112年中正大學光機電整合碩士班-工程數學詳解

國立中正大學112學年度碩士班招生考試

科目:工程數學
系所組別: 機械工程學系光機電整合工程 

解答: $$y''+6y'+5y=0 \Rightarrow y_h=c_1e^{-5t} +c_2e^{-t}\\ \text{Use Fourier transform to solve }y_p :\\\mathcal{F}[y''] +6\mathcal{F}[y']+5 \mathcal{F}[y] =\mathcal{F}[\delta(t-3)] \Rightarrow (-\omega^2+6\omega i+5) \mathcal{F}[y] =e^{-i\omega 3}\\ \Rightarrow \mathcal {F}[y] ={1\over -\omega^2+6\omega i+5}e^{-3\omega i} =-{1\over (\omega-5i)(\omega-i)}e^{-3\omega i} =-{1\over 4i} \left( {1\over \omega-5i}-{1\over \omega -i} \right)e^{-3\omega i} \\=-{1\over 4} \left( {1\over 5+\omega i}-{1\over 1+\omega i} \right)e^{-3\omega i} \Rightarrow y_p(t) =\mathcal{F}^{-1} \left[-{1\over 4} \left( {1\over 5+\omega i}-{1\over 1+\omega i} \right)e^{-3\omega i} \right] \\ =-{1\over 4}\left(e^{-5(t-3)}-e^{-(t-3)} \right) u(t-3) \Rightarrow y(t)=y_h+y_p \\\Rightarrow \bbox[red, 2pt]{y(t)=c_1e^{-5t} +c_2e^{-t}-{1\over 4}\left(e^{-5(t-3)}-e^{-(t-3)} \right) u(t-3)}$$

解答: $$\textbf{(a)}\;y''+4y'=0 \Rightarrow \lambda^2+4\lambda=0 \Rightarrow \lambda=0,-4 \Rightarrow y_h=c_1+c_2e^{-4x} \\ \quad y_p= A\cos x+B\sin x +Cx+D \Rightarrow y_p'= -A\sin x+B\cos x+C \Rightarrow y_p''=-A\cos x-B\sin x \\ \quad \Rightarrow y_p''+4y_p'=(-A+4B)\cos x +(-4A-B)\sin x+4C= 8+34\cos(x)\\\quad \Rightarrow \cases{-A+4B=34\\ -4A-B=0 \\4C=8} \Rightarrow \cases{A=-2\\B=8 \\C=2} \Rightarrow y_p=-2\cos x+8\sin x+2x+D \\ \Rightarrow y=y_h+y_p =c_3+c_2e^{-4x}-2\cos x+8\sin x+2x \Rightarrow y'=-4c_2e^{-4x}+2\sin x+8\cos x+2 \\ \cases{y(0)=c_3+c_2-2=3\\ y'(0)=-4c_2+8+2=2} \Rightarrow \cases{c_2=2\\ c_3=3} \Rightarrow \bbox[red, 2pt]{y=3+2e^{-4x}-2\cos x+ 8\sin x+2x}\\ \textbf{(b)}\;2y^2+ye^{xy} +(4xy+xe^{xy}+ 2y)y'=0 \Rightarrow (2y^2+ye^{xy})dx +(4xy+xe^{xy}+ 2y) dy=0\\\quad \cases{P(x,y) =2y^2+ye^{xy} \\Q(x,y)= 4xy+xe^{xy} + 2y} \Rightarrow \cases{P_y=4y+e^{xy}+ xye^{xy} \\ Q_x=4y+e^{xy} +xye^{xy}} \Rightarrow P_y=Q_x \Rightarrow \text{exact} \\\quad \Phi(x,y) =\int (2y^2+ye^{xy} )\,dx = \int (4xy+xe^{xy}+ 2y)dy \\ \quad \Rightarrow \Phi(x,y)=2xy^2+ e^{xy} +\phi(y) =2xy^2+ e^{xy}+ y^2+ \rho(x ) \\ \quad\Rightarrow \bbox[red, 2pt]{2xy^2+e^{xy}+y^2+ C=0}$$

解答: $$A＝\begin{bmatrix}0 & 2 & -1 \\2 & 3 & -2 \\-1 & -2 & 0 \end{bmatrix} \Rightarrow \det(A-\lambda) =-(\lambda+1)^2(\lambda-5) =0 \Rightarrow \text{ eigenvalues: -1, 5,} \\ \text{ and corresponding eigenvectors: }v_1= \begin{bmatrix}-2 \\1 \\0\end{bmatrix}, v_2 = \begin{bmatrix}1 \\0 \\1\end{bmatrix}, v_3=\begin{bmatrix}-1 \\-2 \\1 \end{bmatrix} \\ \text{Applying Gram-Schmidt process, we have }e_1=\begin{bmatrix}-2/\sqrt 5 \\1/\sqrt 5 \\0\end{bmatrix}, e_2= \begin{bmatrix} 1/\sqrt{30} \\2/\sqrt{30} \\5/ \sqrt{30}\end{bmatrix}, e_3=\begin{bmatrix}-1/\sqrt 6 \\-2/\sqrt 6 \\1/ \sqrt 6\end{bmatrix}\\ \text{Let }P=[e_1 \mid e_2 \mid e_3] =\begin{bmatrix}-2/\sqrt 5 & 1/\sqrt{30} &-1/\sqrt 6  \\1/\sqrt 5 &2/\sqrt{30}  &-2/\sqrt 6  \\0 & 5/\sqrt{30} & 1/\sqrt 6 \end{bmatrix}, \text{ then }PP^T=I \text{ and }\\ \bbox[red, 2pt]{A=\begin{bmatrix}-2/\sqrt 5 & 1/\sqrt{30} &-1/\sqrt 6  \\1/\sqrt 5 &2/\sqrt{30}  &-2/\sqrt 6  \\0 & 5/\sqrt{30} & 1/\sqrt 6 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\0 & -1 & 0 \\0 &0  &5 \end{bmatrix} \begin{bmatrix}-2/\sqrt 5 & 1/\sqrt{5} &0\\1/\sqrt {30} &2/\sqrt{30}  &5/\sqrt {30}  \\-1/\sqrt 6 & -2/\sqrt{6} & 1/\sqrt 6 \end{bmatrix}} \\B=\begin{bmatrix}5 & -4 & -2 \\-4 & 5 & -2 \\-2 & -2 & 8 \end{bmatrix} \Rightarrow \det(B-\lambda I) =-\lambda(\lambda-9)^2 0\Rightarrow \text{ eigenvalues: }0,9, \text{and corresponding }\\ \text{eigenvectors: }v_1= \begin{bmatrix}2 \\2 \\1 \end{bmatrix}, v_2= \begin{bmatrix}-1 \\1 \\0 \end{bmatrix}, v_3 = \begin{bmatrix}-1 \\0 \\2 \end{bmatrix}. \text{Applying Gram-Schmit process, we have }\\ e_1=\begin{bmatrix}2/3 \\2/3 \\1/3 \end{bmatrix}, e_2= \begin{bmatrix}-1/\sqrt 2 \\1/\sqrt 2 \\0 \end{bmatrix}, e_3 = \begin{bmatrix}-\sqrt 2/6 \\-\sqrt 2/6 \\2\sqrt 2/3 \end{bmatrix} \Rightarrow P=[e_1\mid e_2\mid e_3] \\ \Rightarrow \bbox[red, 2pt] {B=\begin{bmatrix}2/3 &-1/\sqrt 2 &-\sqrt 2/6 \\2/3 & 1/\sqrt 2& -\sqrt 2/6\\ 1/3& 0 & 2\sqrt 2/3 \end{bmatrix} \begin{bmatrix}0 &0  &0  \\0 & 9 &0  \\  0& 0 & 9\end{bmatrix} \begin{bmatrix}2/3 &2/3 & 1/3 \\-1/\sqrt 2 & 1/\sqrt 2& 0\\ -\sqrt 2/6& -\sqrt 2/6 & 2\sqrt 2/3 \end{bmatrix}} \\C=\begin{bmatrix}2 & 2 & -2 \\2 & -1 & 4 \\-2 & 4 & -1 \end{bmatrix} \Rightarrow \det(C-\lambda I) =-(\lambda-3)^2(\lambda+6) =0 \Rightarrow \text{eigenvalues: }3,-6\\ \text{and the corresponding eigenvectors: }v_1 = \begin{bmatrix} 2 \\1 \\0 \end{bmatrix} , v_2= \begin{bmatrix}-2 \\0 \\1 \end{bmatrix} ,v_3 =\begin{bmatrix}1 \\-2 \\2 \end{bmatrix}.\\ \text{Applying Gram-Schmit process, we have }e_1=\begin{bmatrix}2/\sqrt 5 \\1/\sqrt 5 \\0 \end{bmatrix} ,e_2=\begin{bmatrix}-2\sqrt 5/15 \\4\sqrt 5/15 \\\sqrt 5/3 \end{bmatrix} ,e_3=\begin{bmatrix}1/3 \\-2/3 \\2/3 \end{bmatrix}. \\ \text{Let }P=[e_1\mid e_2\mid e_3] =\begin{bmatrix}2/\sqrt 5 & -2\sqrt 5/15 & 1/3 \\1/\sqrt 5 & 4\sqrt 5/15 & -2/3 \\0 & \sqrt 5/3  & 2/3\end{bmatrix}, \text{then }\\ \bbox[red, 2pt] {C= \begin{bmatrix}2/\sqrt 5 & -2\sqrt 5/15 & 1/3 \\1/\sqrt 5 & 4\sqrt 5/15 & -2/3 \\0 & \sqrt 5/3  & 2/3\end{bmatrix} \begin{bmatrix}3 & 0 &0 \\0 &3  & 0 \\0 &0  &-6\end{bmatrix} \begin{bmatrix}2/\sqrt 5 & 1/\sqrt 5  & 0 \\-2/\sqrt 5/15 & 4\sqrt 5/15 & \sqrt 5/3 \\1/3 & -2/3  & 2/3\end{bmatrix} }$$

解答: $$\textbf{(a)}\; L\{y''\} +4L\{y\} =s^2Y(s)-sy(0)-y'(0)+4Y(s) =(s^2+4)Y(s)+8=0 \Rightarrow Y(s)=-{8\over s^2+4 } \\\qquad \Rightarrow y(t) =L^{-1}\{Y(s)\} =L^{-1}\left\{-{4\cdot 2\over s^2+2^2} \right\} =-4\sin(2t) \Rightarrow \bbox[red, 2pt]{y(t)=-4\sin(2t)} \\\textbf{(b)}\; L\{y''\} +2L\{y'\} -L\{y\}= s^2Y(s)-6+2sY(s)-Y(s)=0 \Rightarrow Y(s)={6\over s^2+2s-1} \\ \qquad \Rightarrow y(t) =L^{-1}\{Y(s)\} =L^{-1} \left\{{6\over (s+1)^2-2}\right\} = 3\sqrt 2e^{-t} \sinh(\sqrt 2 t) \\\qquad \Rightarrow \bbox[red, 2pt]{y(t)= 3\sqrt 2e^{-t} \sinh(\sqrt 2 t)}$$

解答: $$y''+3y'+2y=0 \Rightarrow \lambda^2+3\lambda +2=0 \Rightarrow (\lambda+2) (\lambda+1)=0 \Rightarrow \lambda=-1,-2\\ \Rightarrow y_h=c_1e^{-x} +c_2e^{-2x} \\ y_p=A\cos t+B\sin t \Rightarrow y_p'=-A\sin t+B\cos t \Rightarrow y_p''=-A\cos t-B\sin t\\ \Rightarrow y_p''+3y_p'+ 2y_p = (A+3B) \cos t+(-3A+B) \sin t = 40\sin t \Rightarrow \cases{A+3B=0\\ -3A+B=40} \\\Rightarrow \cases{A=-12\\ B=4} \Rightarrow y_p= -12\cos t+4 \sin t \Rightarrow y=y_h +y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^{-x} +c_2e^{-2x} -12\cos t+4 \sin t }$$

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