112年台北科大冷凍空調碩士班-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 :1410、 1420
能源與冷凍空調工程系碩士班甲 、乙組
第一節 工程數學 試題

解答：$$xy'={y^2\over x}+y \Rightarrow y'-{1\over x}y={y^2\over x^2}\;為n=2的白努利方程式\\因此取v=y^{1-2}={1\over y} \Rightarrow v'=-{y'\over y^2} \Rightarrow y'=-y^2v'代回原式 \Rightarrow -y^2v'-{1\over x}y={y^2\over x^2}\\ \Rightarrow v'+{1\over x}v=-{1\over x^2} \Rightarrow xv'+v=-{1\over x} \Rightarrow (xv)'=-{1\over x} \\ \Rightarrow xv=-\ln x+C \Rightarrow {x\over y}=-\ln x+C \Rightarrow \bbox[red, 2pt]{y={x\over -\ln x+C}}$$

解答：$$令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2}\\ \Rightarrow x^2y''-5xy'+10y=0 \Rightarrow m(m-1)x^m-5mx^m+10x^m=0 \\ \Rightarrow (m^2-6m+10)x^m=10 \Rightarrow m^2-6m+10=0 \Rightarrow m=3\pm i \\ \Rightarrow y=c_1x^{(3+i)}+c_2x^{(3-i)} =x^3(A\cos(\ln x)+B\sin (\ln x)) \\ \Rightarrow y'=3x^2(A\cos(\ln x)+B\sin(\ln x))+x^3(-{1\over x}A\sin(\ln x) +{1\over x}B\cos(\ln x)) \\ =3x^2(A\cos(\ln x)+B\sin(\ln x))+x^2(B\cos(\ln x)-A\sin(\ln x)) \\ \Rightarrow \cases{y(1)=A=4\\ y'(1)=3A+B=-6} \Rightarrow \cases{A=4\\ B=-18} \Rightarrow \bbox[red,2pt]{y=x^3(4\cos(\ln x)-18 \sin(\ln x))}$$

解答：$$先求齊次解: y''+2y'-3=0 \Rightarrow \lambda^2+2\lambda -3=0 \Rightarrow (\lambda-1)(\lambda+ 3)=0 \Rightarrow \lambda=1,-3\\ \Rightarrow y_h=c_1e^x+ c_2e^{-3x}\\ 假設y_p=ax^2+bx+c+de^{2x} \Rightarrow y_p'=2ax+b+2de^{2x} \Rightarrow y_p''=2a+4de^{2x} \\ \Rightarrow y_p''+2y_p'-3y_p=-3ax^2+(4a-3b)x +2a+2b-3c+5de^{2x} =4x^2-x+11e^{2x} \\ \Rightarrow \cases{-3a=4\\ 4a-3b=-1\\ 2a+2b-3c=0\\ 5d=11} \Rightarrow \cases{a=-4/3\\ b=-13/9\\ c=-50/27\\ d=11/5} \Rightarrow y_p=-{4\over 3}x^2-{13\over 9}x-{50\over 27}+{11\over 5}e^{2x} \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red,2pt]{y=c_1e^x+ c_2e^{-3x}-{4\over 3}x^2-{13\over 9}x-{50\over 27}+{11\over 5}e^{2x}}$$

解答：$$L\{y'' \}+4L\{y' \}+3L\{y \}=L\{ e^t\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+3Y(s)={1\over s-1}\\\Rightarrow s^2Y(s)-2+4sY(s)+3Y(s)={1\over s-1} \Rightarrow (s^2+4s+3)Y(s)={1\over s-1}+2\\ \Rightarrow Y(s)={1\over (s-1)(s+1)(s+3)} +{2\over (s+1)(s+3)}\\={1\over 8(s-1)}-{1\over 4(s+1)}+{1\over 8(s+3)}+{1\over  (s+1)}-{1\over  (s+3)} \\={1\over 8(s-1)}+{3\over 4(s+1)}-{7\over 8(s+3)} \Rightarrow \bbox[red, 2pt]{y(t)={1\over 8}e^t+{3\over 4}e^{-t}-{7\over 8}e^{-3t}}$$

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：$$f(t)=\begin{cases} 0,& t\lt 3\\ t,& t\ge 3\end{cases} \Rightarrow L\{f(t)\} =\int_0^\infty f(t)e^{-st}\,dt =\int_3^\infty te^{-st}\,dt =\left.\left[-{t\over s}e^{-st}-{1\over s^2}e^{-st} \right] \right|_3^\infty \\={3\over s}e^{-3s}-{1\over s^2}e^{-3s} \Rightarrow L\{y''\}+ 4L\{y\}=  s^2Y(s)-sy(0)-y'(0)+4Y(s)=L\{f(t)\} \\ \Rightarrow (s^2+4)Y(s)= e^{-3s}({3\over s}-{1\over s^2}) \Rightarrow {1\over s^2+4}e^{-3s}\cdot {3s-1\over s^2}\\\Rightarrow y(t)=L^{-1}\{e^{-3s}\cdot {3s-1\over s^2(s^2+4)}\} =H(t-3)L^{-1}\{{3s-1\over s^2(s^2+4)} \}(t-3) \\=H(t-3)\left( -{3\over 4}\cos(2(t-3))+{1\over 8}\sin(2(t-3)) +{3\over 4}H(t-3)-{t-3\over 4}\right) \\ \Rightarrow \bbox[red,2pt]{y(t)=\begin{cases}0,& t\lt 3\\ -{3\over 4}\cos(2(t-3))+{1\over 8}\sin(2(t-3))+{3\over 4}-{t-3\over 4}& t\ge 3\end{cases}}$$

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