國立立臺灣科技大學115學年度碩士班招生
系所組別:1200自動化及控制研究所碩士班
科目:工程數學
解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-3xy'+3y=(m(m-1)-3m+3)x^m \\=(m^2-4m+3)x^m=0 \Rightarrow m^2-4m+3=0 \Rightarrow (m-1)(m-3)=0 \Rightarrow m=1,3 \\ \Rightarrow y_h=c_1x+c_2x^3\\ \cases{y_1=x\\ y_2= x^3\\ r(x)=2x^2e^x} \Rightarrow W= \begin{vmatrix} y_1 & y_2\\ y_1'& y_2'\end{vmatrix} = \begin{vmatrix} x& x^3\\ 1& 3x^2\end{vmatrix} =2x^3 \Rightarrow y_p=-y_1 \int {y_2 r\over W}\,dx +y_2 \int {y_1r\over W}\,dx \\ \Rightarrow y_p=-x \int x^2e^x\,dx + x^3 \int e^x\,dx =-xe^x( x^2-2x+2) +x^3e^x =2xe^x(x-1) \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt] {y= c_1x+c_2x^3+ 2xe^x(x-1)}$$
解答:$$L\{y'(t)\} = L\{\cos t\} + L \left\{\int_0^t y(\tau)\cos(t-\tau) \,d\tau\right\} = L\{\cos t\} + L \{y(t)\}\cdot L\{ \cos t\} \\ \Rightarrow sY(s)-1={s\over s^2+1} +Y(s)\cdot {s\over s^2+1} \Rightarrow Y(s)={s^2+s+1\over s^3} ={1\over s}+{1\over s^2}+{1\over s^3} \\ \Rightarrow y(t)=L^{-1}\{Y(s)\} =L^{-1}\{{1\over s}\}+L^{-1}\{{1\over s^2}\}+L^{-1}\{{1\over s^3}\} =1+t+{t^2\over 2} \Rightarrow \bbox[red, 2pt]{y(t)=1+t +{t^2\over 2}}$$
解答:$$\oint_C \mathbf F\cdot d \mathbf r = \iint_S (\nabla \times \mathbf F)\cdot d\mathbf S \\ z=g(x,y)=1-x-y \Rightarrow d\mathbf S = \left( -{\partial z\over \partial x}\mathbf i-{\partial z\over \partial y} \mathbf j+ \mathbf k \right)dA = (\mathbf i+\mathbf j+\mathbf k)dA \\\mathbf F=(y+ \sin x)\mathbf i+ (z^2+\cos y)\mathbf j +x^3 \mathbf k \Rightarrow \nabla\times \mathbf F= \begin{vmatrix} \mathbf i&\mathbf j& \mathbf k\\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\ y+\sin x& z^2+\cos y& x^3\end{vmatrix} \\=(-2z)\mathbf i+(-3x^2) \mathbf j+(-1) \mathbf k =-2(1-x-y)\mathbf i+(-3x^2) \mathbf j+(-1) \mathbf k\\\Rightarrow (\nabla \times \mathbf F) \cdot d\mathbf S = \left( (-2(1-x-y))\mathbf i+(-3x^2) \mathbf j+(-1) \mathbf k \right) \cdot (\mathbf i+\mathbf j+\mathbf k)dA \\= (2x+2y-3x^2-3)dA \Rightarrow \iint_S (\nabla \times \mathbf F)\cdot d\mathbf S =\int_0^1 \int_0^{1-x} (2x+2y-3x^2-3)dydx \\=\int_0^1 (3x^3-4x^2+3x-2)dx = \bbox[red, 2pt]{-{23\over 12}}$$
解答:$$\textbf{(1) } y= \sum_{n=0}^\infty a_n x^{n+r}, a_0\ne 0 \Rightarrow y'= \sum_{n=0}^\infty (n+r)a_n x^{n+r-1} \Rightarrow y''=\sum_{n=0}^\infty (n+r)(n+r -1)a_n x^{n+r-2} \\ \Rightarrow \sum_{n=0}^\infty 2(n+r)(n+r-1)a_n x^{n+r} +\sum_{n=0}^\infty 3(n+r)a_n x^{n+r} -\sum_{n=0}^\infty a_n x^{n+r+2}-\sum_{n=0}^\infty a_n x^{n+r} \\ \quad = \sum_{n=0}^\infty [2(n+r)(n+r-1) +3(n+r)-1]a_n x^{n+r} -\sum_{n=0}^\infty a_n x^{n+r+2} =0 \\ \text{Setting the coefficient of }x^r \text{ to zero: }a_0[2r(r-1)+3r-1] =0 \Rightarrow \text{ indicial equation: }\bbox[red, 2pt]{2r^2+r-1=0} \\ \Rightarrow (2r-1)(r+1)=0 \Rightarrow \bbox[red, 2pt]{r={1\over 2},-1} \\\textbf{(2) } \text{larger root }r={1\over 2} \Rightarrow \sum_{n=0}^\infty [2(n+r)(n+r-1)+ 3(n+r)-1]a_nx^{n+r} -\sum_{n=0}^\infty a_nx^{n+r+2}=0 \\ \Rightarrow [2(k+r)(k+r-1) +3(k+r)-1]a_k-a_{k-2} =0 \quad (k=n+2 \Rightarrow n=k-2) \\ \Rightarrow k(2k+3)a_k-a_{k-2} =0 \quad (r={1\over 2}) \Rightarrow a_k={a_{k-2} \over k(2k+3)} \Rightarrow \bbox[red, 2pt]{\cases{a_n=\displaystyle {a_{n-2} \over n(2n+3)}, n\ge 2 \\a_1=0}}$$
解答:$$\mathbf F=(x^3+e^{yz})\mathbf i+ (y^3+ \sin(xz))\mathbf j + (z^3+ \ln(x^2+y^2+1))\mathbf k \Rightarrow \nabla \cdot \mathbf F =3x^2+3y^2+3z^2 \\ \Rightarrow \iint_S \mathbf F\cdot \mathbf n d\mathbf S = \iiint_E (\nabla \cdot \mathbf F)\,dV = \iiint_E 3(x^2+y^2+z^2)\,dV \\= \int_0^{2\pi} \int_0^\pi \int_0^a 3\rho^2\cdot \rho^2 \sin \phi\,d\rho\,d\phi\,d \theta =3 \left( \int_0^{2 \pi}d \theta \right) \left( \int_0^\pi \sin \phi\,d\phi \right) \left( \int_0^a \rho^4\,d\rho \right) \\=3\cdot 2\pi \cdot 2\cdot {a^2\over 5}= \bbox[red, 2pt]{12\pi a^5\over 5}$$
解答:$$\textbf{(1) }A = \begin{bmatrix}5&-4& 4\\12& -11&12\\ 4&-4&5 \end{bmatrix} \Rightarrow \det(A-\lambda I) =-(\lambda-1)^2(\lambda+3)=0 \Rightarrow \lambda=1,-3\\ \lambda_1=-3 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}8&-4& 4\\ 12&-8& 12\\ 4&-4& 8 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} =0 \Rightarrow \cases{2x-y+z=0\\ 3x-2y+3z=0} \\\qquad \Rightarrow v= z \begin{bmatrix}1\\ 3\\1 \end{bmatrix} \Rightarrow \text{ Choosing z=1, we get }v_1= \begin{bmatrix}1\\3\\1 \end{bmatrix} \\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}4&-4& 4\\ 12&-12 &12\\ 4& -4& 4 \end{bmatrix} \begin{bmatrix}x\\ y\\z \end{bmatrix}=0 \Rightarrow y=x+z \\ \qquad \Rightarrow v= \begin{bmatrix}x\\ x+z\\ z \end{bmatrix} =x \begin{bmatrix}1\\1\\0 \end{bmatrix} +z \begin{bmatrix}0\\1\\1 \end{bmatrix} \Rightarrow \text{Choosing x=1 and z=0, we get }v_2= \begin{bmatrix}1\\1\\0 \end{bmatrix}\\ \qquad \text{And choosing x=0 and z=1, we get }v_3= \begin{bmatrix} 0\\1\\ 1\end{bmatrix} \\\bbox[red, 2pt]{ \text{Eigenvalues: -3, 1 and the corresponding eigenvctors: }\begin{bmatrix}1\\3\\1 \end{bmatrix}, \begin{bmatrix}1\\1\\0 \end{bmatrix},\begin{bmatrix} 0\\1\\ 1\end{bmatrix}}\\ \textbf{(2) }X(t)= c_1e^{\lambda_1 t}v_1 + c_2 e^{\lambda_2 t}v_2 + c_3 e^{\lambda_2 t}v_3 \Rightarrow \bbox[red, 2pt]{X(t)=c_1e^{-3 t} \begin{bmatrix}1\\3\\1 \end{bmatrix} + c_2 e^{ t} \begin{bmatrix}1\\1\\ 0 \end{bmatrix} + c_3 e^{ t} \begin{bmatrix}0\\1\\1 \end{bmatrix}} \\\textbf{(3) } \cases{P=[v_1\; v_2 \; v_3] \\D= \begin{bmatrix}\lambda_1&0 &0\\ 0& \lambda_2 & 0\\ 0&0& \lambda_2 \end{bmatrix}} \Rightarrow A=PDP^{-1} \Rightarrow \bbox[red, 2pt]{ A= \begin{bmatrix}1& 1& 0\\ 3& 1& 1\\ 1& 0& 1 \end{bmatrix} \begin{bmatrix}-3&0&0\\0& 1& 0\\0& 0& 1 \end{bmatrix} \begin{bmatrix}-1&1&-1\\ 2&-1&1\\1&-1& 2 \end{bmatrix}}$$
解答:$$f(x) \sim {a_0\over 2}+ \sum_{n=1}^\infty (a_n \cos nx+ b_n \sin nx) \\ a_0={1\over \pi} \int_{-\pi}^\pi f(x)\,dx = {1\over \pi} \int_0^\pi x\,dx ={\pi\over 2} \\a_n ={1\over \pi}\int_0^\pi x\cos(nx)\,dx ={1\over n^2\pi} \left( (-1)^n-1 \right) \\ b_n ={1\over \pi} \int_0^\pi x\sin(nx) \,dx = {1\over n}(-1)^{n+1} \\ \Rightarrow \bbox[red, 2pt]{f(x) \sim {\pi\over 4}+ \sum_{n=1}^\infty ({1\over n^2\pi} \left( (-1)^n-1 \right) \cos nx+ {1\over n}(-1)^{n+1} \sin nx)} \\ F(x) = \int_{-\pi}^x f(t)\,dt = \int_{-\pi}^0 0\,dt+ \int_0^x t\,dt ={x^2\over 2} \Rightarrow F(x)= \begin{cases} 0,& -\pi\le x\le 0\\ x^2/2 & 0\lt x\le \pi\end{cases} \\ F(x) \sim {A_0\over 2}+ \sum_{n=1}^\infty (A_n\cos (nx)+ B_n \sin(nx)) \\ A_0 = {1\over \pi} \int_0^\pi {x^2\over 2}\,dx ={\pi^2\over 6} \\ A_n ={1\over \pi} \int_0^\pi {x^2\over 2} \cos(nx)\,dx = {1\over n^2}(-1)^n \\ B_n= {1\over \pi} \int_0^\pi {x^2\over 2}\sin(nx)\,dx = -{\pi\over 2n}(-1)^n+ {1\over n^3\pi}((-1)^n-1) \\ \Rightarrow F(x) = \bbox[red, 2pt]{\int_{-\pi}^x f(t)\,dt = {\pi^2\over 12}+ \sum_{n=1}^\infty \left( {1\over n^2}(-1)^n \cos(nx) + (-{\pi\over 2n}(-1)^n+ {1\over n^3\pi}((-1)^n-1)) \sin(nx)\right)}$$
解答:$$\textbf{(1)} p(t)= \sqrt t \sin{\pi t\over 6}\mathbf i+ \sqrt t \cos{\pi t\over 6} \mathbf j+ \sqrt t \mathbf k \\ \Rightarrow v(t)=p'(t) ={d\over dt} (\sqrt t \sin{\pi t\over 6})\mathbf i+ {d\over dt}(\sqrt t \cos{\pi t\over 6}) \mathbf j+ {d\over dt}(\sqrt t) \mathbf k \\ \Rightarrow v(t) = \left( {1\over 2\sqrt t}\sin{\pi t\over 6}+{\pi \sqrt t\over 6}\cos{\pi t\over 6} \right)\mathbf i + \left( {1\over 2\sqrt t}\cos {\pi t\over 6}-{\pi\sqrt t\over 6} \sin{\pi t\over 6} \right) \mathbf j+ {1\over 2\sqrt t} \mathbf k \\ u(t)= |v(t)| = \sqrt{\left( {1\over 2\sqrt t}\sin{\pi t\over 6}+{\pi \sqrt t\over 6}\cos{\pi t\over 6} \right)^2 +\left( {1\over 2\sqrt t}\cos {\pi t\over 6}-{\pi\sqrt t\over 6} \sin{\pi t\over 6} \right)^2 + \left( {1\over 2\sqrt t} \right)^2} \\ \Rightarrow \bbox[red, 2pt]{u(t)= \sqrt{{1\over 2t}+{\pi^2 t\over 36}} \; \text{m/s}} \Rightarrow \bbox[red, 2pt]{\cases{u(3)=\sqrt{{1\over 6}+{\pi^2\over 12}} \\u(3)=\sqrt{{1\over 12}+{\pi^2\over 6}} \\u(9)=\sqrt{{1\over 18}+{\pi^2\over 4}} }}\\ \textbf{(2) } \begin{array}{l} t& \sqrt t & \pi t/6 & x(t)& y(t) \\\hline0& 0& 0^\circ& 0& 0\\ 2 & \sqrt 2 & 60^\circ & \sqrt 6/2 & \sqrt 2/2\\ 4& 2 & 120^\circ & \sqrt 3& -1\\ 6& \sqrt 6& 180^\circ &0&-\sqrt 6 \\ 8& 2\sqrt 2& 240^\circ&-\sqrt 6& -\sqrt 2 \\10& \sqrt{10} & 300^\circ & -\sqrt{30}/2 & \sqrt{10}/2 \\ 12 & 2\sqrt 3& 360^\circ &0 & 2\sqrt 3 \\\hline\end{array}$$
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解題僅供參考，碩士班歷年試題及詳解
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