國立中山大學113學年度碩士班招生考試
科目名稱: 工程數學【資訊工程礝士班乙組】
解答:A=[10−2050−204]⇒det(A−λI)=−λ3+10λ2−15λ=−λ(λ−5)2=0⇒ eigenvalues: 0,5
解答:B=[−405−335−122]⇒[B∣I]=[−405100−335010−122001]R1/(−4)→R1→[10−54−1400−335010−122001]R1+R3→R3,3R1+R2→R2→[10−54−14000354−34100234−1401]R2/3→R2→[10−54−140001512−141300234−1401]−2R2+R3→R3→[10−54−140001512−1413000−11214−231]−12R3→R3→[10−54−140001512−14130001−38−12](−5/12)R3+R2→R2,(5/4)R3+R1→R1→[100−410−150101−35001−38−12]⇒B−1=[−410−151−35−38−12]

解答:x'+2x=e^{-t} \Rightarrow \text{ integrating factor }I(t) =e^{2t} \Rightarrow e^{2t}x'+ 2xe^{2t}=e^t \Rightarrow (e^{2t}x)'=e^t \\ \Rightarrow e^{2t}x= \int e^t\,dt = e^t+c_1 \Rightarrow x=e^{-t}+c_1e^{-2t} \Rightarrow x(0)=1+c_1={3\over 4} \Rightarrow c_1=-{1\over 4} \\ \Rightarrow \bbox[red, 2pt]{x=e^{-t}-{1\over 4}e^{-2t}}

解答:\textbf{5.1}
\textbf{5.2}\; a_k={1\over 2} \int_0^4 e^{-t} \cos({k\pi t\over 2})\,dt =\left. \left[ {e^{-t}k\pi \sin({k\pi t\over 2})-2e^{-t}\cos( {k\pi t\over 2}) \over k^2\pi^2 +4}\right] \right|_0^4 ={2(1-e^{-4})\over k^2 \pi^2+4} \\ \\\qquad \Rightarrow \bbox[red, 2pt]{a_k={2(1-e^{-4})\over k^2 \pi^2+4},k=0,1,2,\dots}

解答:\cases{y_1'=y_1+y_2+ 5\cos t\\ y_2'=3y_1-y_2-5\sin t} \Rightarrow \begin{bmatrix}y_1' \\y_2' \end{bmatrix} =\begin{bmatrix}1 & 1 \\3 & -1 \end{bmatrix} \begin{bmatrix}y_1 \\y_2 \end{bmatrix}+\begin{bmatrix} 5\cos t \\-5\sin t \end{bmatrix}\equiv \mathbf y'=A\mathbf y+\mathbf g \\ A= \begin{bmatrix}1 & 1 \\3 & -1 \end{bmatrix} \Rightarrow \text{eigenvalues of }A \text{ are 2,-2, and eigenvectors: } \begin{bmatrix}1\\ 1 \end{bmatrix}, \begin{bmatrix}1\\ -3 \end{bmatrix} \\ \Rightarrow \mathbb y_h=c_1e^{2t}\begin{bmatrix}1\\ 1 \end{bmatrix} +c_2e^{-2t} \begin{bmatrix}1\\ -3 \end{bmatrix} \Rightarrow \mathbf Y=\begin{bmatrix}e^{2t} & e^{-2t} \\e^{2t} & -3e^{-2t} \end{bmatrix} \Rightarrow \mathbf Y^{-1}= \begin{bmatrix}\frac{3}{4}e^{-2t} & \frac{1}{4} e^{-2t} \\\frac{1}{4}e^{2t} & -\frac{1}{4}e^{2t} \end{bmatrix} \\ \Rightarrow \mathbf Y^{-1}\mathbf g= \begin{bmatrix}\frac{3}{4}e^{-2t} & \frac{1}{4} e^{-2t} \\\frac{1}{4}e^{2t} & -\frac{1}{4}e^{2t} \end{bmatrix} \begin{bmatrix} 5\cos t \\-5\sin t \end{bmatrix} =\begin{bmatrix} {15\over 4} e^{-2t} \cos t -{5\over 4}e^{-2t}\sin t \\ {5\over 4}e^{2t}\cos t+ {5\over 4}e^{2t} \sin t \end{bmatrix} \\ \Rightarrow \int \mathbf Y^{-1}\mathbf g\,dt =\begin{bmatrix} {5\over 4}e^{-2t}(-\cos t+\sin t) \\{1\over 4}e^{2t}(\cos t+ 3\sin t) \end{bmatrix} \\\Rightarrow \mathbf y_p=\mathbf Y \int \mathbf Y^{-1}\mathbf g\,dt = \begin{bmatrix}e^{2t} & e^{-2t} \\e^{2t} & -3e^{-2t} \end{bmatrix} \begin{bmatrix} {5\over 4}e^{-2t}(-\cos t+\sin t) \\{1\over 4}e^{2t}(\cos t+ 3\sin t) \end{bmatrix} =\begin{bmatrix}-\cos t+2\sin t \\-2\cos t-\sin t \end{bmatrix}\\ \Rightarrow \mathbf y=\mathbf y_h+\mathbf y_p =c_1e^{2t}\begin{bmatrix}1\\ 1 \end{bmatrix} +c_2e^{-2t} \begin{bmatrix} 1\\ -3 \end{bmatrix} +\begin{bmatrix}-\cos t+2\sin t \\-2\cos t-\sin t \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\cases{y_1(t) =c_1e^{2t}+ c_2e^{-2t}-\cos t+2\sin t\\ y_2(t)=c_1e^{2t} -3c_2e^{-2t}-2\cos t-\sin t}}
==================== END ======================
解題僅供參考,其他歷年試題及詳解
沒有留言:
張貼留言