108年台綜大轉學考-工程數學D09詳解

臺灣綜合大學系統108學年度學士班轉學生聯合招生考試

科目名稱：工程數學
類組代碼：D09

解答：$$\mathbf{(a)}\;常微分/二階/二次/非線性\\ \mathbf{(b)}\; 常微分/一階/一次/非線性\\\mathbf{(c)}\; 偏微分/二階/三次/非線性 \\\mathbf{(d)}\; 常微分/二階/一次/線性 \\\mathbf{(e)}\;常微分/一階/一次/線性$$

解答：$$\mathbf{(a)}\;y'=-{\sin x \cos y\over \cos x\sin y} \Rightarrow -{\sin y\over \cos y}dy = {\sin x\over \cos x}\,dx \Rightarrow -\int \tan y\, dy = \int \tan x\,dx \\\Rightarrow \ln |\cos y| =\ln |\sec x|+C_1 \Rightarrow |\cos y| = C_2|\sec x| \Rightarrow  \bbox[red, 2pt]{y =\pm \cos^{-1} (C_2 \sec x)} \\\mathbf{(b)}\; xy'+2y=\cos x \Rightarrow y'+{2\over x}y ={1\over x}\cos x \Rightarrow 積分因子I(x)=e^{\int (2/x)dx} =x^2 \\ \Rightarrow I(x)y'+I(x){2\over x}y = I(x){1\over x}\cos x \Rightarrow x^2y' +2xy =x\cos x \Rightarrow (x^2y)'=x\cos x\\ \Rightarrow x^2y = \int x\cos x\,dx = x\sin x+\cos x+C \Rightarrow y={1\over x}\sin x+{1\over x^2} \cos x+{C\over x^2}\\ 最後將y(\pi/2)=0 代入\Rightarrow 0= {2\over \pi }\sin {\pi \over 2}+ {4\over \pi^2} \cos {\pi \over 2} +{4\over \pi^2}C ={2\over \pi}+{4\over \pi^2}C \Rightarrow C=-{\pi\over 2}\\ \Rightarrow \bbox[red, 2pt]{ y= {1\over x}\sin x+{1\over x^2} \cos x-{\pi \over 2x^2}} $$

解答：$$\mathbf{(a)}\;e^{2x}\cos(2x)和e^{2x}\sin(2x)為齊次解\Rightarrow 特徵方程式的解為\lambda=2\pm 2i \Rightarrow \lambda^2-4\lambda +8=0\\ \Rightarrow 二階ODE:\bbox[red, 2pt]{y''-4y'+8y=0}; \\又假設對所有的x，滿足ae^{2x}\cos(2x)+be^{2x}\sin(2x)=0，則\cases{x=0 \Rightarrow a=0\\ x=\pi/2 \Rightarrow b=0}\\ 因此e^{2x}\cos(2x)和e^{2x}\sin(2x)為\bbox[red,2pt]{線性獨立} \\\mathbf{(b)}\;y= C_1e^{2x}\cos(2x) + C_2e^{2x}\sin(2x) \Rightarrow y'= 2e^{2x} ((C_1+C_2)\cos (2x)+ (C_2-C_1)\sin (2x))\\將\cases{y(0)=1/2\\ y'(0)=-1/2} 代入，可得\cases{C_1=1/2\\ 2(C_1+C_2) = -1/2} \Rightarrow C_2=-3/4\\ \Rightarrow \bbox[red, 2pt]{ y=e^{2x}({1\over 2}\cos(2x) -{3\over 4}\sin(2x))}$$

解答：$$先求齊次解，即y''+5y'-6y=0 \Rightarrow \lambda^2+5 \lambda-6=0 \Rightarrow (\lambda+6)(\lambda-1)=0\\ \Rightarrow \lambda = 1,-6 \Rightarrow y_h= C_1e^x +C_2e^{-6x}\\ 接著令y_p= Axe^x \Rightarrow y_p'=Axe^x +Ae^x \Rightarrow y_p''=Axe^x +2Ae^x \Rightarrow y_p''+5y_p'-6y_p = 7Ae^x=4e^x\\ \Rightarrow A={4\over 7} \Rightarrow y_p={4\over 7}xe^x \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red,2pt]{y= C_1e^x +C_2e^{-6x} +{4\over 7}xe^x}$$

解答：$$3y''+6y'+3y = 3t \Rightarrow \mathcal L\{3y''+6y'+3y \} =L\{3t\}\\ \Rightarrow 3(s^2Y(s)- sy(0)-y'(0)) +6(sY(s)-y(0))+3 Y(s)={3\over s^2} \\ \Rightarrow 3s^2Y(s)+ 3s-3 +6sY(s)+6+3Y(s)={3\over s^2} \Rightarrow (3s^2+6s+3)Y(s)={3\over s^2}-3-3s \\ \Rightarrow Y(s)={1\over s^2(s+1)^2}  -{1\over s+1 } ={-2\over s}+{1\over s^2} +{1\over s+1} +{1\over (s+1)^2} \\ \Rightarrow y(t)=\mathcal L^{-1}\{Y(s)\}= \mathcal L^{-1}\{{-2\over s}+{1\over s^2} +{1\over s+1} +{1\over (s+1)^2} \} \\ \Rightarrow \bbox[red,2pt]{y=-2+t+ e^{-t}+ te^{-t}}$$
 

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