113年清華工程與系統碩士班-工程數學詳解

 國立清華大學 113 學年度碩士班考試入學試題

系所班組別:工程與系統科學系碩士班 乙組(0530)
考試科目(代碼):工程數學(3001)

解答:$$\textbf{(a) }x{dy\over dx}-2y=x^4e^x \Rightarrow {1\over x^2}{dy\over dx}-{2\over x^3}y=xe^x \Rightarrow \left( {y\over x^2} \right)'=xe^x \Rightarrow {y\over x^2} =\int xe^x\,dx=xe^x-e^x+c_1 \\ \quad \Rightarrow \bbox[red, 2pt]{y=x^3e^x-x^2e^x+c_1x^2} \\\textbf{(b) } x{d^2y\over dx^2}+{dy\over dx}= \left( x{dy\over dx} \right)'=-9x^2 \Rightarrow x{dy\over dx }=\int -9x^2\,dx =-3x^3+c_1 \Rightarrow {dy\over dx}=-3x^2+{c_1\over x} \\\quad \Rightarrow y= \int \left( -3x^2+{c_1\over x} \right)\,dx =-x^3+c_1\ln|x|+c_2 \Rightarrow \cases{y(1)= -1+c_2=7\\ y(2)=-8+c_1\ln 2+c_2=2\ln 2} \\\quad \Rightarrow \cases{c_1=2\\ c_2=8} \Rightarrow \bbox[red, 2pt]{y=-x^3+2\ln x+8} \quad \text{(Since our domain involves positive values of x (from  }\\x=1 \text{ to } x=2\text{)we can drop the absolute value in the natural logarithm.)} \\ \textbf{(c) }{dy\over dx}=2x+y-2 \Rightarrow {dy\over dx}-y=2x-2 \Rightarrow e^{-x}{dy\over dx}-e^{-x}y= \left( ye^{-x} \right)'=e^{-x}(2x-2) \\\quad \Rightarrow ye^{-x}= \int e^{-x}(2x-2) \,dx =-2xe^{-x}+c_1 \Rightarrow y=-2x+c_1e^x \Rightarrow y(0) =c_1=2 \\ \quad \Rightarrow\bbox[red, 2pt] {y=-2x+2e^x}$$

解答:$$y(x) = \sum_{n=0}^\infty a_nx^{n+r} \Rightarrow y'= \sum_{n=0}^\infty (n+r)a_nx^{n+r-1} \Rightarrow y''= \sum_{n=0}^\infty (n+r)(n+r -1)a_nx^{n+r-2} \\ \text{Substitute these into the original equation : }xy''+2y'-xy=0 \\ \Rightarrow x\sum_{n=0}^\infty (n+r)(n+r -1)a_nx^{n+r-2}+ 2 \sum_{n=0}^\infty (n+r)a_nx^{n+r-1}-x  \sum_{n=0}^\infty a_nx^{n+r}=0 \\ \Rightarrow  \sum_{n=0}^\infty (n+r) (n+r-1)a_nx^{n+r-1} -\sum_{n=0}^\infty a_n x^{n+r+1}=0 \\ \Rightarrow r(r+1)a_0x^{r-1} +(r+1)(r+2)a_1x^r + \sum_{k=2}^\infty [(k+r)(k+r+1)a_k-a_{k-2}] x^{k+r-1} =0  \\ \Rightarrow \text{indicial equation: }r(r+1)=0 \Rightarrow r_1=0, r_2=-1 \\ r=-1 \Rightarrow \cases{k=1\Rightarrow (-1+1)(-1+2)a_1 =0 \Rightarrow 0\cdot a_1=0 \Rightarrow a_1 \text{ is arbitrary} \\ k\ge 2 \Rightarrow (k-1)(k)a_k-a_{k-2}=0 \Rightarrow a_k=\displaystyle {a_{k-2} \over k(k-1)}} \\ \qquad \Rightarrow \cases{k=2 \Rightarrow a_2=a_0/2\\ k=3 \Rightarrow a_3=a_1/6\\ k=4 \Rightarrow a_4=a_0/24 \\ k=5 \Rightarrow a_5=a_1/120 \\ k=6 \Rightarrow a_6=a_0/720}\\ \Rightarrow y(x)=x^{-1} \left( a_0+a_1x +{a_0\over 2}x^2+{a_1\over 6}x^3+{a_0\over 24}x^4+ {a_1\over 120}x^5 +{a_0\over 720}x^6+ \cdots \right) \\\qquad =a_1 \left( 1+{x^2\over 6}+{x^4\over 120}+{x^6\over 5040}+ \cdots \right)+ a_0 \left( {1\over x}+{x\over 2}+{x^3\over 24}+{x^5\over 720}+ \cdots \right) \\ \Rightarrow \bbox[red, 2pt]{ \cases{y_1(x)= 1+\displaystyle {x^2\over 6}+{x^4\over 120}+{x^6\over 5040}+ \cdots \\ y_2(x)= \displaystyle {1\over x}+{x\over 2}+{x^3\over 24}+{x^5\over 720}+ \cdots}}$$

解答:$$L\{{dy\over dt}\}+ L \left\{ \int_0^t y(\tau) \,d\tau\right\} =L\{ u(t-1)\} -L\{u(t-2)\} \Rightarrow sY(s)-0+{Y(s) \over s}={e^{-s}\over s} -{e^{-2s} \over s} \\ \Rightarrow Y(s) ={e^{-s} -e^{-2s} \over s^2+1} \Rightarrow y(t)= L^{-1}\{Y(s)\} = L^{-1} \left\{ {e^{-s} \over s^2+1}\right\}- L^{-1} \left\{ {e^{-2s} \over s^2+1}\right\} \\ \Rightarrow \bbox[red, 2pt]{y(t)= u(t-1) \sin(t-1) -u(t-2)\sin(t-2)}$$

解答:$$\textbf{(a) }M = \begin{bmatrix} 5 & 6 & -3 \\ -2 & -3 & 2 \\ 2 & 2 & 0 \end{bmatrix} \Rightarrow \det(M)= 2 \begin{vmatrix} 6 & -3 \\ -3 & 2 \end{vmatrix} - 2 \begin{vmatrix} 5 & -3 \\ -2 & 2 \end{vmatrix} + 0 \begin{vmatrix} 5 & 6 \\ -2 & -3 \end{vmatrix} \\= 2(12 - 9) - 2(10 - 6) + 0= \bbox[red, 2pt]{-2} \\ \cases {C_{11} = +(0 - 4) = -4 \\C_{12} = -(0 - 4) = 4 \\C_{13} = +(-4 - (-6)) = 2 \\C_{21} = -(0 - (-6)) = -6 \\C_{22} = +(0 - (-6)) = 6 \\C_{23} = -(10 - 12) = 2 \\C_{31} = +(12 - 9) = 3 \\C_{32} = -(10 - 6) = -4 \\C_{33} = +(-15 - (-12)) = -3} \Rightarrow C = \begin{bmatrix}C_{11}& C_{12} &C_{13} \\C_{21}& C_{22} & C_{23} \\ C_{31}& C_{32}& C_{33} \end{bmatrix} = \begin{bmatrix} -4 & 4 & 2 \\ -6 & 6 & 2 \\ 3 & -4 & -3 \end{bmatrix} \\ \Rightarrow M^{-1} = {1\over \det(M)} C^T = \bbox[red, 2pt]{\begin{bmatrix}2& 3& -3/2\\ -2& -3& 2\\ -1& -1& 3/2 \end{bmatrix}} \\\textbf{(b) } \det(M-\lambda I) = -(\lambda-1)(\lambda+1)(\lambda-2) =0 \Rightarrow \lambda= \bbox[red, 2pt]{1,-1,2} \\ \lambda_1=1 \Rightarrow (M-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}4 & 6 & -3 \\-2 & -4 & 2 \\2 & 2 & -1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} = 0 \Rightarrow \cases{x_1=0\\ 2x_2=x_3} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}0\\ 1/2\\ 1 \end{bmatrix} \Rightarrow \text{ choosing }v_1= \bbox[red, 2pt]{\begin{bmatrix}0\\1\\2 \end{bmatrix} }\\ \lambda_2=-1 \Rightarrow (M-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}6 & 6 & -3 \\-2 & -2 & 2 \\2 & 2 & 1 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} = 0 \Rightarrow  \cases{x_1+x_3=0\\ x_3=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}-1\\1\\0 \end{bmatrix} \Rightarrow \text{ choosing }v_2= \bbox[red, 2pt]{\begin{bmatrix}-1\\1\\0 \end{bmatrix} }\\ \lambda_3=2 \Rightarrow (M-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix}3 & 6 & -3 \\-2 & -5 & 2 \\2 & 2 & -2 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} = 0 \Rightarrow \cases{x_1=x_3\\ x_2=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix}1\\0\\1 \end{bmatrix} \Rightarrow \text{ choosing }v_3= \bbox[red, 2pt]{\begin{bmatrix}1\\0\\1 \end{bmatrix}}$$

解答:$$u(x,t)= X(x)T(t) \Rightarrow XT''=a^2 X''T \Rightarrow {T''\over a^2T} ={X''\\\over X} =-\lambda \Rightarrow \cases{X''+\lambda X=0\\ T''+a^2\lambda T=0} \\ \cases{u(0,t) =X(0)T(t) =0\\ u(\pi,t) =X(\pi) T(t)=0} \Rightarrow \cases{X(0) =0\\ X(\pi)=0} \\ \textbf{Solving for }X''+\lambda X=0: \text{If }\lambda \le 0, \text{ we only get the trivial solution }X=0\\ \quad \text{If }\lambda = k^2 \gt 0, \text{ then }X= A\cos(kx)+B\sin(kx) \Rightarrow X(0)=A=0 \Rightarrow X(\pi) =B\sin (k\pi)=0 \\ \qquad \Rightarrow k=n \Rightarrow X_n=\sin(nx), n=1,2,\dots \\\textbf{Solving for }T''+a^2\lambda T=0: \lambda=k^2=n^2 \Rightarrow T_n =C_n \cos(ant) +D_n \sin (ant) \\ \quad \left.  {\partial u\over \partial t} \right|_{t=0} =0 \Rightarrow T'(0)=0 \Rightarrow T_n'= -anC_n \sin(ant)+anD_n \cos(ant) \Rightarrow T_n'(0)= anD_n=0 \\ \quad \Rightarrow D_n=0 \Rightarrow T_n= C_n \cos(ant),n=1,2,\dots\\ u(x,t) =\sum_{n=1}^\infty B_n \sin(nx) \cos(ant) \Rightarrow u(x,0)={\pi\over 2}x =\sum_{n=1}^\infty B_n \sin(nx)  \text{ fro }0\lt x\lt \pi \\ \Rightarrow B_n= {2\over \pi}\int_0^{\pi} {\pi\over 2}x \sin(nx)\,dx = \int_0^\pi x\sin(nx)\,dx ={\pi\over n}(-1)^{n+1} \\ \Rightarrow \bbox[red, 2pt]{u(x,t)= \sum_{n=1}^\infty {\pi\over n}(-1)^{n+1} \sin(nx) \cos(ant)}$$

解答:$$\text{Let }U(x,\omega) = \mathcal F_c\{u(x,y)\} = \int_0^\infty u(x,y) \cos(\omega y) \,dy \Rightarrow \cases{\mathcal F_c\{\displaystyle {\partial^2 u\over \partial x^2}\} ={d^2 U\over dx^2} \\ \mathcal F_c\{ \displaystyle {\partial^2 u\over \partial y^2}\} = -\omega^2 U(x,\omega)-u_y(x,0)} \\ \Rightarrow \mathcal F_c\{u_{xx}+ u_{yy}\} ={d^2U\over dx^2}-\omega^2 U=0 \Rightarrow U(x,\omega) =A(\omega)e^{-\omega x} +B(\omega) e^{\omega x} \\ \text{We are looking for a steady-state temperature on a domain where }x\to \infty, \text{ the temperature }\\\text{must ramin bounded} \Rightarrow  B(\omega )=0 \Rightarrow U(x,\omega) =A(\omega)e^{-\omega x} \\ u(0,y) = \begin{cases} 30, & 0\le y\le \pi\\ 0,& y\gt \pi\end{cases} \Rightarrow \mathcal F_c\{u(0,y)\}= U(0,\omega) =\int_0^\pi 30 \cos(\omega y)\,dy ={30\sin(\pi \omega) \over \omega} \\ \Rightarrow A(\omega)={30 \sin(\pi \omega) \over \omega} \Rightarrow U(x,\omega) ={30 \sin(\pi \omega) \over \omega}  e^{-\omega x} \\ \Rightarrow u(x,y) = \mathcal F_c^{-1} \{U(x,\omega)\} ={2\over \pi} \int_0^\infty {30\sin(\pi \omega)\over \omega} e^{-\omega x}\cos(\omega y)\,d\omega \\={30\over \pi} \left( \int_0^\infty {e^{-\omega x} \sin((\pi +y)\omega) \over \omega}\,d\omega +\int_0^\infty {e^{-\omega x} \sin((\pi -y)\omega) \over \omega}\,d\omega \right) \\ \Rightarrow \bbox[red, 2pt]{u(x,y)={30\over \pi}   \left[  \tan^{-1} {\pi+y\over x} + \tan^{-1} {\pi-y\over x}\right]  }$$

解答:$$u(r,t) =R(r)T(t) \Rightarrow RT'=k(R''T+{1\over r}R'T) \Rightarrow {T'\over kT} ={R''+{1\over r}R' \over R} =-\lambda^2\\ \text{Solving for }{T'\over kT}=-\lambda^2 \Rightarrow T'+\lambda^2kT=0 \Rightarrow T(t)=Ce^{-k\lambda^2 t}\\ \text{Solving for }{R''+{1\over r}R' \over R} =-\lambda^2 \Rightarrow r^2R''+rR'+ \lambda^2r^2 R=0 \Rightarrow R(r)=AJ_0(\lambda r)+BY_0(\lambda r) \\ \text{Since th etemperature must be finite at }r=0, \text{ and }Y_0(0)=-\infty, \text{ we set }B=0 \\ u(c,t)=0 \Rightarrow R(c)=0 \Rightarrow J_0(\lambda c)=0 \Rightarrow \lambda_n={\alpha_n\over c}, \text{ where }\alpha_n \text{ is the n-th positive root of }J_0(x)=0 \\ \Rightarrow \bbox[red, 2pt]{u(r,t)= \sum_{n=1}^\infty  c_n J_0({\alpha_n\over c}r) e^{-k\alpha_n^2 t/c^2}} \\ u(r,0) =f(r) \Rightarrow \bbox[red, 2pt]{c_n={2\over c^2J_1^2(\alpha_n)} \int_0^c rf(r)J_0({\alpha_n\over c}r)\,dr}$$

解答:$$\textbf{(a) }\mathbf F=(2x+y) \mathbf i+(x+y+z) \mathbf j+(y+2z) \mathbf k \Rightarrow \nabla \times \mathbf F = \begin{vmatrix} \mathbf i& \mathbf j& \mathbf k \\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\ 2x+y& x+y+z& y+2z\end{vmatrix} =0 \\ \text{By Stokes' Theorem, }\oint_C \mathbf F\cdot d \mathbf r = \iint_S (\nabla \times \mathbf F) \cdot d\mathbf S =\bbox[red, 2pt] 0 \\\textbf{(b) } \omega=z-i \Rightarrow f(z)={1\over (z-i)^2 (z+2i)} ={1\over \omega^2(\omega+3i)} ={1\over \omega^3(1+{3i\over \omega})} \\={1\over \omega^3} \sum_{n=0}^\infty \left( -{3i\over \omega} \right)^n =  \sum_{n=0}^\infty  {(3i)^n\over \omega^{n+3}}  \Rightarrow \bbox[red, 2pt]{f(z) = \sum_{n=0}^\infty{(-3i)^n \over (z-i)^{n+3}}}$$

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解題僅供參考，碩士班歷年試題及詳解