114年高科大電子碩士班-微分方程詳解

 國立高雄科技大學114學年度碩士班招生考試

系所別:電子工程系碩士班(建工校區)
組 別:電信與系統組
考科代碼:2011
考科:微分方程

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

解答: $$\cases{P(x,y)=4y\\ Q(x,y)=3x} \Rightarrow u'={P_y-Q_x\over Q}u={1\over 3x}u \Rightarrow \text{integrating factor }u=(3x)^{1/3} \\ \Rightarrow \cases{uP=4y(3x)^{1/3}\\ uQ=(3x)^{4/3}} \Rightarrow (uP)_y=4(3x)^{1/3} =(uQ)_x \Rightarrow \int 4y(3x)^{1/3}\,dx =\int (3x)^{4/3}dy\\ \Rightarrow y(3x)^{4/3}=c_1 \Rightarrow \bbox[red, 2pt]{y=c_1(3x)^{-4/3}}$$

解答: $$v={1\over y} \Rightarrow v'=-{1\over y^2}y' \Rightarrow y'=-y^2v' \Rightarrow -y^2v'+y=y^2 \Rightarrow -v'+ v=1 \\ \Rightarrow v'-v=-1 \Rightarrow e^{-x}v'-e^{-x}v=-e^{-x} \Rightarrow (e^{-x}v)'=-e^{-x }\Rightarrow e^{-x}v=e^{-x}+c_1 \\ \Rightarrow v={1\over y}=1+c_1e^x \Rightarrow \bbox[red ,2pt]{y={1\over 1+c_1e^x}}$$

解答: $$y'=3x^2+2x \Rightarrow y=x^3+x^2+c_1 \Rightarrow 14=8+4+c_1 \Rightarrow c_1=2 \\ \Rightarrow \bbox[red, 2pt]{y=c^3+x^2+2}$$

解答: $$y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda+2=0 \Rightarrow (\lambda-2)(\lambda-1)=0 \Rightarrow \lambda=1,2 \\ \Rightarrow \bbox[red, 2pt]{y=c_1e^x +c_2e^{2x}}$$

解答: $$y''-6y'+9y=0 \Rightarrow \lambda^2-6\lambda+9=0 \Rightarrow (\lambda-3)^2 =0 \Rightarrow \lambda=3 \Rightarrow y_h=c_1e^{3x}+ c_2xe^{3x}\\ y_p=Ax^2e^{3x} \Rightarrow y_p'=2Axe^{3x}+3Ax^2e^{3x} \Rightarrow y_p'' =2Ae^{3x}+ 12Axe^{3x}  + 9Ax^2e^{3x} \\ \Rightarrow y_p''-6y_p'+9y_p= 2Ae^{3x}=e^{3x} \Rightarrow A={1\over 2} \Rightarrow y_p={1\over 2}x^2e^{3x} \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{3x}+ c_2xe^{3x} +{1\over 2}x^2e^{3x}}$$

解答: $$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow y'''=m(m-1)(m-2)x^{m-3} \\ \Rightarrow x^3y'''-x^2y''+xy'=[m(m-1)(m-2)-m(m-1)+ m]x^m=0\\ \Rightarrow m(m-2)^2=0 \Rightarrow m=0,2 \Rightarrow y_h= c_1+ c_2x^2 +c_3x^2 \ln x \\ y_p=Ax^{-2} \Rightarrow y_p'=-2Ax^{-3} \Rightarrow y_p''=6Ax^{-4} \Rightarrow y_p'''=-24Ax^{-5} \\ \Rightarrow x^3y_p'''-x^2y_p''+xy_p'=-32Ax^{-2} =x^{-2} \Rightarrow A=-{1\over 32} \Rightarrow y_p=-{1\over 32}x^{-2} \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1+ c_2x^2 +c_3x^2 \ln x-{1\over 32}x^{-2}}$$

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解題僅供參考，其他碩士班試題及詳解