113年臺北科大能源碩士班-工程數學詳解

國立臺北科技大學113學年度碩士班招生考試

系所組別: 能源與冷凍空調工程系碩士班甲、乙組
第一節 工程數學

解答:$$y'={y(x-1)^2 \over y+3} \Rightarrow {y+3\over y}dy= (x-1)^2dx \Rightarrow \int \left(1+{3\over y}\right)\,dy=\int (x-1)^2 \,dx \\ \Rightarrow y+3\ln y={1\over 3}(x-1)^3+c_1  \Rightarrow y^3e^y=c_2e^{(x-1)^3/3} \\ y(3)=-1 \Rightarrow -e^{-1}=c_2e^{8/3} \Rightarrow c_2=-e^{-11/3} \Rightarrow \bbox[red, 2pt]{y^3e^y =-e^{((x-1)^3-11)/3}}$$

解答:$$y''+3y'+2y=0 \Rightarrow \lambda^2+3\lambda+2=0 \Rightarrow(\lambda +2)(\lambda+1)=0 \Rightarrow \lambda=-1,-2\\ \Rightarrow y_h=c_1e^{-x} +c_2e^{-2x} \\ y_p=Ax^2+Bx+C \Rightarrow y_p'=2Ax+B \Rightarrow y_p''=2A \\ \Rightarrow y_p''+3y_p'+2y_p= 2Ax^2 +(6A+2B)x+ (2A+3B+2C)=-2x^2+3 \\ \Rightarrow \cases{2A=-2\\ 6A+2B=0\\ 2A+3B+2C=3} \Rightarrow \cases{A=-1\\ B=3\\ C=-2} \Rightarrow y_p=-x^2+3x-2 \Rightarrow y=y_h+y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1e^{-x} +c_2e^{-2x}-x^2+3x-2}$$

解答:$$y=\sum_{n=0}^\infty a_n x^n \Rightarrow y''= \sum_{n=0}^\infty n(n-1)a_n x^{n-2} = \sum_{n=0}^\infty (n+2)(n+1) a_{n+2}x^n\\ \Rightarrow x^2y = \sum_{n=0}^\infty  a_nx^{n+2} =\sum_{n=0}^\infty  a_{n-2}x^{n} \;(a_k=0,k\lt 0)\\ \Rightarrow y''+x^2y= \sum_{n=0}^\infty \left( (n+2)(n+1)a_{n+2}+ a_{n-2}\right)x^n= 0 \Rightarrow (n+2)(n+1)a_{n+2}+ a_{n-2}=0\\ \cases{n=-2 \Rightarrow 0\cdot a_0=0\\ n=-1 \Rightarrow 0\cdot a_1=0 \\ n=0 \Rightarrow 2a_2=0\\ n=1 \Rightarrow 6a_3=0} \Rightarrow \cases{a_0\; \text{arbitrary} \\a_1\; \text{arbitrary} \\a_2=0\\ a_3=0} \text{ and } a_{n}=-{a_{n-4} \over n(n-1)}, n\ge 4\\ \Rightarrow y=a_0+ a_1x-\sum_{n=4}^\infty {a_{n-4} \over n(n-1)}x^n,a_2=a_3=0 \\ \Rightarrow \bbox[red, 2pt]{y=a_0\left( 1-{1\over 12}x^4+{1 \over 672}x^3+ \cdots \right) +a_1\left( x-{1\over 20}x^5+{1\over 1440}x^9+ \cdots \right)}$$

解答:$$L\{y'\}-4 L\{y\}=L\{1\} \Rightarrow sY(s)-1-4Y(s)={1\over s} \Rightarrow Y(s)={1\over s(s-4)}+ {1 \over s-4} \\ \Rightarrow y(t)=L^{-1} \{Y(s)\} =L^{-1} \left\{{1\over s(s-4)}+ {1 \over s-4} \right\}  ={5\over 4}L^{-1} \left\{{1\over s-4} \right\} -{1\over 4} L^{-1} \left\{{1\over s} \right\} \\={5\over 4}e^{4t}-{1\over 4} \Rightarrow \bbox[red, 2pt]{y= {5\over 4}e^{4t}-{1\over 4}}$$

解答:$$ L^{-1} \left\{{4 \over s^2+4s+20} \right\}  = L^{-1} \left\{{4 \over (s+2)^2+4^2} \right\}  = \bbox[red, 2pt]{e^{-2t} \sin(4t) }$$

解答:$$f(t)=2t^2+ \int_0^t f(t-\tau) e^{-\tau}\,d \tau \Rightarrow L\{f(t)\}=L\{2t^2 \}+L\{f(t)\}L\{e^{-t}\}\\ \Rightarrow F(s)={4\over s^3}+F(s)\cdot {1\over s+1} \Rightarrow {s\over s+1}F(s)={4\over s^3} \Rightarrow F(s)={4(s+1) \over s^4} \\ \Rightarrow f(t)=L^{-1}\{F(s)\}= L^{-1}\left\{ {4(s+1) \over s^4} \right\} = L^{-1}\left\{ {4s \over s^4} +{4\over s^4}\right\} =4\cdot {t^2\over 2}+ 4\cdot {t^3\over 6} \\ \Rightarrow \bbox[red, 2pt]{f(t)=2t^2+{2\over 3}t^3}$$
 

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解題僅供參考,其他歷年試題及詳解