國立臺北大學109學年度日間學士班轉學生招生
學制系級:通訊工程學系日間學士班3年級
科目:工程數學
解答:(1){u=excosyv=exsiny⇒{ux=excosyuy=−exsinyvx=exsinyvy=excosy⇒{ux=vyuy=−vx⇒f(z)analytic(2)f(z)=z2=(x+iy)2=x2−y2+2xyi⇒{u=x2−y2v=2xy⇒{ux=2xuy=−2yvx=2yvy=2x⇒{ux=vyuy=−vx⇒f(z)analytic
解答:(1)r2+6r+5=0⇒(r+5)(r+1)=0⇒r=−5,−1⇒{y1=e−5xy2=e−x⇒Wronskian =|y1y2y′1y′2|=|e−5xe−x−5e−5x−e−x|=4e−6x(2)y=C1e−5x+C2e−x⇒y′=−5C1e−5x−C2e−x⇒y″
解答:先求齊次解,即y''-4y'+4y=0 \Rightarrow r^2-4r+4=0 \Rightarrow r=2 \Rightarrow y_h=C_1e^{2x}+ C_2xe^{2x}\\ 令\cases{y_1= e^{2x}\\ y_2=xe^{2x}} \Rightarrow w(y_1,y_2)= y_1y_2'-y_2y_1'= e^{4x} \Rightarrow y_p = -y_1 \int {y_2 \cdot r(x)\over w}\,dx + y_2 \int{ y_1\cdot r(x)\over w}\,dx \\=-e^{2x} \int{1\over x^3}\,dx +xe^{2x} \int {1\over x^4}\,dx ={e^{2x}\over 2x^2} -{1\over 3x^2}e^{2x} ={1\over 6x^2}e^{2x} \Rightarrow y=y_h+y_p\\ \Rightarrow \bbox[red,2pt]{y= C_1e^{2x}+ C_2xe^{2x}+{1\over 6x^2}e^{2x}}
解答:\mathbf{(1)}\;r(t)=\begin{cases}1,& 0\lt t\lt 1\\ 0, &\text{otherwise} \end{cases} \Rightarrow r(t)=u(t)-u(t-1) \Rightarrow L\{r(t)\}= L\{u(t)\}-L\{u(t-1)\} \\ \qquad ={1\over s}-{e^{-s}\over s} =\bbox[red,2pt]{{1\over s}(1-e^{-s})} \\\mathbf{(2)}\; L\{ y''\}+L\{y' \}= L\{r(t) \} \Rightarrow s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0) ={1\over s}(1-e^{-s}) \\\qquad \Rightarrow (s^2+1)Y(s)={1\over s}(1-e^{-s}) \Rightarrow Y(s)={1\over s(s^2+1)}(1-e^{-s}) =({1\over s}-{s\over s^2+1})(1-e^{-s}) \\ \qquad \Rightarrow y=L^{-1}\left( ({1\over s}-{s\over s^2+1})(1-e^{-s}) \right) =L^{-1}\{ {1\over s}(1-e^{-s})\} -L^{-1}\{{s\over s^2+1} \}+L^{-1}\{{s\over s^2+1}e^{-s} \} \\\qquad \Rightarrow \bbox[red,2pt]{y(t)=r(t)-\cos(s)+u(t-1)\cos(t-1)}
解答:f為奇函數且週期為2\pi \Rightarrow f(x)=\begin{cases} \pi/2, & 0\lt x\le \pi/2\\ \pi-x, & \pi/2\lt x\lt \pi \\ -\pi/2, & -\pi/2 \lt x\le 0\\ -x-\pi,& -\pi\lt x\lt -\pi/2\end{cases}\\ \Rightarrow a_n=0且b_n={1\over \pi }\int_{-\pi}^\pi f(x)\sin nx \,dx= {1\over \pi }\left(\int_{-\pi}^{-\pi/2} (-x-\pi)\sin nx\,dx +\int_{-\pi/2}^0 (-{\pi\over 2}\sin nx\,dx \right)\\ \quad +{1\over \pi }\left(\int_0^{\pi/2} {\pi\over 2}\sin nx\,dx +\int_{\pi/2}^\pi (\pi-x)\sin nx\,dx \right) \\ = {1\over \pi }\left({2\over n^2}\sin(n\pi/2) +{\pi\over n}\cos(n\pi/2)+{2\pi \over n}\sin^2(n\pi/4) \right) = {1\over \pi}\left( {2\over n^2} \sin(n\pi/2) +{\pi\over n}\right) \\={1\over n}+{2\over n^2\pi} \sin(n\pi/2) \Rightarrow \text{Fourier series} = \bbox[red,2pt]{\sum_{n=1}^\infty \left({1\over n}+{2\over n^2\pi} \sin(n\pi/2)\right)\sin(nx)}
解答:\mathbf{f}=[x^2+y^2,3xy+z^2, 4z^3] \Rightarrow \cases{f_x= 2x\\ f_y=3x\\ f_z= 12z^2} \Rightarrow \text{div }\mathbf f= f_x+ f_y +f_z= 5x+12z^2\\ \text{curl }\mathbf f=\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\{\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\ x^2+y^2 & 3xy+z^2 & 4z^3\end{vmatrix} =0\mathbf i +0\mathbf j+3y\mathbf k-2y \mathbf k-0\mathbf j-2z\mathbf i = [-2z,0,y] \\ 因此\bbox[red,2pt]{\cases{\text{div }\mathbf f= 5x+12z^2\\ \text{curl }\mathbf f= [-2z,0,y]} }
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