114年台北科大製造科技碩士班-微分方程詳解

國立臺北科技大學114學年度碩士班招生考試

系所組別:1201 製造科技研究所
第一節 微分方程 試題

解答: $$y'={2x-e^x \sin y\over e^x \cos y+1} \Rightarrow (2x-e^x \sin y)dx-(e^x \cos y+1)dy =0 \Rightarrow \cases{P(x,y)=2x-e^x \sin y\\ Q(x,y) =-e^x \cos y-1} \\ \Rightarrow \cases{P_y=-e^x\cos y\\ Q_x=-e^x \cos y} \Rightarrow P_y=Q_x \Rightarrow \text{ Exact} \Rightarrow \Phi(x,y)= \int (2x-e^x \sin y)\,dx = \int (-e^x \cos y-1)\,dy \\ \Rightarrow \Phi(x,y)= x^2-e^x \sin y + \phi(y) =-e^x \sin y-y+ \rho(x)  \Rightarrow \bbox[red, 2pt]{x^2-e^x \sin y-y+c_1=0}$$

解答: $$y_p= A+Bx+Ce^{-2x} \Rightarrow y'=B-2Ce^{{-2x}} \Rightarrow y''=4Ce^{-2x} \\ \Rightarrow y_p''+4y_p= 4A+ 4Bx+8Ce^{-2x} =x+2e^{-2x} \Rightarrow \cases{A=0\\ B=1/4\\ C=1/4} \\\Rightarrow \text{ a particular solution: }\bbox[red, 2pt]{y= {1\over 4}x+{1\over 4}e^{-2x}}$$

解答: $$y''+2y'+y=0 \Rightarrow \lambda^2+2\lambda +1=0 \Rightarrow (\lambda+1)^2=0 \Rightarrow \lambda=-1 \Rightarrow y_h= c_1e^{-x} +c_2xe^{-x}\\ y_p=A\cos x+B\sin x \Rightarrow y_p'=-A\sin x+B\cos x \Rightarrow y_p''=-A\cos x-B\sin x \\ \Rightarrow y_p''+2y_p'+y_p = 2B\cos x-2A\sin x= 2\cos x \Rightarrow \cases{A=0 \\ B=1} \Rightarrow y_p=\sin x \\ \Rightarrow y=y_h+ y_p \Rightarrow y=c_1e^{-x} +c_2xe^{-x} +\sin x \Rightarrow y'=(c_2-c_1)e^{-x} -c_2xe^{-x}+ \cos x \\ \Rightarrow \cases{y(0)= c_1= 0\\ y'(0)=c_2-c_1+1=0 }\Rightarrow \cases{c_1=0\\ c_2=-1} \Rightarrow \bbox[red, 2pt]{y=-xe^{-x}+ \sin x}$$

解答: $$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow x^2y''-5xy'+10y=(m^2-6m+10)x^m =0 \\ \Rightarrow m^2-6m+10=0 \Rightarrow m=3\pm i \Rightarrow y=x^3(c_1 \cos(\ln x)+ c_2 \sin(\ln x)) \\ \Rightarrow y'=3x^2(c_1 \cos(\ln x)+ c_2 \sin(\ln x)) + x^2(-c_1\sin(\ln x)+ c_2\cos(\ln x)) \\ \Rightarrow \cases{y(1) = c_1=4\\ y'(1)= 3c_1+c_2=-6} \Rightarrow \cases{c_1=4\\ c_2=-18} \Rightarrow \bbox[red, 2pt]{y=x^3(4\cos(\ln x)-18 \sin (\ln x))}$$

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

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解題僅供參考，其他碩士班試題及詳解