109年北科大電機碩士班-工程數學詳解

 國立臺北科技大學109學年度碩士班招生考試

系所組別:2131 電機工程系碩士班丙組
第一節 工程數學 試題(選考)

解答: $$\textbf{(a) }y=a-x \Rightarrow {dy\over dx }=-1 =(x+y)(x+y-2)=a(a-2) \Rightarrow a^2-2a+1=0 \Rightarrow \bbox[red, 2pt]{a=1} \\\textbf{(b) }{dy\over dx }=(x+y)(x+y-2)  \Rightarrow \text{Riccati's equation} \Rightarrow y=S(x)+{1\over v} =1-x+{ 1\over v} \\\quad \Rightarrow y'=-1-{v'\over v^2} =(x+y)(x+y-2)= (1+{1\over v})({1\over v}-1) \Rightarrow -1-{v'\over v^2} ={1\over v^2}-1\\ \quad \Rightarrow v'=-1 \Rightarrow v=-x+c_1 \Rightarrow \bbox[red, 2pt]{y=1-x+{1\over c_1-x}}$$

解答: $$x^2y''-2xy'+2y=0 \Rightarrow \text{Let }y=x^m, \text{ then }y'=mx^{m-1}, y''=m(m-1)x^{m-2} \\ \Rightarrow m(m-1)x^m-2mx^m+2x^m = (m^2-3m+2)x^m=0 \Rightarrow m^2-3m+2=0 \Rightarrow m=1,2\\ \Rightarrow y_h= c_1x+ c_2x^2 \\\text{Suppose that }\cases{y_1=x\\ y_2=x^2} \Rightarrow \cases{y_1'=1\\ y_2'=2x} \Rightarrow W(x,y)= \begin{vmatrix}y_1& y_2\\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix}x & x^2\\ 1& 2x \end{vmatrix} = x^2 \\ \text{Using variations of parameters, }y_p=-x \int {x^2 \cdot{1\over x^2}\cos {1\over x} \over x^2}\,dx +x^2 \int {x\cdot {1\over x^2}\cos({1\over x})\over x^2}\,dx \\ \Rightarrow y_p=-x \int {\cos {1\over x}\over x^2}\,dx +x^2 \int {\cos{1\over x}\over x^3} \,dx =-x(-\sin {1\over x}) +x^2\left(-{\sin{1\over x}+x \cos{1\over x} \over x} \right) \\\Rightarrow  y_p=-x^2\cos {1\over x} \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1x+ c_2x^2 -x^2 \cos {1\over x}}$$

解答: $$\cases{2x_1+ 2x_2+x_2'=\delta(t-3)  \\ x_1'-x_2=0  }   \Rightarrow  \cases{L\{2x_1+ 2x_2+x_2' \}= L\{\delta(t-3) \}  \\ L\{x_1'-x_2\}=0  } \\ \Rightarrow \cases{2X_1(s) +2X_2(s)+sX_2(s) =e^{-3s} \\ sX_1(s) -X_2(s)=0} \Rightarrow 2X_1(s)+ 2sX_1(s)+s^2X_1(s)= e^{-3s} \\ \Rightarrow X_1(s)= {1\over s^2+ 2s+2} e^{-3s} \Rightarrow x_1(t) =L^{-1}\{ X_1(s)\} =L^{-1} \left\{ {1\over s^2+ 2s+2} e^{-3s} \right\} \\=L^{-1} \left\{ {1\over (s+1)^2+1} e^{-3s} \right\}  \Rightarrow \bbox[red, 2pt] {x_1(t)=H(t-3)e^{-t+3} \sin(t-3)} \\ X_2(s)=sX_1(s) ={s\over s^2+ 2s+2} e^{-3s} \Rightarrow x_2(t) =L^{-1}\{ X_2(s)\} =L^{-1} \left\{ {s\over s^2+ 2s+2} e^{-3s} \right\} \\ \Rightarrow \bbox[red, 2pt] {x_2(t)= H(t-3)e^{-t+3}\left( \cos(t-3) -\sin(t-3)\right)}$$

解答: $$A=\begin{bmatrix} 1 & 1 & 0 \\-1 & 1 & 0 \\0 & 0 & 1\end{bmatrix} \Rightarrow \det(A-\lambda I) = -(\lambda-1)(\lambda^2-2\lambda+2) =0 \Rightarrow \lambda=1,1\pm i\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow x_1=x_2=0\\ \qquad \Rightarrow v= x_3\begin{pmatrix}0\\ 0\\ 1 \end{pmatrix}, \text{ choose }v_1= \begin{pmatrix}0\\ 0\\ 1 \end{pmatrix}\\ \lambda_2= 1-i \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}i & 1 & 0 \\-1 & i & 0 \\0 & 0 & i \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0  \Rightarrow \cases{x_1=ix_2\\ x_3=0} \\\qquad \Rightarrow v=x_2 \begin{pmatrix}i \\ 1\\ 0 \end{pmatrix} , \text{ choose }v_2= \begin{pmatrix}i \\ 1\\ 0 \end{pmatrix} \\ \lambda_3=1+i \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix}-i & 1 & 0 \\-1 & -i & 0 \\0 & 0 & -i \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0  \Rightarrow \cases{x_1+ix_2=0\\ x_3=0} \\\qquad \Rightarrow v=x_2 \begin{pmatrix}-i \\ 1\\ 0 \end{pmatrix} , \text{ choose }v_3= \begin{pmatrix}-i \\ 1\\ 0 \end{pmatrix} \\ \Rightarrow \text{eigenvectors: }v_1,v_2,v_3 =\bbox[red, 2pt]{\begin{pmatrix}0\\ 0\\ 1 \end{pmatrix}, \begin{pmatrix}i \\ 1\\ 0 \end{pmatrix}, \begin{pmatrix}-i \\ 1\\ 0 \end{pmatrix}}$$

解答: $$\textbf{(a) }\text{Let }\cases{f_1=e^{-2x} \\f_2=e^{2x}}  \text{ and }\{g_1,g_2\} \text{ denote the orthogonal basis. We apply the Gram-Schmidt process.}\\ g_1=f_1=e^{-2x} \Rightarrow \cases{f_2\cdot g_1= \int_0^1 1\,dx =1 \\ ||g_1||^2 = g_1\cdot g_1 = \int_0^1 e^{-4x}\,dx={1\over 4}(1-e^{-4})} \\ \Rightarrow   g_2 =f_2-{f_2\cdot g_1\over ||g_1||^2}g_1 =e^{2x} -{4\over  1-e^{-4}}e^{-2x} \Rightarrow \{g_1,g_2\}=\bbox[red, 2pt]{\left\{e^{-2x}, e^{2x} -{4\over  1-e^{-4}}e^{-2x}\right\}} \\\textbf{(b) } f(x)=x \Rightarrow \cases{f\cdot g_1 = \int_0^1 xe^{-2x}\,dx = {1\over 4}(1-3e^{-2}) \\ f\cdot g_2 =\int_0^1 (xe^{2x}-{4x\over 1-e^{-4}}e^{-2x}) \,dx ={1\over 4}(1+e^2-{4e^2(e^2-3) \over e^4-1})\\ g_1\cdot g_1= \int_0^1 e^{-4x}dx ={1\over 4}(1-e^{-4}) \\ g_2\cdot g_2 = \int_0^1 (e^{2x}-{4\over 1-e^{-4}}e^{-2x})^2\,dx= {1-18e^4+e^8 \over 4(e^4-1)}} \\ \Rightarrow f(x) \approx{f\cdot g_1 \over g_1\cdot g_1}g_1 +{f\cdot g_2\over g_2\cdot g_2}g_2 ={1-3e^{-2}\over 1-e^{-4}}g_1+{ 1+e^2 -{4e^2(e^2-3)\over e^4-1}\over e^4-1}g_2 \\ \Rightarrow \bbox[red, 2pt]{f(x)\approx {1-3e^{-2}\over 1-e^{-4}}g_1+{e^6-3e^4+11e^2-1 \over (e^4-1)^2}g_2}$$

解答: $$\textbf{(a) }\mathbf v=\begin{bmatrix} v_1\\ v_2\\ v_3\end{bmatrix} \Rightarrow A\mathbf v = \begin{bmatrix} \cos(\alpha) & -\sin(\alpha) & 0\\ \sin(\alpha) & \cos(\alpha) & 0\\ 0& 0 & 1\end{bmatrix} \begin{bmatrix} v_1\\ v_2\\ v_3\end{bmatrix}=  \begin{bmatrix} v_1 \cos (\alpha) -v_2\sin(\alpha)\\ v_1\sin(\alpha)+ v_2\cos(\alpha)\\ v_3\end{bmatrix} =\begin{bmatrix} u_1\\ u_2\\ u_3\end{bmatrix} =\mathbf u \\\quad \Rightarrow \begin{bmatrix} u_1\\ u_2 \end{bmatrix} =\begin{bmatrix} \cos(\alpha) & -\sin(\alpha)  \\ \sin(\alpha) & \cos(\alpha)  \end{bmatrix} \begin{bmatrix} v_1\\ v_2 \end{bmatrix} \Rightarrow \mathbf u= \text{ rotating }\mathbf v \text{ about }z \text{ axis through angle }\alpha\quad \bbox[red, 2pt]{QED}\\ \textbf{(b) }A=\begin{bmatrix} \cos(\alpha) & -\sin(\alpha) & 0\\ \sin(\alpha) & \cos(\alpha) & 0\\ 0& 0 & 1\end{bmatrix} =\begin{bmatrix} B & 0\\   0&   1\end{bmatrix} ,\text{ where }B= \begin{bmatrix} \cos(\alpha) & -\sin(\alpha)  \\ \sin(\alpha) & \cos(\alpha)  \end{bmatrix}  \text{ is a rotation matrix}\\ \Rightarrow A^{20} =\begin{bmatrix} B^{20} & 0\\   0&   1\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} \cos(20\alpha) & -\sin(20\alpha) & 0\\ \sin(20 \alpha) & \cos(20\alpha) & 0\\ 0& 0 & 1\end{bmatrix}}$$

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解題僅供參考，碩士班歷年試題及詳解