114年中央環工碩士班-工程數學詳解

 國立中央大學114學年度碩士班考試入學

系所: 環境工程研究所碩士班 甲組(一般生)
環境工程研究所碩士班 乙組(一般生)
科目: 工程數學

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow x^2y''+2xy'-6y =m(m-1)x^m+ 2mx^m-6x^m= (m^2+m-6)x^m=0 \\ \Rightarrow m^2+m-6=0 \Rightarrow (m+3)(m-2)=0 \Rightarrow m=2,-3 \Rightarrow y=c_1 x^2+c_2x^{-3} \\ \Rightarrow y'=2c_1 x-3c_2x^{-4} \Rightarrow \cases{y(1)= c_1+c_2=0.5\\ y'(1)=2c_1-3c_2=1.5} \Rightarrow \cases{c_1= 0.6\\ c_2=-0.1} \Rightarrow \bbox[red, 2pt]{y=0.6x^2-0.1x^{-3}}$$

解答:$$L\{y''\}+9 L\{y\}=10 L\{e^{-t}\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+9Y(s)={10\over s+1} \\ \Rightarrow (s^2+9)Y(s)={10\over s+1} \Rightarrow Y(s)={10\over (s^2+9) (s+1)} ={1-s\over s^2+9}+{1\over s+1} \\ \Rightarrow y(t)= L^{-1}\{Y(s)\} =L^{-1}\left\{{1-s\over s^2+9} \right\} +L^{-1}\left\{{1\over s+1} \right\} ={1\over 3}\sin(3t)-\cos(3t)+e^{-t} \\ \Rightarrow \bbox[red, 2pt]{y(t) ={1\over 3}\sin(3t)-\cos(3t)+e^{-t}}$$

解答:$$\vec V=2x\vec i-8y\vec j+4z\vec k \Rightarrow \text{div }\vec V={\partial \over \partial x}(2x)+ {\partial \over \partial y}(-8y)+ {\partial \over \partial z}(4z) =2-8+4= \bbox[red, 2pt]{-2}$$

解答:$$\vec r(t)= 4t\vec i+4t^2\vec j+4t^3\vec k \Rightarrow \vec r'(t)=4\vec i+ 8t\vec j+12t^2\vec k \Rightarrow \vec r''(t)=8\vec j+24t\vec k \\ \Rightarrow r'(t)\times \vec r''(t)=96t^2 \vec i-96t\vec j+32\vec k \Rightarrow \kappa ={||r'(t)\times r''(t)|| \over ||r'(t)||^3} ={||96t^2 \vec i-96t\vec j+32\vec k|| \over ||4\vec i+ 8t\vec j+12t^2\vec k||^3} \\= {\sqrt{9216t^4+9216t^2+1024} \over (16+64t^2+144t^4)^{3/2}} ={32\sqrt{9t^4+9t^2+1} \over 64(9t^4+4t^2+1)^{3/2}} = \bbox[red, 2pt]{\sqrt{9t^4+9t^2+1} \over  (9t^4+4t^2+1)^{3/2}}$$

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解題僅供參考,其他碩士班試題及詳解