國立成功大學115學年度碩士班招生考試試題
系所:資源工程學系
科目:工程數學
解答:$$\textbf{(a) }{dy\over dx}=-{1\over y}\sin x \Rightarrow \int y\,dy=\int-\sin x\,dx \Rightarrow {1\over 2}y^2=\cos x+c_1 \Rightarrow y^2=2\cos x +c_2 \\ \quad \Rightarrow y= \pm \sqrt{2\cos x+c_2} \Rightarrow y(0)=\pm \sqrt{2+c_2}=1 \Rightarrow c_2=-1 \Rightarrow \bbox[red, 2pt]{y=\sqrt{2\cos x-1}} \\ \textbf{(b) }y''+y'+y=0 \Rightarrow r^2+r+1=0 \Rightarrow r={-1\pm \sqrt 3i\over 2} \\ \quad \Rightarrow y_h= e^{-x/2} (c_1\cos (\sqrt 3x/2) +c_2 \sin(\sqrt 3x/2)) \\ \quad y_p=Ax+B \Rightarrow y_p'=A \Rightarrow y_p''=0 \Rightarrow y_p''+y_p'+y_p=Ax+A+B= x \Rightarrow \cases{A=1\\ B=-1} \\ \quad \Rightarrow y_p=x-1 \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= e^{-x/2} \left( c_1\cos (\sqrt 3x/2) +c_2 \sin(\sqrt 3x/2) \right)+x-1}$$
解答:$$L\{{dy\over dt}\} +L\{y\} =L\{\delta(t-1)\} \Rightarrow sY(s)-1+Y(s)=e^{-s} \Rightarrow Y(s)={e^{-s}+1\over s+1} \\ \Rightarrow y(t)= L^{-1} \left\{Y(s) \right\} =L^{-1} \left\{{e^{-s}\over s+1}+{1\over s+1} \right\}\Rightarrow \bbox[red, 2pt]{y(t) =u(t-1)e^{-(t-1)}+e^{-t}}$$
解答:$$\textbf{(a) }f=xz-yz \Rightarrow \nabla f=[f_x,f_y,f_z] =[z,-z,x-y] \Rightarrow \nabla f(0,3,1)=\bbox[red, 2pt]{[1,-1,-3]} \\\textbf{(b) }\cases{ \text{div}(v) ={\partial \over \partial x}(4z) + {\partial \over \partial y}(2y) +{\partial \over \partial z}(x-z) = 0+2-1=1\\ \text{div}(w)= {\partial \over \partial x}(y^2) + {\partial \over \partial y}(y^2-x^2) +{\partial \over \partial z}(2z^2) =0+2y+4z=2y+4z} \Rightarrow \bbox[red, 2pt] {\cases{\text{div}(v)=1\\ \text{div}(w)=2y+4z}} \\\textbf{(c) } \text{curl}(v) = \begin{vmatrix} \mathbf i& \mathbf j& \mathbf k\\ {\partial \over \partial x}& {\partial \over \partial y}& {\partial \over \partial z} \\ 4z& 2y&x-z\end{vmatrix} =0\mathbf i+0\mathbf k+4\mathbf j-0\mathbf k-\mathbf j-0\mathbf i =3\mathbf j \Rightarrow \bbox[red, 2pt]{\text{curl}(v)= [0,3,0]}$$
解答:$$\textbf{(1) } \text{RREF}(A) = \begin{bmatrix} 1 & 0 & 0\\0 & 1 & - \frac{1}{3}\\0 & 0 & 0\end{bmatrix} \Rightarrow \text{rank}(A)=2 \Rightarrow \text{rank}(A)+\text{nullity}(A)=3 \\\qquad \Rightarrow \text{nullity}(A)=3-2=1 \Rightarrow \bbox[red, 2pt]{\cases{\text{rank}(A)=2\\ \text{nullity}(A)=1}} \\\textbf{(2) } [A\mid B] = \begin{bmatrix} 5 & -3 & 1 & 1\\2 & 3 & -1 & 3\\8 & 9 & -3 & 7 \end{bmatrix} \Rightarrow \text{RREF}([A\mid B]) = \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 1 & - \frac{1}{3} & 0\\0 & 0 & 0 & 1\end{bmatrix} \Rightarrow \text{rank}([A\mid B])=3\\\quad \Rightarrow \text{rank}([A\mid B]) \gt \text{rank}(A) \Rightarrow \bbox[red, 2pt] 0\text{ solution}$$
解答:$$\textbf{(1)} I=F\left\{ {d^2f\over dx^2}\right\} =\int_0^b f''(x)\sin(\beta x) \,dx \\\cases{u=\sin(\beta x) \\dv=f''(x)\,dx } \Rightarrow \cases{du=\beta \cos(\beta x)\,dx\\ v=f'(x)} \Rightarrow I= \left. \left[ f'(x) \sin(\beta x) \right] \right|_0^b - \int_0^b \beta f'(x)\cos(\beta x)\,dx \\=f'(b)\sin({n\pi\over b}\cdot b)-0-\beta \int_0^bf'(x)\cos(\beta x)\,dx =-\beta \int_0^bf'(x)\cos(\beta x)\,dx \\ \cases{u=\cos(\beta x) \\dv=f'(x)\,dx} \Rightarrow \cases{du=-\beta\sin(\beta x)\, dx\\ v=f(x)} \Rightarrow I=-\beta \left( \left. \left[ f(x)\cos (\beta x) \right] \right|_0^b +\int_0^b \beta f(x)\sin(\beta x)\,dx\right) \\ =-\beta \left( 0 +\int_0^b \beta f(x)\sin(\beta x)\,dx\right)=-\beta^2 \int_0^b \sin(\beta x)\,dx =-\beta^2 F\{f(x)\} \\\Rightarrow F\left\{ {d^2f\over dx^2}\right\} =-\beta^2 F\{f(x)\} \quad \bbox[red, 2pt]{QED.} \\\textbf{(2)} {d^2 f\over dx^2}+f=1 \Rightarrow F\left\{ {d^2 f\over dx^2}\right\}+ F\{f\}= F\{1\} \Rightarrow -\beta^2 F\{f\}+ F\{f\}= \int_0^b \sin(\beta x)\,dx\\ \Rightarrow (1-\beta^2)F\{f\} ={1\over \beta}(1-(-1)^n) \Rightarrow F\{f\} =F(\beta) ={1\over \beta(1-\beta^2)}(1-(-1)^n) \\ b=1\Rightarrow \beta={n\pi\over b}=n\pi \Rightarrow F(\beta) ={1\over n\pi(1-n^2\pi^2)} (1-(-1)^n) \\ \Rightarrow f(x) ={2\over b} \sum_{n=1}^\infty F(\beta)\sin(\beta x)\Rightarrow \bbox[red, 2pt]{f(x) =2 \sum_{n=1}^\infty {1\over n\pi(1-n^2\pi^2)} (1-(-1)^n) \sin(n\pi x)}$$
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解題僅供參考，碩士班歷年試題及詳解
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