111年高科大電子工程碩士班-微分方程詳解

 

國立高雄科技大學111學年度碩士班招生考試

系所別: 電子工程(建工校區), 組別:乙組
考科:微分方程, 考科代碼: 3011

解答: $$\textbf{1.}\;{dy\over dx}=2x \Rightarrow \int 1\,dy = \int 2x\,dx \Rightarrow y=x^2+c_1\\ y(1)=0 \Rightarrow 0=1+c_1 \Rightarrow c_1=-1 \Rightarrow \bbox[red, 2pt]{y=x^2-1} \\\textbf{2.}\; y'-2y=1 \Rightarrow {1\over 2y+1}dy = 1\,dx \Rightarrow {1\over 2} \ln(2y+1)=x+c_1 \Rightarrow 2y+1=e^{2(x+c_1)} \\ \Rightarrow y={1\over 2}(c_2e^{2x}-1) \Rightarrow \bbox[red, 2pt]{y=c_3e^{2x}-{1\over 2}} \\\textbf{3.}\; 令\cases{P(x,y)=2xy+1\\ Q(x,y)=x^2-2} \Rightarrow P_y=2x= Q_x \Rightarrow \text{Exact} \\ \qquad \Rightarrow \Phi(x,y)=\int P\,dx =\int Q\,dy  \Rightarrow \Phi(x,y)=\int (2xy+1)\,dx = \int (x^2-2)\,dy\\\qquad  \Rightarrow x^2y+x+\rho(y)=x^2y-2y+ \phi(x) \Rightarrow \Phi(x,y)= \bbox[red, 2pt]{x^2y+x-2y+c_1=0} \\\textbf{4.}\; y''-6y'+9y=0 \Rightarrow y_h=c_1e^{3x} +c_2xe^{3x}\\\qquad y_p=Ax^2e^{3x} \Rightarrow y_p'=2Axe^{3x}+ 3Ax^2e^{3x} \Rightarrow y_p''= 2Ae^{3x}+12Axe^{3x} + 9Ax^2e^{3x} \\ \qquad \Rightarrow y_p''-6y_p'+9y_p =2Ae^{3x}=e^{3x} \Rightarrow A={1\over 2} \Rightarrow y_p={1\over 2}x^2e^{3x} \\\qquad \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y =c_1e^{3x} +c_2xe^{3x}+ {1\over 2}x^2e^{3x}} \\\textbf{5.}\; y''-3y'+2y=0 \Rightarrow y_h=c_1e^{x} +c_2e^{2x}\\\qquad y_p=Ae^{3x} \Rightarrow y_p'=3Ae^{3x}  \Rightarrow y_p''= 9Ae^{3x}  \Rightarrow y_p''-3y_p'+2y_p =2Ae^{3x}=e^{3x} \Rightarrow A={1\over 2} \\\qquad \Rightarrow y_p={1\over 2} e^{3x}  \Rightarrow y=y_h+y_p \Rightarrow y =c_1e^{x} +c_2e^{2x}+ {1\over 2}e^{3x} \\ \qquad \Rightarrow y'=c_1e^x+ 2c_2e^{2x} +{3\over 2}e^{3x} \Rightarrow \cases{y(0)= c_1+c_2 +{1\over 2}=0\\ y'(0)=c_1+ 2c_2 +{3\over 2}= 0} \Rightarrow \cases{c_1=1/2\\ c_2=-1} \\ \qquad \Rightarrow \bbox[red, 2pt]{y={1\over 2}e^x -e^{2x}+{1\over 2}e^{3x}}$$

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