國立臺北科技大學 l11學 年度碩 士班招生考試
系所組別 :2401、 2402、 2403光 電工程系碩士班
第一節 工程數學 試題
解答:δ=[01/201/200001]=[01−1011100][10001/2000−1/2][0011/21/20−1/21/20]≡PDP−1⇒cosδ=P[cos1000cos(1/2)000cos(1/2)]P−1=[cos(1/2)000cos(1/2)000cos1]⇒sinδ=P[sin1000sin(1/2)000−sin(1/2)]P−1=[0sin(1/2)0sin(1/2)0000sin1]因此eiδ=cosδ+isinδ=[cos(1/2)000cos(1/2)000cos1]+[0isin(1/2)0isin(1/2)0000isin1]=[cos(1/2)isin(1/2)0isin(1/2)cos(1/2)000isin1+cos1]解答:L{y″
解答:y''+4y=0 \Rightarrow y_h=c_1\cos(2x) +c_2\sin(2x)\\ 令\cases{y_1= \cos(2x)\\ y_2=\sin(2x)} \Rightarrow W=\begin{vmatrix} y_1& y_2 \\ y_1' & y_2'\end{vmatrix} =\begin{vmatrix} \cos(2x)& \sin(2x) \\ -2\sin(2x) & 2\cos(2x) \end{vmatrix} =2 \\ \Rightarrow y_p =-\cos(2x) \int{ \sin(2x)5 x e^{-x}\over 2}dx +\sin(2x) \int{ \cos(2x) 5xe^{-x}\over 2}dx \\ =-{5\over 2} \cos(2x)\left({1\over 25}e^{-x}((3-5x)\sin(2x) -2(5x+2)\cos(2x)) \right)\\ \qquad +{5\over 2}\sin(2x)\left( {1\over 25}e^{-x}(( 10x+4)\sin(2x)+ (3-5x)\cos(2x)) \right) \\={1\over 10}e^{-x}(10x+4) \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos(2x) +c_2\sin(2x)+{1\over 10}e^{-x}(10x+4)}

解答:(x^3+17){dy\over dx} =x^2 y \Rightarrow {1\over y}dy={x^2\over x^3+17}dx \Rightarrow \ln y={1\over 3}\ln(x^3+17)+c_1 \\ \Rightarrow \bbox[red, 2pt]{y=c_2(x^3+17)^{1/3}}

解答:f(x)為\text{Gaussian distribution} \Rightarrow \int_{-\infty}^\infty f(x)\,dx =1 \Rightarrow \int_{-\infty}^\infty e^{-x^2/2a}\,dx =1 \\\Rightarrow a=1/2\pi \Rightarrow f(x)=e^{-\pi x^2 }\\假設\mathcal F(f(x))= F(\omega),現在f(x)=e^{-x^2/2a } \Rightarrow f'(x)=-{x\over a}f(x) \\\Rightarrow \mathcal F(f'(x))=\mathcal F(-{x\over a}f(x)) \Rightarrow i\omega F(\omega)=-{1\over a} iF'(\omega) \Rightarrow F'(\omega)=-a\omega F(\omega)\\ \Rightarrow\bbox[red, 2pt]{ F(\omega)= e^{-a\omega^2/2}= e^{-\omega^2/4\pi}}\\ 註:f(x)為\text{Gaussian distribution},初始值F(0)=0
解答:\cases{x=r\cos\theta \\ y=r\sin \theta} \Rightarrow \int_{-4}^4 \int_0^{\sqrt{16-x^2}} exp(-(x^2+y^2))\,dydx = \int_0^\pi \int_0^4 r e^{-r^2}\,drd\theta \\ =\int_0^\pi \left. \left[ -{1\over 2} e^{-r^2}\right] \right|_0^4 d\theta=\int_0^\pi -{1\over 2}e^{-16}+{1\over 2}\,d\theta = \bbox[red, 2pt]{{\pi\over 2}\left( 1-e^{-16}\right)}
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解題僅供參考,其他歷年試題及詳解
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