115年台北科技大學機電碩士班甲組-工程數學詳解

 國立臺北科技大學115學年度碩士班招生考試

機械工程系機電整合碩士班甲組
科目:工程數學

解答:$$\textbf{(一) }y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-xy'+y=x^m(m(m-1)-m+1)=0 \\ \qquad \Rightarrow x^m(m^2-2m+1)=x^m(m-1)^2=0 \Rightarrow (m-1)^2=0 \Rightarrow m=1 \Rightarrow y=c_1x+ c_2x\ln x \\ \qquad \Rightarrow y'=c_1+c_2\ln x+c_2 \Rightarrow \cases{y(1)=c_1=2\\ y'(1)=c_1+c_2=1} \Rightarrow \cases{c_1=2\\ c_2=-1} \Rightarrow \bbox[red, 2pt]{y=2x-x\ln x} \\ \textbf{(二) }\cases{M(x,y)=-2xy\\ N(x,y)=3x^2-y^2} \Rightarrow M\,dx+ N\,dy=0 \Rightarrow \cases{M_y=-2x\\ N_x=6x} \Rightarrow M_y \ne N_x \Rightarrow \text{ Not Exact} \\ \qquad \Rightarrow  (N_x-M_y)/M=8x/(-2xy)=-{4\over y} \Rightarrow \text{integration factor }\mu(y)= e^{\int-(4/y)\,dy }= y^{-4} \\ \qquad \Rightarrow \cases{\mu M=-2xy^{-3} \\ \mu N=3x^2y^{-4}-y^{-2}} \Rightarrow (\mu M)_y=6xy^{-4}= (\mu N)_x \Rightarrow \text{ Exact} \\ \qquad \Rightarrow \psi(x,y)= \int (-2xy^{-3})\,dx  = \int (3x^2y^{-4}-y^{-2})\,dy \Rightarrow  \psi=-x^2y^{-3}+ \rho(y) =-x^2y^{-3} +y^{-1} + \phi(x) \\ \qquad \Rightarrow \psi(x,y)= \bbox[red, 2pt]{-x^2y^{-3}+y^{-1}=C}$$

解答:$$\textbf{(一) }F(s)= \ln  {s+2\over s+3}= \ln(s+2) - \ln(s+3) \Rightarrow F'(s)={1\over s+2}-{1\over s+3}\\ 利用L\{-tf(t)\} = F'(s) \Rightarrow -tf(t)=L^{-1}\{F'(s)\}= L^{-1} \left\{{1\over s+2}-{1\over s+3} \right\} =e^{-2t}-e^{-3t} \\ \Rightarrow f(t)={e^{-2t}-e^{-3t}\over -t} \Rightarrow \bbox[red, 2pt]{f(t)={e^{-3t}-e^{-2t} \over t}} \\ \textbf{(二) }G(s)= {1\over (s+1)^3} \Rightarrow g(t)=L^{-1}\{ G(s)\} ={t^2\over 2}e^{-t} \\\qquad 再利用L^{-1}\{e^{-as} G(s)\}= g(t-a)u(t-a) \Rightarrow L^{-1} \{ e^{-3s}G(s)\} =g(t-3)u(t-3) \\ \qquad \Rightarrow \bbox[red, 2pt]{f(t)={1\over 2}(t-3)^2e^{-(t-3)}u(t-3)}$$

解答:$$\textbf{(一) } A = \begin{bmatrix} -5 & 9 \\ -6 & 10 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2-5\lambda+4= (\lambda-1)(\lambda-4)=0 \Rightarrow 特徵值\lambda_1=1, \lambda_2=4 \\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow  \begin{bmatrix} -6 & 9 \\ -6 & 9 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}  \Rightarrow 2x_1=3x_2 \\ \qquad \Rightarrow v= x_2 \begin{bmatrix}3/2\\ 1 \end{bmatrix} \Rightarrow 取v_1= \begin{bmatrix}3\\ 2 \end{bmatrix} \\ \lambda_2=4 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow  \begin{bmatrix} -9 & 9 \\ -6 & 6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow x_1=x_2 \\ \qquad \Rightarrow v=x_2 \begin{bmatrix}1\\1 \end{bmatrix} \Rightarrow 取v_2= \begin{bmatrix}1\\1 \end{bmatrix} \\\Rightarrow \bbox[red, 2pt] {特徵值:1,4及特徵向量: \begin{bmatrix}3\\2 \end{bmatrix}, \begin{bmatrix}1\\1 \end{bmatrix}} \\ \textbf{(二) }\cases{P= [v_1\; v_2] = \begin{bmatrix}3& 1\\2& 1 \end{bmatrix} \\D= \begin{bmatrix}\lambda_1& 0\\0& \lambda_2 \end{bmatrix} = \begin{bmatrix}1& 0\\0 & 4\end{bmatrix}} \Rightarrow P^{-1} =   \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix} \Rightarrow A= PDP^{-1} \\\qquad \Rightarrow \bbox[red, 2pt]{A=   \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix} } \\ \textbf{(三) }A^{30}= PD^{30}P^{-1}  =   \begin{bmatrix} 3 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 4^{30} \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix} \Rightarrow  \bbox[red, 2pt]{A^{30} = \begin{bmatrix} 3 - 2 \cdot 4^{30} & -3 + 3 \cdot 4^{30} \\ 2 - 2 \cdot 4^{30} & -2 + 3 \cdot 4^{30} \end{bmatrix} }$$

解答:$$\textbf{(一) }向量  \mathbf{F} = (e^x \sin y)\mathbf{i} + (\cos y + xz)\mathbf{j} + (z^2 xy)\mathbf{k} = \langle P,Q,R\rangle , 其中\cases{P=e^x\sin y\\ Q=\cos y+xz\\ R=z^2xy} \\ \quad \textbf{1. } \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} = \bbox[red, 2pt]{e^x\sin y -\sin y+2zxy} \\\quad \mathbf{2. } \nabla \times \mathbf{F} = \left\vert{} \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{matrix} \right\vert{} = \left\vert{} \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^x \sin y & \cos y + xz & z^2 xy \end{matrix} \right\vert{}  = \bbox[red, 2pt]{(xz^2 - x)\mathbf{i} - yz^2\mathbf{j} + (z - e^x \cos y)\mathbf{k} } \\ \textbf{(二) } \varphi(x,y,z) = x \cos(x+y+z) \Rightarrow  \nabla \varphi = \frac{\partial \varphi}{\partial x}\mathbf{i} + \frac{\partial \varphi}{\partial y}\mathbf{j} + \frac{\partial \varphi}{\partial z} \\\qquad = \langle \cos(x+y+z) - x\sin(x+y+z), \ -x\sin(x+y+z), \ -x\sin(x+y+z) \rangle = \langle A,B,C \rangle \\ \Rightarrow  \nabla \times (\nabla \varphi) = \left\vert{} \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A & B & C \end{matrix} \right\vert{} = \left( {\partial C\over \partial y}-{\partial B\over \partial z} \right) \mathbf i+ \left( {\partial C\over \partial y} -{\partial B\over \partial z}\right) \mathbf j+ \left( {\partial A\over \partial z}-{\partial C\over \partial x} \right) \mathbf k \\ = \left(   [-x\cos(x+y+z)] - [-x\cos(x+y+z)]  \right) \mathbf i\\ \qquad + \left(   [-\sin(x+y+z) - x\cos(x+y+z)] - [-\sin(x+y+z) - x\cos(x+y+z)] \right) \mathbf j \\ \qquad + \left(  [-\sin(x+y+z) - x\cos(x+y+z)] - [-\sin(x+y+z) - x\cos(x+y+z)]  \right)\mathbf k \\\qquad = 0\mathbf i+0\mathbf j+ 0\mathbf k =0  \Rightarrow  \nabla \times (\nabla \varphi) = 0\; \bbox[red, 2pt]{故得證} \\\textbf{(三) }將 f(x) 視為定義在 (-2, 2) 上的奇函數 \Rightarrow  f(x) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) ，其中  L = 2 \\ \quad \Rightarrow b_n= {2\over L} \int_0^L f(x)\sin {n\pi x\over L}\,dx =\int_0^2 x\sin{n\pi x\over 2}\,dx=   \left[ -\frac{2x}{n\pi} \cos\left(\frac{n\pi x}{2}\right) \right]_{0}^{2} +   \frac{2}{n\pi} \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_{0}^{2}  \\=  -\frac{4}{n\pi} (-1)^n + 0 = \frac{4(-1)^{n+1}}{n\pi} \Rightarrow \cases{b_1=4/\pi \\ b_2=-2/\pi \\ b_3=4/3\pi} \\ \quad \Rightarrow  \bbox[red, 2pt]{f(x) = \frac{4}{\pi} \sin\left(\frac{\pi x}{2}\right) - \frac{2}{\pi} \sin(\pi x) + \frac{4}{3\pi} \sin\left(\frac{3\pi x}{2}\right) + \cdots }$$

解答:$$u(x,t) =X(x)T(t) \Rightarrow a^2X''T =XT'' \Rightarrow {X''\over X}={T''\over a^2T} =-\lambda \Rightarrow \cases{X''+ \lambda X=0\\ T''+a^2 \lambda T=0}\\ 邊界條件:\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}\\ 考慮X''+\lambda X=0 \\ 若\lambda \le 0 \Rightarrow X=0 \Rightarrow u(x,t)=0為明顯解\\ 若\lambda\gt 0, 取\lambda=\alpha^2 (\alpha\gt 0) \Rightarrow X''+\alpha^2X=0 \Rightarrow X=c_1\cos(\alpha x)+ c_2\sin(\alpha x) \\ \quad \Rightarrow 邊界條件: X(0)=c_1=0 \Rightarrow X(L)=c_2\sin \alpha L =0 \Rightarrow \alpha L= n\pi \Rightarrow \alpha_n={n\pi\over L} \\ \quad \Rightarrow \lambda_n= \left( {n\pi \over L} \right)^2 \Rightarrow X_n(x)= \sin {n\pi x\over L}, n=1,2,3,\dots \\ 將\lambda_n代入T''+ a^2 \lambda T=0 \Rightarrow  T_n''(t) + a^2 \left(\frac{n\pi}{L}\right)^2 T_n(t) = 0 \Rightarrow T_n(t)= A_n \cos {an\pi t\over L}+ B_n \sin{an\pi t\over L} \\ \Rightarrow u(x,t)= \sum_{n=1}^\infty X_n(x)T_n( t)  = \sum_{n=1}^{\infty} \left[ A_n \cos\left(\frac{an\pi t}{L}\right) + B_n \sin\left(\frac{an\pi t}{L}\right) \right] \sin\left(\frac{n\pi x}{L}\right)  \\ \Rightarrow u(x,0)=  \sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L}\right) = f(x) \Rightarrow  A_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx  \\ \Rightarrow  \frac{\partial u}{\partial t} = \sum_{n=1}^{\infty} \left[ -A_n \left(\frac{an\pi}{L}\right) \sin\left(\frac{an\pi t}{L}\right) + B_n \left(\frac{an\pi}{L}\right) \cos\left(\frac{an\pi t}{L}\right) \right] \sin\left(\frac{n\pi x}{L}\right) \\ \Rightarrow  \left.\frac{\partial u}{\partial t}\right\vert{}_{t=0} = \sum_{n=1}^{\infty} B_n \left(\frac{an\pi}{L}\right) \sin\left(\frac{n\pi x}{L}\right) = g(x) \Rightarrow B_n \left(\frac{an\pi}{L}\right) = \frac{2}{L} \int_{0}^{L} g(x) \sin\left(\frac{n\pi x}{L}\right) dx \\ \Rightarrow  B_n = \frac{2}{an\pi} \int_{0}^{L} g(x) \sin\left(\frac{n\pi x}{L}\right) dx  \\ \Rightarrow \bbox[red, 2pt]{ u(x,t) = \sum_{n=1}^{\infty} \left[ A_n \cos\left(\frac{an\pi t}{L}\right) + B_n \sin\left(\frac{an\pi t}{L}\right) \right] \sin\left(\frac{n\pi x}{L}\right) }, 其中\\ \qquad \bbox[red, 2pt]{ A_n = \frac{2}{L} \int_{0}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx,B_n = \frac{2}{an\pi} \int_{0}^{L} g(x) \sin\left(\frac{n\pi x}{L}\right) dx}$$

解答:$$u(x,y,t) =X(x)Y(y) T(t) \Rightarrow XYT'= k(X''YT+ XY''T) \Rightarrow {T'\over kT}={X''\over X}+{Y''\over Y}=-\lambda\\ 令{X''\over X}=-\mu_1^2, {Y''\over Y} =-\mu_2^2 , 其中 \lambda =\mu_1^2 +\mu_2^2\\ 邊界條件\cases{u(0,y,t) =X(0)Y(y)T(t) =0\\ u(1,y,t) =X(1)Y(t)T(t) =0} \Rightarrow \cases{X(0)=0\\ X(1)=0}, 同理可得\cases{Y(0)=0\\ Y(1)=0} \\ 考慮{X''\over X}=-\mu_1^2 \Rightarrow Xm(x) = \sin(m\pi x) \Rightarrow \mu_1=m\pi, m=1,2,3,\dots\\ 考慮{Y''\over Y}=-\mu_2^2 \Rightarrow Y_n(y)=\sin(n\pi y) \Rightarrow \mu_2= n\pi, n=1,2,3,\dots \\\Rightarrow  T'(t) + k(m^2\pi^2 + n^2\pi^2)T(t) = 0 \Rightarrow T_{mn}(t)= e^{-k\pi^2(m^2+n^2)t} \\\Rightarrow u(x,y,t) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{mn} \sin(m\pi x) \sin(n\pi y) e^{-k\pi^2(m^2+n^2)t} \\ \Rightarrow  u(x,y,0) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{mn} \sin(m\pi x) \sin(n\pi y) = x(1-x^2)y(1-y)  \\ \Rightarrow  A_{mn} = \frac{4}{1 \cdot 1} \int_{0}^{1} \int_{0}^{1} x(1-x^2)y(1-y) \sin(m\pi x) \sin(n\pi y) \,dx \,dy  \\ =  4 \left( \int_{0}^{1} (x - x^3) \sin(m\pi x) \,dx \right) \left( \int_{0}^{1} (y - y^2) \sin(n\pi y) \,dy \right) \\ = 4 \times \left( \frac{6(-1)^{m+1}}{m^3\pi^3} \right) \times \left( \frac{2(1 - (-1)^n)}{n^3\pi^3} \right)  = \frac{48 (-1)^{m+1} (1 - (-1)^n)}{m^3 n^3 \pi^6}  \\ \Rightarrow \bbox[red, 2pt]{ u(x,y,t) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left[ \frac{48 (-1)^{m+1} (1 - (-1)^n)}{m^3 n^3 \pi^6} \right] \sin(m\pi x) \sin(n\pi y) e^{-k\pi^2(m^2+n^2)t} }$$

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解題僅供參考，碩士班歷年試題及詳解