中央土木碩士班-工程數學詳解

 國立中央大學114學年度碩士班考試入學

系所: 土木工程學系碩士班 力學與結構工程組(一般生)
科目: 工程數學

解答:$$\textbf{(1)} my''+cy'+ky=r(t) \Rightarrow y''+3y'+2y=4 \cos t \\ \quad \Rightarrow \lambda^2+3\lambda+2=0 \Rightarrow (\lambda+2) (\lambda+1)=0 \Rightarrow \lambda=-2,-1 \Rightarrow y_h= c_1e^{-2t}+c_2 e^{-t} \\ \quad y_p=A\cos t+ B\sin t \Rightarrow y_p'=-A\sin t+B\cos t \Rightarrow y_p''=-A\cos t-B\sin t \\\quad \Rightarrow y_p''+3y_p' +2y_p = (A+3B)\cos t+(-3A+B)\sin t=4\cos t \Rightarrow \cases{A +3B=4\\ B= 3A} \Rightarrow \cases{A=2/5\\ B=6/5}\\\Rightarrow y_p={2\over 5} \cos t+{6\over 5}\sin t  \Rightarrow y=y_h+ y_p \Rightarrow y=c_1 e^{-2t} +c_2 e^{-t}+ {2\over 5} \cos t+{6\over 5}\sin t \\ \Rightarrow y'=-2c_1e^{-2t} -c_2e^{-t}-{2\over 5}\sin t+{6\over 5}\cos t \Rightarrow \cases{y(0)=c_1+c_2+2/5=0\\ y'(0)=-2c_1-c_2+6/5=0} \\ \Rightarrow \cases{c_1= 8/5 \\ c_2=-2} \Rightarrow  \bbox[red, 2pt]{y={8\over 5}e^{-2t}-2 e^{-t}+ {2\over 5} \cos t+{6\over 5}\sin t} \\\textbf{(2)} y_p=Ate^{-t} \Rightarrow y_p'=Ae^{-t}-Ate^{-t} \Rightarrow y_p''=-2Ae^{-t}+Ate^{-t} \\\quad \Rightarrow y_p''+3y_p'+2y_p =Ae^{-t} =e^{-t} \Rightarrow A=1 \Rightarrow y_p=te^{-t} \Rightarrow y=y_h+ y_p \\\quad \Rightarrow y=c_1e^{-2t}+c_2 e^{-t}+ te^{-t} \Rightarrow y'=-2c_1e^{-2t}-c_2e^{-t}+e^{-t}-te^{-t} \\\quad \Rightarrow \cases{y(0)=c_1+c_2=0\\ y'(0)=-2c_1-c_2+1=0} \Rightarrow \cases{c_1=1\\ c_2=-1} \Rightarrow \bbox[red, 2pt]{y= e^{-2t}- e^{-t}+ te^{-t}}$$

解答:$$A= \begin{bmatrix} 1 & -2 \\2 & -3\end{bmatrix} \Rightarrow \det(A-\lambda I )= (\lambda+1)^2=0 \Rightarrow \lambda=-1 \\ \Rightarrow (A-\lambda I)v=0 \Rightarrow \begin{bmatrix} 2 & -2 \\2 & -2\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow x_1=x_2 \Rightarrow v=x_2 \begin{pmatrix} 1\\ 1\end{pmatrix}, \text{ choose }v=\begin{pmatrix} 1\\ 1\end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvalue: -1, eigenvector: } \begin{pmatrix} 1\\ 1\end{pmatrix} } \\ \text{Because we only have one eigenvector for a repeated eigenvalue, we need to find a generalized}\\\text{ eigenvector }u \text{ by solving }(A−λI)u=v \text{ i.e.,}\Rightarrow \begin{bmatrix} 2 & -2 \\2 & -2\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} = \begin{bmatrix}1\\ 1\end{bmatrix} \Rightarrow x_1-x_2={1\over 2}, \\ \text{ we choose }u=\begin{pmatrix} 1/2\\ 0\end{pmatrix} \Rightarrow \mathbf y(t)=c_1e^{-t}\begin{pmatrix} 1\\ 1\end{pmatrix}  +c_2e^{-t} \left( t\begin{pmatrix} 1\\ 1\end{pmatrix} +\begin{pmatrix} 1/2\\ 0 \end{pmatrix} \right) \\ \Rightarrow \bbox[red, 2pt]{\begin{bmatrix} y_1(t) \\ y_2(t)\end{bmatrix} =c_1 \begin{bmatrix} e^{-t}\\ e^{-t}\end{bmatrix}+ c_2 \begin{bmatrix} (t+1/2)e^{-t}\\ te^{-t}\end{bmatrix}}$$

解答:$$\det(A-\lambda I)=-(\lambda-1)(\lambda-4)(\lambda-7)=0 \Rightarrow \lambda=1,4,7 \\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 3 & 2 & -2 \\2 & 4 & 0 \\-2 & 0 & 2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ 2x_2+x_3=0} \\\qquad \Rightarrow v= x_3 \begin{bmatrix} 1\\ -1/2\\ 1\end{bmatrix}, \text{ choose }v_1= \begin{bmatrix} 1\\ -1/2\\ 1\end{bmatrix} \\ \lambda_2=4 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} 0 & 2 & -2 \\2 & 1 & 0 \\-2 & 0 & -1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{2x_1+x_3=0\\ x_2=x_3} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix} -1/2\\ 1\\ 1\end{bmatrix}, \text{ choose }v_2= \begin{bmatrix} -1/2\\ 1\\ 1\end{bmatrix} \\ \lambda_3= 7 \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix} -3 & 2 & -2 \\2 & -2 & 0 \\-2 & 0 & -4\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_1+2x_3=0\\ x_2+2x_3=0} \\ \qquad \Rightarrow v= x_3 \begin{bmatrix} - 2\\ -2\\ 1\end{bmatrix}, \text{ choose }v_3= \begin{bmatrix} - 2\\ -2\\ 1\end{bmatrix} \\ \Rightarrow P=[v_1 v_2 v_3] = \begin{bmatrix}1 & \frac{-1}{2} & -2 \\\frac{-1}{2} & 1 & -2 \\1 & 1 & 1 \end{bmatrix} \Rightarrow A= P \begin{bmatrix} 1& 0& 0\\ 0& 4& 0\\ 0& 0& 7\end{bmatrix} P^{-1} \Rightarrow A^{10} =P \begin{bmatrix} 1& 0& 0\\ 0& 4^{10}& 0\\ 0& 0& 7^{10}\end{bmatrix} P^{-1} \\ \Rightarrow    \bbox[red, 2pt]{A^{10}={1\over 9}\begin{bmatrix}4+4^{10}+4\cdot 7^{10}& -2+2\cdot 4^{10}+4\cdot 7^{10} & 4-2\cdot 4^{10}-2\cdot 7^{10} \\ -2+2\cdot 4^{10}+4\cdot 7^{10} & 1+4^{11}+4 \cdot 7^{10}& -2+4^{11}-2 \cdot 7^{10} \\ 4-2\cdot 4^{10}-2\cdot 7^{10}& -2+4^{11}-2\cdot 7^{10} & 4+4^{11}+7^{10}\end{bmatrix}}$$

解答:$$\cases{\text{rank}(A)=\text{dim}(C(A)) = \text{dim}(R(A))  \\\text{rank}(A^T)=\text{dim}(C(A^T)) = \text{dim}(R(A^T))} \text{ and }\cases{R(A) =C(A^T) \\ C(A)=R(A^T)}\\, \text{where }C(A)= \text{column space of }A, R(A) = \text{column space of }A \\ \Rightarrow \text{rank}(A) =\text{rank}(A^T) \Rightarrow \text{rank}(AB) =\text{rank}((AB)^T) =\text{rank}(B^TA^T)\quad \bbox[red, 2pt]{QED.}$$

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