112年中央大學土木工程碩士班-工程數學詳解

 國立中央大學112學年度碩士班考試入學試題

所別: 土木工程研究所
科目:工程數學

解答:$$y(x)=u(x)e^{2x} \Rightarrow y'=u'e^{2x} +2ue^{2x} \Rightarrow y''=u''e^{2x}+ 4u'e^{2x}  + 4ue^{2x} \\ \Rightarrow (x+2)(u''e^{2x}+ 4u'e^{2x}  + 4ue^{2x})-(2x+5)(u'e^{2x} +2ue^{2x})+2ue^{2x}=(x+1)e^x \\ \Rightarrow (x+2)e^{2x}u'' + (2x+3)e^{2x}u'  =(x+1)e^x \Rightarrow u''+{2x+3\over x+2}u'={x+1\over x+2}e^{-x} \\ \Rightarrow \text{integration factor }I(x)=e^{\int {2x+3 \over x+2}dx} =e^{2x-\ln(x+2)}\\ \Rightarrow e^{2x-\ln(x+2)}u''+{2x+3\over x+2} e^{2x-\ln(x+2)}u'={x+1\over x+2}e^{x-\ln(x+2)} \\ \Rightarrow \left(e^{2x-\ln(x+2)}u' \right)' ={x+1\over x+2}e^{x-\ln(x+2)} \Rightarrow e^{2x-\ln(x+2)}u'=e^{x-\ln(x+2)} +c_1 \\ \Rightarrow u' = e^{-x}+c_1 e^{ \ln(x+2)-2x}  \Rightarrow u=-e^{-x}+ c_1(-{1\over 4})(2x+5)e^{-2x}+ c_2 \\ \Rightarrow y=ue^{2x}= -e^x +c_3(2x+5) +c_2e^{2x} \Rightarrow y'=-e^x+2c_3+ 2c_2e^{2x} \\ \Rightarrow \cases{y(0)= -1+5c_3 +c_2=1 \\y'(0)= -1+2c_3+2c_2=1} \Rightarrow \cases{c_2=3/4\\ c_3=1/4} \\ \Rightarrow \bbox[red, 2pt]{y={3\over 4}e^{2x}-e^x +{3\over 2}x+{5\over 4}}$$

解答:$$P_n(x)={1\over 2^n n!}\cdot \frac{\text{d}^n}{\text{d}x^n} (x^2-1)^n \Rightarrow P_5(x)={1\over 2^5\cdot 5!}\cdot \frac{\text{d}^5}{\text{d}x^5} (x^2-1)^5 \\={1\over 2^5\cdot 5!}\cdot \frac{\text{d}^5}{\text{d}x^5} (x^{10}-5x^8+10x^6-10x^4+5x^2-1) \\={1\over 2^5\cdot 5!}\cdot \frac{\text{d}^4}{\text{d}x^4} (10x^9-40x^7+60 x^5-40x^3 +10x) \\{1\over 2^5\cdot 5!}\cdot \frac{\text{d}^3}{\text{d}x^3} (90x^8-280x^6+300 x^4-120x^2 +10) \\={1\over 2^5\cdot 5!}\cdot \frac{\text{d}^2}{\text{d}x^2} (720x^7-1680x^5+1200 x^3-240x) \\= {1\over 2^5\cdot 5!}\cdot \frac{\text{d} }{\text{d}x} (5040x^6-8400x^4+3600 x^2-240) \\={1\over 2^5\cdot 5!} (30240x^5-33600x^3+7200 x ) =\bbox[red, 2pt]{{63\over 8}x^5-{35\over 4}x^3+ {15\over 8}x}$$

解答:$$my''+cy'+ ky=f(t) \Rightarrow y''+4y'+4y=e^{-2t} \Rightarrow L\{y''\}+ 4L\{y'\}+ 4L\{ y\} =L\{e^{-2t}\} \\ \Rightarrow s^2Y(s)+4sY(s)+ 4Y(s)={1\over s+2} \Rightarrow Y(s)={1\over (s+2)^3} \\ \Rightarrow y(t)= L^{-1}\{ Y(s)\} =L^{-1} \left\{ {1\over (s+2)^3}\right\} =e^{-2t} L^{-1}\left\{ {1\over s^3} \right\} ={1\over 2}t^2e^{-2t} \Rightarrow\bbox[red, 2pt]{ y(t)={1\over 2}t^2e^{-2t}}$$

解答:$$A=\begin{bmatrix}4 & 2&-2 \\2 & 5& 0 \\ -2& 0 & 3 \end{bmatrix} \Rightarrow A^2= \left[ \begin{matrix}24 & 18 & -14 \\18 & 29 & -4 \\-14 & -4 & 13 \end{matrix} \right] \Rightarrow A^4= \left[ \begin{matrix} 1096 & 1010 & -590 \\1010 & 1181 & -420 \\-590 & -420 & 381 \end{matrix} \right] \\ \Rightarrow A^5= A^4A= \bbox[red, 2pt]{ \left[ \begin{matrix} 7584 & 7242 & -3962 \\7242 & 7925 & -3280 \\-3962 & -3280 & 2323 \end{matrix} \right]}$$

解答:$$\bbox[red, 2pt]{\text{Green's theorem in the plane is a special case of Stokes' theorem.} }$$

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解題僅供參考,其他歷年試題及詳解