112年台北科大機電整合碩士班乙-工程數學詳解

國立 臺北科技 大學 112學 年度碩 士班招 生考試

系所組別 :1120機械工程系機電整合碩士班乙組
第一節 工程數學 試題

解答： $$\mathbf{1.}\; (2xy)dx+(x^2-1)dy=0 \Rightarrow {2x\over 1-x^2}dx={1\over y}dy \Rightarrow \int{2x\over 1-x^2} \,dx =\int {1\over y}\,dy\\ \Rightarrow -\ln(x^2-1)+c_1=\ln y \Rightarrow \bbox[red, 2pt]{y= {c_2\over x^2-1}}\\\mathbf{2.}\; xy'+y=x^2y^2 \Rightarrow y'+{1\over x}y=xy^2\;\text{Bernoullii Equation}\\ \Rightarrow 取u={1\over y} \Rightarrow u'=-{y'\over y^2} 代入原式 \Rightarrow -u'y^2+{y\over x}=xy^2 \Rightarrow u'-{1\over x}\cdot {1\over y}=-x \Rightarrow u'-{u\over x}=-x\\ \Rightarrow 積分因子I(x)=e^{\int{-1/x}\,dx}={1\over x} \Rightarrow {u'\over x}-{u\over x^2}=-1 \Rightarrow ({u\over x})'=-1 \Rightarrow {u\over x}=-x+c_1\\ \Rightarrow u={1\over y}=-x^2+c_1x \Rightarrow y={1\over c_1x-x^2} \Rightarrow \bbox[red,2pt]{y={1\over c_2x+x^2}}$$

解答： $$\mathbf{1.}\; 令y=x^m \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow 4x^2y''+17y=4m(m-1)x^m+17x^m=(4m^2-4m+17)x^m=0\\ \Rightarrow 4m^2-4m+17=0 \Rightarrow m={1\over 2}\pm 2i \Rightarrow y=c_1x^{1/2+2i}+c_2x^{1/2-2i} =c_1\sqrt x\cdot x^{2i}+ c_2\sqrt x\cdot x^{-2i} \\ =c_1\sqrt x e^{2i\ln x}+ c_2\sqrt x e^{-2i\ln x}\Rightarrow \bbox[red, 2pt]{y=\sqrt x(A\cos(2\ln x)+ B\sin (2\ln x))}\\ \mathbf{2.}\;先求齊次解，y''+3y'+2y=0 \Rightarrow \lambda^2+3\lambda +2=0 \Rightarrow \lambda=-2,-1 \Rightarrow y_h=c_1e^{-2x}+c_2 e^{-x} \\ 令y_p=ax^2+bx+c \Rightarrow y_p'=2ax+b \Rightarrow y_p''=2a \\\Rightarrow y''+3y'+2y=2ax^2+ (6a+2b)x+2a+3b+2c=2x^2 \Rightarrow \cases{2a=2\\ 6a+2b=0\\ 2a+3b+2c=0} \\ \Rightarrow \cases{a=1\\ b=-3\\c=7/2} \Rightarrow y_p=x^2-3x+{7\over 2} \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{-2x} +c_2e^{-x}+x^2-3x+{7\over 2}}$$

解答： $$\mathbf{1.}\; L\{t\cos(3t)\}=-{d\over ds}L\{\cos(3t)\} =-{d\over ds}({s\over s^2+3^2})=-{1\over s^2+3^2}+{2s^2\over (s^2+3^2)^2 }=\bbox[red,2pt]{s^2-3^2\over (s^2+3^2)^2} \\\mathbf{2.}\; L\{y''\}+ 3L\{y' \}+2 L\{y\}=2L\{u(t-1)\} \\\quad \Rightarrow s^2Y(s)-sy(0)-y'(0)+3(sY(s)-y(0))+2Y(s)=2\cdot {e^{-s}\over s} \\ \quad \Rightarrow Y(s)={2e^{-s} \over (s^2+3s+2)s} \Rightarrow y(t)=L^{-1}\left( {2e^{-s} \over (s^2+3s+2)s}\right) \\ =L^{-1}\left( e^{-s}\left(-{2\over s+1}+{1\over s+2}+{1\over s} \right)\right)=u(t-1)\left( -2e^{-(t-1)}+ e^{-2(t-1)}+1\right) \\\Rightarrow \bbox[red, 2pt]{y=u(t-1)\left( -2e^{-(t-1)}+ e^{-2(t-1)}+1\right)}$$

解答： $$\mathbf{1.}\;A=\begin{bmatrix}-4 & -6 \\3 & 5 \end{bmatrix} \Rightarrow \det(A-\lambda I)=(\lambda+1)(\lambda-2) =0 \Rightarrow \lambda=-1,2\\  \lambda_1=-1 \Rightarrow (A-\lambda_1 I)x=0 \Rightarrow \begin{bmatrix}-3 & -6 \\3 & 6 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix}=0 \Rightarrow x_1=-2x_2,取v_1=\begin{bmatrix}-2 \\1 \end{bmatrix}\\ \lambda_2=2\Rightarrow (A-\lambda_2 I)x=0 \Rightarrow \begin{bmatrix}-6 & -6 \\3 & 3 \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \end{bmatrix}=0 \Rightarrow x_1=-x_2,取v_2= \begin{bmatrix}-1 \\1 \end{bmatrix}\\ \quad 因此特徵值為\bbox[red,2pt]{-1,2}，相對應的特徵向量為\bbox[red, 2pt]{\begin{bmatrix}-2 \\1 \end{bmatrix}, \begin{bmatrix}-1 \\1 \end{bmatrix}}\\ \mathbf{2.}A=PDP^{-1} =\begin{bmatrix}-2 & -1 \\1 & 1 \end{bmatrix} \begin{bmatrix}-1 & 0 \\0 & 2 \end{bmatrix}\begin{bmatrix}-2 & -1 \\1 & 1 \end{bmatrix}^{-1} = \bbox[red, 2pt]{\begin{bmatrix}-2 & -1 \\1 & 1 \end{bmatrix} \begin{bmatrix}-1 & 0 \\0 & 2 \end{bmatrix}\begin{bmatrix}-1 & -1 \\1 & 2 \end{bmatrix}} \\ \mathbf{3.}\;A^{30} =(PDP^{-1})^{30} =PD^{30}P{-1} =\begin{bmatrix}-2 & -1 \\1 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 \\0 & 2^{30} \end{bmatrix}\begin{bmatrix}-1 & -1 \\1 & 2 \end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix}2-2^{30} & 2-2^{31} \\-1+2^{30} & -1+2^{31} \end{bmatrix}}$$

解答： $$\mathbf{1.}\;f(t)=\begin{cases}-1,& -\pi\lt t\lt 0\\ 1,& 0\lt t\lt \pi \end{cases} \Rightarrow f(t)為奇函數\Rightarrow a_n=0,n=0,1,2,...\\ b_n={1\over \pi}\int_{-\pi}^\pi f(x)\sin (nx)\,dx ={1\over \pi}\int_{-\pi}^0 -\sin (nx)\,dx +{1\over \pi}\int_{0}^\pi \sin (nx)\,dx  \\= {1\over \pi} \left(\left.\left[ {1\over n}\cos(nx) \right] \right|_{-\pi}^0+\left.\left[ -{1\over n}\cos(nx)\right] \right|_{0}^\pi\right) ={2\over n\pi}(1-(-1)^n)=\begin{cases} 4/n\pi,& n是奇數\\ 0,& n是偶數\end{cases}\\ \Rightarrow \bbox[red,2pt]{\cases{a_n=0,n=0,1,2,\dots\\ b_n=\cases{4/n\pi,n是偶數\\ 0,n是奇數},n\in\mathbb N}} \\\mathbf{2.}\;F(\omega)=\int_{-\infty}^\infty f(t)e^{-j\omega t}\,dt =\int_{-\infty}^1 e^{2(t-1)}\;e^{-j\omega t}\,dt+\int_{1}^\infty e^{-2(t-1)}\;e^{-j\omega t}\,dt \\= e^{-2}\int_{-\infty}^1 e^{(2-j\omega)t} \,dt+e^2\int_{1}^\infty e^{-(2+j\omega)t} \,dt =e^{-2}\left. \left[ {1\over 2-j\omega}e^{(2-j\omega)t}\right] \right|_{-\infty}^1+ e^{2}\left. \left[ {1\over -2-j\omega}e^{-(2+j\omega)t} \right] \right|_{1}^\infty \\={1\over 2-j\omega}e^{-j\omega}-{1\over -2-j\omega}e^{-j\omega} =e^{-j\omega}\left({1\over 2-j\omega}+{1\over 2+j\omega}\right) =\bbox[red,2pt]{e^{-j\omega}\cdot {4\over 4+\omega^2}}$$
 

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