\\4 \end{bmatrix}\\ \Rightarrow \text{eigenbasis: }\{v_1,v_2,v_3\} = \bbox[red, 2pt]{  \left\{ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ 4 \end{bmatrix} \right\} } \\ P=[v_1\; v_2\; v_3] , D= \begin{bmatrix}\lambda_1& 0& 0\\0& \lambda_2 & 0\\ 0&0& \lambda_3  \end{bmatrix}\Rightarrow A=PDP^{-1} \\ \Rightarrow \bbox[red, 2pt]{A=   \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 3 \\ 1 & 3 & 4 \end{bmatrix}  \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}  \begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 3 \\ 1 & 3 & 4 \end{bmatrix}^{-1}}$$

解答:$$ y^{(v)} - 5y''' + 4y' = 0 \Rightarrow  r^5 - 5r^3 + 4r = 0  \Rightarrow  r(r - 1)(r + 1)(r - 2)(r + 2) = 0 \Rightarrow r=0, \pm 1,\pm  2 \\ \Rightarrow  y(x) = C_1 + C_2e^x + C_3e^{-x} + C_4e^{2x} + C_5e^{-2x} \Rightarrow  y'(x) = C_2e^x - C_3e^{-x} + 2C_4e^{2x} - 2C_5e^{-2x} \\ \Rightarrow y''(x) = C_2e^x + C_3e^{-x} + 4C_4e^{2x} + 4C_5e^{-2x} \Rightarrow y'''(x) = C_2e^x - C_3e^{-x} + 8C_4e^{2x} - 8C_5e^{-2x}\\ \Rightarrow y^{(iv)}(x) = C_2e^x + C_3e^{-x} + 16C_4e^{2x} + 16C_5e^{-2x}\\ \Rightarrow \cases{y(0) = 3\\ y'(0) = -5\\ y''(0) = 11\\ y'''(0) = -23\\ y^{(iv)}(0) = 47} \Rightarrow \cases{ C_1 + C_2 + C_3 + C_4 + C_5 = 3\\ C_2 - C_3 + 2C_4 - 2C_5 = -5\\ C_2 + C_3 + 4C_4 + 4C_5 = 11 \\ C_2 - C_3 + 8C_4 - 8C_5 = -23 \\C_2 + C_3 + 16C_4 + 16C_5 = 47} \Rightarrow \cases{C_1 = 1\\ C_2 = 0\\ C_3 = -1\\ C_4 = 0\\ C_5 = 3 } \\ \Rightarrow \bbox[red, 2pt]{y(x) = 1 - e^{-x} + 3e^{-2x}}$$

解答:$$\mathcal F\{f(x)\} = \int_0^\infty f(x)\cos (\omega x)\,dx =  \int_{0}^{1} 2 \cos(\omega x) dx + \int_{1}^{2} (-3) \cos(\omega x) dx + \int_{2}^{\infty} 0 \cdot \cos(\omega x) dx \\= 2 \left. \left[ {\sin(\omega x)\over \omega} \right] \right|_0^1 -3 \left. \left[ {\sin(\omega x) \over \omega} \right] \right|_1^2+0 ={2\sin \omega\over \omega}+{-3\sin(2\omega)+ 3\sin \omega\over \omega} = \bbox[red, 2pt]{5\sin \omega-3 \sin(2\omega) \over \omega}$$

解答:$$f(t) = \begin{cases}4t,& 0\lt t \lt 1 \\ 8,&  t\gt 1\end{cases} \Rightarrow f(t)=4t(u(t)-u(t-1))+8u(t-1)=4t+ (8-4t)u(t-1) \\ \Rightarrow L\{y''\}+ L\{3y'\}+ L\{2y\}= L\{f(t)\} \Rightarrow s^2Y(s)+3sY(s)+2Y(s)= \frac{4}{s^2} + e^{-s} \left( \frac{4}{s} - \frac{4}{s^2} \right)  \\ \Rightarrow (s^2+3s+2)Y(s)= \frac{4}{s^2} + 4e^{-s} \frac{s-1}{s^2} \Rightarrow  Y(s) = \frac{4}{s^2(s+1)(s+2)} + e^{-s} \frac{4(s-1)}{s^2(s+1)(s+2)} \\ = \left(   -\frac{3}{s} + \frac{2}{s^2} + \frac{4}{s+1} - \frac{1}{s+2}  \right) + e^{-s} \left(  \frac{5}{s} - \frac{2}{s^2} - \frac{8}{s+1} + \frac{3}{s+2}  \right) \\ \Rightarrow y(t)=L^{-1} \{Y(s)\} = \left(  -3 + 2t + 4e^{-t} - e^{-2t}  \right) + \left[   5 - 2(t-1) - 8e^{-(t-1)} + 3e^{-2(t-1)}  \right] u(t-1) \\ \Rightarrow  \bbox[red, 2pt]{y(t) = -3 + 2t + 4e^{-t} - e^{-2t} + \left[ 7 - 2t - 8e^{-(t-1)} + 3e^{-2(t-1)} \right] u(t-1) }$$

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