111年台科大材料碩士班-工程數學詳解

國立臺灣科技大學111學年度碩士班招生試題

系所組別:材料科學與工程
科目:工程數學

解答: $$\cases{P(x,y)=y^2+xy^3\\ Q(x,y)= 5y^2-xy+y^3\sin y} \Rightarrow \cases{P_y= 2y+3xy^2\\ Q_x=-y} \Rightarrow \text{Not exact}\\- {P_y-Q_x\over P}=-{3y+3xy^2\over y^2+xy^3} =-{3\over y} \Rightarrow u'=-{3\over y} u \Rightarrow u={1\over y^3} \\ \Rightarrow \cases{uP={1\over y}+x \\ uQ={5\over y}-{x\over y^2}+ \sin y} \Rightarrow \cases{(uP)_y=-{1\over y^2}\\ (uQ)_x =-{1\over y^2}} \Rightarrow \Phi(x,y)= \int uP\,dx = \int uQ\,dy\\ \Rightarrow \Phi(x,y)={x\over y}+{1\over 2}x^2+ \phi(y)= 5\ln y+{x\over y}-\cos y+ \rho(x) \\ \Rightarrow \Phi= \bbox[red, 2pt]{{x\over y}+{1\over 2}x^2+ 5\ln y-\cos y+ C=0}$$

解答: $$先求齊次解,x^2y''+ xy'-y=0, 令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''= m(m-1) x^{m-2} \\ \Rightarrow x^2y''+xy'-y= m(m-1)x^m+mx^m-x^m=(m^2-1)x^m=0 \\ \Rightarrow m^2-1=0 \Rightarrow m=\pm 1 \Rightarrow y=c_1x+ {c_2 \over x}\\ 令\cases{y_1=x\\ y_2=1/x} \Rightarrow W=\begin{vmatrix} y_1 & y_2 \\ y_1'& y_2'\end{vmatrix} =\begin{vmatrix} x & 1/x \\1 & -1/x^2\end{vmatrix} =-{2\over x} \\ \Rightarrow y_p =-x \int {(1/x)\cdot (1/x^2(x+1))\over -2/x}\,dx + {1\over x}\int {x/x^2(x+1)\over -2/x}\,dx\\ ={x\over 2}\int {1\over x^2(x+1)}\,dx -{1\over 2x}\int{1\over x+1}\,dx ={x\over 2}\left( -{1\over x}-\ln x+ \ln(x+1)\right)-{1\over 2x}\ln(x+1) \\=-{1\over 2}-{x\over 2}\ln x+{x\over 2}\ln(x+1)-{1\over 2x}\ln(x+1) \Rightarrow y=y_h+y_p \\ \Rightarrow \bbox[red, 2pt]{y=c_1x + {c_2\over x}-{1\over 2}-{x\over 2}\ln x+{x\over 2}\ln(x+1)-{1\over 2x}\ln(x+1)}$$

解答: $$\textbf{(1)}\;{3(s^2+4) \over s(s^2+4s+8)} ={3\over 2}\cdot {1\over s}+{3s-12\over 2(s^2+4s+4)} \\={3\over 2}\cdot {1\over s}+{3\over 2}\cdot {s+2\over (s+2)^2+2^2} -9\cdot {1\over (s+2)^2+2^2} \\ \Rightarrow L^{-1}\left\{ {3(s^2+4) \over s(s^2+4s+8)}\right\} ={3\over 2}+{3\over 2}e^{-2t}\cos(2t)-9e^{-2t}\cdot {1\over 2}\sin(2t) \\= \bbox[red, 2pt]{{3\over 2}+{3\over 2}e^{-2t}(\cos(2t)-3\sin(2t))} \\\textbf{(2)}\; L\{\cos(3t)\} ={s\over s^2+3^2} \Rightarrow L\{t\cos(3t)\} ={s^2-3^2\over (s^2+3^2)^2} ,又L\{\sin(3t)\} ={3\over s^2+3^2} \\ \Rightarrow L\{ \sin(3t)-3t\cos(3t)\} ={3\over s^2+3^2} -{3(s^2-3^2)\over (s^2+3^2)^2} ={54\over (s^2+3^2)^2} \\ \Rightarrow L\{ e^{-2t}(\sin(3t)-3t\cos(3t))\} = {54\over ((s+2)^2+3^2)^2}  =9\cdot {6\over (s^2+4s+13)^2} \\ \Rightarrow L^{-1}\left\{ {6\over (s^2+4s+13)^2}\right\} = \bbox[red, 2pt]{{1\over 9}e^{-2t} (\sin(3t)- 3t\cos(3t))}$$

解答: $$t=-{1\over 2}\ln(1-x^2) \Rightarrow \frac{\text{d}t}{\text{d}x} ={x\over 1-x^2} \Rightarrow \frac{\text{d}^2t}{\text{d}x^2} ={1+x^2\over (1-x^2)^2}\\ 又\frac{\text{d}y}{\text{d}x} =\frac{\text{d}y}{\text{d}t} \frac{\text{d}t}{\text{d}x} =\frac{\text{d}y}{\text{d}t}\cdot {x\over 1-x^2}\\ \Rightarrow \frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}^2y}{\text{d}t^2}(\frac{\text{d}t}{\text{d}x})^2 + \frac{\text{d}y}{\text{d}t} \frac{\text{d}^2t}{\text{d}x^2} = \frac{\text{d}^2y}{\text{d}t^2}({x\over 1-x^2})^2 +\frac{\text{d}y}{\text{d}t}{1+x^2\over (1-x^2)^2} \\原式x(1-x^2)^2 \frac{\text{d}^2y}{\text{d}x^2}-(1-x^2)^2\frac{\text{d}y}{\text{d}x}  +x^3y \\= x(1-x^2)^2\left( \frac{\text{d}^2y}{\text{d}t^2}({x\over 1-x^2})^2 +\frac{\text{d}y}{\text{d}t}{1+x^2\over (1-x^2)^2} \right)-(1-x^2)^2\frac{\text{d}y}{\text{d}t}\cdot {x\over 1-x^2} +x^3y\\ =x^3 \frac{\text{d}^2y}{\text{d}t^2}+ 2x^3 \frac{\text{d}y}{\text{d}t}+x^3 y=0 \Rightarrow \frac{\text{d}^2y}{\text{d}t^2}+ 2\frac{\text{d}y}{\text{d}t}+ y=0 \\ \Rightarrow y=c_1e^{-t} +c_2te^{-t} =c_1 \sqrt{1-x^2}-{1\over 2}c_2 \sqrt{1-x^2}\ln(1-x^2) \\ \Rightarrow \bbox[red,2pt]{y=c_1 \sqrt{1-x^2}+ c_3 \sqrt{1-x^2} \ln(1-x^2)}$$

解答: $$取u(x,y)=X(x)Y(y) 則\\ B.C.: \cases{u(0,y)=0 \\ u(a,y)=0\\ u(x,0)=0\\ u(x,b)=f(x)} \Rightarrow \cases{X(0)Y(y)=0\\ X(a)Y(y)=0 \\ X(x)Y(0)=0 \\ X(x)Y(b)=f(x)} \Rightarrow \cases{X(0)=0\\ X(a)=0 \\ Y(0)=0} \\ 而原式 \frac{\partial^2 u}{\partial x^2} +\frac{\partial^2 u}{\partial y^2} =0 \Rightarrow X''Y+ XY''=0 \Rightarrow {X''\over X}=-{Y''\over Y}=k\\ 先討論{X''\over X}=k\\ \textbf{Case I: }k=0. X''=0 \Rightarrow X=c_1x+c_2 \Rightarrow \cases{X(0)=0\\ X(a)=0} \Rightarrow \cases{c_2=0\\ c_1a+c_2=0} \\\qquad \Rightarrow c_1=c_2=0 \Rightarrow X=0\\  \textbf{Case II: }k\gt 0. 假設k=\lambda^2 (\lambda \gt 0) \Rightarrow X^2-\lambda^2 X=0 \Rightarrow X=c_1e^{\lambda x} +c_2 e^{-\lambda x} \\\qquad \Rightarrow \cases{X(0)=0\\ X(a)=0} \Rightarrow \cases{c_1+ c_2=0 \\ c_1e^{a\lambda} +c_2e^{-a\lambda}=0 } \Rightarrow c_1(e^{2a\lambda}-1)=0 \Rightarrow c_1=0\\\qquad  \Rightarrow c_2=0 \Rightarrow X=0\\   \textbf{Case III: }k\lt 0.假設k=-\lambda^2 (\lambda \gt 0) \Rightarrow X^2+ \lambda^2 X=0 \Rightarrow X= c_1\cos (\lambda x) + c_2 \sin (\lambda x)\\ \qquad \cases{X(0)=c_1=0\\ X(a)=c_2 \sin(\lambda a)=0} \Rightarrow \sin(\lambda a)=0 \Rightarrow \lambda a=n\pi \Rightarrow \lambda ={n \pi \over a} \Rightarrow X= c_2 \sin{n\pi x\over a} \\ Y''-\lambda^2 Y=0 \Rightarrow Y= c_3 \sinh{\lambda y} +c_4 \cosh{\lambda y}, \\\qquad 又Y(0)=0 \Rightarrow c_4=0 \Rightarrow Y= c_3 \sinh{\lambda y} \Rightarrow u=c_2c_3 \sin{n\pi x\over a} \sinh{n\pi y\over a} \\ \Rightarrow u(x,y)= A_n \sin{n\pi x\over a} \sinh{n \pi y\over a} \Rightarrow u(x,y) =\sum_{n=1}^\infty A_n \sin{n\pi x\over a} \sinh{n \pi y\over a} \\ \Rightarrow u(x,b)=\sum_{n=1}^\infty A_n  \sinh{n b\pi \over a}\sin{n\pi x\over a} =f(x) \Rightarrow A_n  \sinh{n b\pi \over a} ={2\over a} \int_0^a f(x) \sin{n\pi x\over a}dx\\ \Rightarrow \bbox[red, 2pt]{u(x,y) =\sum_{n=1}^\infty A_n \sin{n\pi x\over a} \sinh{n \pi y\over a}, 其中A_n={2\over a \sinh(nb\pi/a)} \int_0^a f(x) \sin{n\pi x\over a}dx}$$

解答: $$\int f_1 f_2\,dx = \int_{\pi/4}^{5\pi/4} e^x \sin x\,dx = \left. \left[{1\over 2}e^x (\sin x-\cos x) \right] \right|_{\pi/4}^{5\pi/4} =0 \\ \Rightarrow \bbox[red, 2pt]{\text{Yes}}, f_1 \text{ and }f_2 \text{are orthogonal to each other on}[\pi/4,5\pi/4]$$

解答: $$f(x,y)=x^2-4y^2 \Rightarrow \nabla f=(f_x,f_y)= (2x,-8y) \Rightarrow \nabla f(2,4)= \bbox[red, 2pt]{(4,-32)}$$

解答: $$A=\begin{bmatrix} 1& -1\\ 1& 1 \end{bmatrix} \Rightarrow \det(A- \lambda I) =\lambda^2-2\lambda+2=0 \Rightarrow \lambda =1\pm i\\ \lambda_1=1-i \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} i& -1\\ 1& i \end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_1=-ix_2 \\\quad \Rightarrow v=\begin{bmatrix} -i x_2\\ x_2\end{bmatrix}, 取v_1= \begin{bmatrix} -i \\ 1\end{bmatrix} \\ \lambda_2=1+i \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -i& -1\\ 1& -i \end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_1=ix_2 \\\quad \Rightarrow v=\begin{bmatrix} i x_2\\ x_2\end{bmatrix}, 取v_2= \begin{bmatrix} i \\ 1\end{bmatrix}\\ \Rightarrow \text{eigenvalues: } \bbox[red, 2pt]{1-i,1+i} \text{ and eigenvectors: } \bbox[red, 2pt]{\begin{bmatrix} -i \\ 1\end{bmatrix} , \begin{bmatrix} i \\ 1\end{bmatrix}}$$
 

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