110年北科大電機碩士班-工程數學詳解

 國立臺北科技大學110學年度碩士班招生考試

系所組別:電機工程碩士班丙組
科目:工程數學

解答:$$v(x)={1\over y} \Rightarrow v'=-{y'\over y^2} \Rightarrow y'=-y^2 v'= -{v'\over v^2} \\ \Rightarrow -{v'\over v^2}-2{1\over v}=-{1\over v^2}e^{3x} \Rightarrow v'+2v=e^{3x} \Rightarrow e^{2x}v'+2e^{2x}v=e^{5x} \Rightarrow(e^{2x} v)'=e^{5x} \\ \Rightarrow e^{2x}v={1\over 5}e^{5x}+c_1 \Rightarrow v={1\over y}={1\over 5}e^{3x}+{c_1\over e^{2x}} \Rightarrow y={1\over {1\over 5}e^{3x}+{c_1\over e^{2x}}} \\ \Rightarrow \bbox[red, 2pt]{y={5e^{2x}\over e^{5x}+c_2}}$$

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow m(m-1)x^{m-2} \\ \Rightarrow x^2y''-5xy'+9y= m(m-1)x^m-5mx^m+9x^m =(m^2-6m+9)x^m=0\\ \Rightarrow m^2-6m+9=(m-3)^2=0 \Rightarrow m=3 \Rightarrow y_h=c_1x^3+c_2x^3\ln x\\ \text{Applying variation of parameters, let } \cases{ y_1=x^3\\ y_2=x^3\ln x \\r(x)=2x+\ln x/x^2} \\\Rightarrow W= \begin{vmatrix} y_1& y_2\\ y_1' &y_2' \end{vmatrix} =\begin{vmatrix} x^3& x^3\ln x\\ 3x^2 & 3x^2\ln x+x^2 \end{vmatrix} =x^5 \\ \Rightarrow y_p =-x^3\int {x^3\ln x\cdot (2x+\ln x/x^2)\over x^5}\,dx + x^3\ln x\int { x^3(2x+\ln x/x^2)\over x^5}\,dx \\ \qquad =-x^3\left( (\ln x)^2-{(\ln x)^2 \over 3x^3}-{2\ln x\over 9x^3}-{2\over 27x^3}\right) +x^3\ln x\left( 2\ln x-{\ln x\over 3x^3}-{1\over 9x^3}\right) \\\qquad = x^3(\ln x)^2+{1\over 9}\ln x+{2\over 27} \Rightarrow y=h_h+ y_p \\ \Rightarrow \bbox[red, 2pt]{ y= c_1x^3 +c_2x^3\ln x+ x^3(\ln x)^2+{1\over 9}\ln x+{2\over 27}}$$

解答:$$f(t)=\begin{cases}0 & t\lt 1\\2t-3 & 1\le t\lt 3 \\ 0 & t\ge 3\end{cases} \Rightarrow f(t)=(u(t-1)-u(t-3))(2t-3) \\ \Rightarrow L\{f(t)\} =-{3\over s}e^{-3s}-{2\over s^2}e^{-3s}-{1\over s}e^{-s}+{2\over s^2}e^{-s}\\ L\{y''(t)\} +2L\{y'(t)\} +2L\{ y(t)\} =s^2Y(s)-1+2sY(s)+2Y(s) =(s^2+2s+2)Y(s)-1 \\ \Rightarrow =(s^2+2s+2)Y(s)-1=-{3\over s}e^{-3s}-{2\over s^2}e^{-3s}-{1\over s}e^{-s}+{2\over s^2}e^{-s} \\ \Rightarrow Y(s)={1\over (s+1)^2+1^2}\left(1 -{3\over s}e^{-3s}-{2\over s^2}e^{-3s}-{1\over s}e^{-s}+{2\over s^2}e^{-s}\right) \\ \Rightarrow y(t)=L^{-1}\{ Y(s)\}= e^{-t}\sin(t)- u(t-3)\left(t-{1\over 2}e^{-t}(3\sin t+\cos t) -{5\over 2}\right) \\\qquad +u(t-1)\left(t+{1\over 2}e^{-t}(\sin t+3\cos t)-{5\over 2} \right)\\ \Rightarrow \bbox[red, 2pt]{y(t)=e^{-t}\sin(t)+ (u(t-1)-u(t-3))(t-{5\over 2})\\\qquad \qquad +{1\over 2}e^{-t}(u(t-1)(\sin t+3\cos t)+u(t-3)(3\sin t+\cos t))}$$

解答:$$\begin{bmatrix}1 & 2 & 3 \\4 & a & 6\\ 2& 4& b \end{bmatrix} \begin{bmatrix}x_1   \\x_2\\ x_3 \end{bmatrix} =\begin{bmatrix}1 \\2 \\ c \end{bmatrix} \equiv A\mathbf x=\mathbf b\\ \textbf{(1)}\; \det(A)= (a-8)(b-6) \ne 0 \Rightarrow a\ne 8,b\ne 6 \Rightarrow \text{unique solution: } \bbox[red, 2pt]{\cases{a\ne 8\\ b\ne 6\\ c\in \mathbb R}}\\ \textbf{(2)} \; \text{many solution: } \cases{{1\over 2}={2\over 4}={3\over b}={1\over c} \\ {4\over 2}={a\over 4}={6\over b}={2\over c}} \Rightarrow \bbox[red, 2pt]{\cases{(a\in \mathbb R,b=6,c=2)\\ (a=8,b=3,c=1)}}\\ \textbf{(3)}\; \text{no solution: }\cases{{1\over 2}={2\over 4}={3\over b}\ne {1\over c} \\ {4\over 2}={a\over 4}={6\over b}\ne {2\over c}} \Rightarrow \bbox[red, 2pt]{\cases{(a\in \mathbb R,b=6,c\ne 2)\\(a=8,b=3,c\ne 1 }}$$

解答:$$A=\begin{bmatrix}1 & 1 \\-1 & 1 \end{bmatrix} \Rightarrow A^{-1}=\begin{bmatrix}\frac{1}{2} & \frac{-1}{2} \\\frac{1}{2} & \frac{1}{2} \end{bmatrix} \Rightarrow A^{-2}= \begin{bmatrix}0 & \frac{-1}{2} \\\frac{1}{2} & 0 \end{bmatrix} \Rightarrow A^{-4}= \begin{bmatrix}\frac{-1}{4} & 0 \\0 & \frac{-1}{4} \end{bmatrix} =-{1\over 4}I_2 \\ \Rightarrow A^{-8}= {1\over 2^4}I_2 \Rightarrow A^{-16}={1\over 2^8}I_2 \Rightarrow A^{-64}={1\over 2^{32}}I_2 \\ \Rightarrow A^{-86}=A^{-64}\cdot A^{-16}\cdot A^{-4} \cdot A^{-2} =-{1\over 2^{42}}I_2 \cdot A^{-2}= \bbox[red, 2pt]{\begin{bmatrix}0 & 1/2^{43} \\-1/2^{43} & 0 \end{bmatrix}}$$

解答:$$\text{Given a 2x2 matrix }\begin{bmatrix}x_1 & x_2 \\x_3 & x_4 \end{bmatrix} , \text{there exists }a,b,c,d \text{ such that} \\a\begin{bmatrix}1 & 1 \\-1 & 1 \end{bmatrix} +b\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix} +c \begin{bmatrix}1 & 0 \\1 & 1 \end{bmatrix} +d\begin{bmatrix}1 & 0 \\0 & -1 \end{bmatrix} =\begin{bmatrix}x_1 & x_2 \\x_3 & x_4 \end{bmatrix} \\ \Rightarrow \cases{a+b+c+d=x_1\\ a+b=x_2\\ -a+c=x_3\\ a+b+c-d=x_4} \Rightarrow \begin{bmatrix}1 & 1& 1& 1 \\1 & 1 & 0 & 0\\ -1 & 0& 1& 0 \\ 1 & 1& 1& -1 \end{bmatrix} \begin{bmatrix}a \\ b \\c \\ d \end{bmatrix} =\begin{bmatrix}x_1 \\ x_2 \\x_3 \\ x_4 \end{bmatrix}\\ \Rightarrow \begin{bmatrix}a \\ b \\c \\ d \end{bmatrix} =\begin{bmatrix}1 & 1& 1& 1 \\1 & 1 & 0 & 0\\ -1 & 0& 1& 0 \\ 1 & 1& 1& -1 \end{bmatrix}^{-1} \begin{bmatrix}x_1 \\ x_2 \\x_3 \\ x_4 \end{bmatrix}=\begin{bmatrix}\frac{1}{2} & -1 & -1 & \frac{1}{2} \\\frac{-1}{2} & 2 & 1 & \frac{-1}{2} \\\frac{1}{2} & -1 & 0 & \frac{1}{2} \\\frac{1}{2} & 0 & 0 & \frac{-1}{2} \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\x_3 \\ x_4 \end{bmatrix} \\ \Rightarrow \cases{a=x_1/2-x_2-x_3+x_4/2 \\b=-x_1/2+2x_2+x_3-x_4/2\\ c=x_1/2-x_2+x_4/2\\ d=x_1/2-x_4/2} \Rightarrow \text{Any 2x2 matrix can be the combination of vectors in }V\\ \Rightarrow \bbox[red, 2pt]{Yes! }\text{V form a basis. }$$

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