115年成功大學機械碩士班-工程數學詳解

國立成功大學115學年度碩士班招生考試試題

系所:機械工程學系
科目:工程數學

解答:$${dy\over dx}=y\ln|x| \Rightarrow \int{1\over y}\,dy=\int \ln|x|\,d x \Rightarrow \ln|y|= x\ln|x|-x+c_1 \\ \Rightarrow y= e^{x\ln|x|}\cdot e^{-x}\cdot e^{c_1} ={|x|^x\over e^x}e^{c_1} \Rightarrow \bbox[red, 2pt]{y=c_2 \left( {|x|\over e} \right)^x}$$

解答:$$y''+y=0 \Rightarrow r^2+1=0 \Rightarrow r=\pm i \Rightarrow y_h= c_1 \cos x+c_2\sin x  \\ y_p =x(A\cos x+B\sin x) \Rightarrow y_p'=(A\cos x+B\sin x)+x(-A\sin x+B\cos x) \\ \Rightarrow y_p''=2B\cos x-2A\sin x-x(A\cos x+B\sin x) \Rightarrow y_p''+y_p= 2B\cos x-2A\sin x= \cos x-\sin x \\ \Rightarrow \cases{2B=1\\ 2A=1} \Rightarrow \cases{A=1/2\\ B=1/2} \Rightarrow y_p= {1\over 2}x(\cos x+ \sin x) \Rightarrow y=y_h+y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1\cos x+c_2\sin x+{1\over 2}x(\cos x+\sin x)}$$

解答:$$C_1: r(x)=(x,0,1) \Rightarrow \cases{dr=(1,0,0) dx \\F= (0,2x,0)} \Rightarrow \int_{C_1} F\cdot dr =0 \\C_2: r(y=(2,y,1-y) \Rightarrow \cases{dr=(0,1,-1)dy\\ F=((1-y)^2 y \cos(2y), 2(1-y)^2 (1+ \cos(2y)), 2(1-y)y\sin(2y))} \\ \Rightarrow \int_{C_2}F\cdot dr = \int_0^1[2(1-y)^2(1+\cos 2y)-2(1-y)y\sin(2y)] \,dy  \\ C_3: r(x)=(x,1,0) \Rightarrow \cases{dr=(1,0,0)dx \\F=(0,0,0)} \Rightarrow \int_{C_3}F\cdot dr=0 \\ C_4: r(y)=(0,y,1-y) \Rightarrow \cases{dr=(0,1,-1)dy\\ F=((1-y)^2y,0,0)} \Rightarrow \int_{C_4}F\cdot dr=0 \\ \Rightarrow I= \int_C F\cdot dr =\int_{C_1} F\cdot dr + \int_{C_2} F\cdot dr + \int_{C_3} F\cdot dr +\int_{C_4} F\cdot dr  =0+\int_{C_2} F\cdot dr +0+0 \\ \Rightarrow I= \int_0^1[2(1-y)^2(1+\cos 2y)-2(1-y)y\sin(2y)] \,dy \\= \underbrace{\int_0^1 2(1-y)^2 dy}_{I_1} +\underbrace{\int_0^1 [2(1-y)^2 \cos(2y)-2y(1-y)\sin (2y)] dy}_{I_2}   \\ \Rightarrow I_1= \left. \left[ -{2\over 3}(1-y)^3 \right] \right|_0^1 ={2\over 3} \\ \Rightarrow I_2= \int_0^1 {d\over dy} \left( (1-y)^2\sin(2y) \right)\,dy = \left. \left[ (1-y)^2\sin(2y) \right] \right|_0^1=0 \\ \Rightarrow I= \int_C F\cdot dr =\bbox[red, 2pt]{2\over 3}$$

解答:$$\textbf{(1) } \begin{bmatrix}F_n\\ F_{n-1} \end{bmatrix} =A \begin{bmatrix}F_{n-1}\\F_{n-2} \end{bmatrix} = \begin{bmatrix}1& 1\\ 1& 0 \end{bmatrix} \begin{bmatrix}F_{n-1}\\F_{n-2} \end{bmatrix}  \Rightarrow A=\bbox[red, 2pt]{\begin{bmatrix}1& 1\\ 1& 0 \end{bmatrix}} \\\textbf{(2) }\det(A-\lambda I) = \lambda^2-\lambda -1=0 \Rightarrow \lambda_1= {1+\sqrt 5\over 2}, \lambda_2={1-\sqrt 5\over 2}\\ \quad (A-\lambda I)v=0 \Rightarrow \begin{bmatrix}1-\lambda & 1\\1& -\lambda \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_1=\lambda x_2 \Rightarrow v= \begin{bmatrix}\lambda\\ 1 \end{bmatrix} \\ \Rightarrow \text{eigenvector:} \begin{bmatrix}\lambda_1\\ 1 \end{bmatrix}, \begin{bmatrix}\lambda_2\\ 1 \end{bmatrix} \Rightarrow \begin{bmatrix}\lambda_1& \lambda_2\\ 1& 1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}{1+\sqrt 5\over 2} & {1-\sqrt 5\over 2}\\1& 1 \end{bmatrix}}$$

解答:$$\iint_R xe^{y^2} \,dA =\int_0^4 \int_0^{\sqrt y} xe^{y^2}\,dx\,dy =\int_0^4 {1\over 2}y e^{y^2}\,dy = \left. \left[ {1\over 4}e^{y^2} \right] \right|_0^4= \bbox[red, 2pt]{{1\over 4} \left( e^{16}-1 \right)}$$

解答:$$L\{f(t)\} = \int_0^\infty f(t)e^{-st}\,dt = \int_0^2 e^{-st}\,dt + \int_2^3 -3e^{-st}\,dt + \int_3^\infty t^2 e^{-st} \,dt \\= \left. \left[ -{1\over s}e^{-st} \right] \right|_0^2 + \left. \left[ {3\over s}e^{-st} \right] \right|_2^3 + \left. \left[ -{1\over s}t^2 e^{-st}-{2\over s^2}te^{-st}-{2\over s^3}e^{-st} \right] \right|_3^\infty\\ = {1\over s} \left( 1-e^{-2s} \right)+{3\over s} \left( e^{-3s}-e^{-2s} \right)+{e^{-3s}} \left( {9\over s}+{6\over s^2}+ {2\over s^3} \right) \\=\bbox[red, 2pt]{{1\over s}-{4\over s}e^{-2s}+ \left( {12\over s}+{6\over s^2}+{2\over s^3} \right)e^{-3s}}$$

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解題僅供參考，碩士班歷年試題及詳解