112年台北科大機電整合碩士班-工程數學詳解

 國立臺北科技大學 112學年度碩士班招生考試

系所組別 :1111、 l112機械工程系機電整合碩士班甲組
第一節 工程數學 試題

解答： $$先求齊次解,y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda+2=0 \Rightarrow (\lambda-2)(\lambda-1)=0 \Rightarrow \lambda=2,1\\ \Rightarrow y_h=c_1e^x+c_2e^2x\\ 再利用參數變換法求特定解,令\cases{y_1=e^x\\ y_2=e^{2x}\\ r(x)=-{e^{3x}\over e^x+1}} \Rightarrow \cases{y_1'=e^x\\ y_2'=2e^2x}\Rightarrow W=\begin{vmatrix}y_1& y_2\\  y_1'& y_2' \end{vmatrix} =\begin{vmatrix} e^x& e^{2x}\\ e^x & 2e^{2x} \end{vmatrix}=e^{3x} \\ \Rightarrow y_p=-y_1 \int {y_2r(x)\over W}dx +y_2\int {y_1r(x)\over W}dx =-e^x \int {-e^{2x}\over e^x+1}\,dx +e^{2x}\int {-e^x\over e^x+1}\,dx\\= e^x\left( e^x-\ln(e^x+1)\right)- e^{2x} \ln(e^x+1)  =e^{2x} -e^x(1+e^x)\ln(e^x+1)\\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1e^x+c_2e^{2x}  -e^x(1+e^x)\ln(e^x+1)}$$

解答： $$L\{y''\}+2L\{y'\}+10L\{y\}=L\{t\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+2(sY(s)-y(0))+10Y(s)={1\over s^2}\\ \Rightarrow (s^2+2s+10)Y(s)-s-3={1\over s^2} \Rightarrow Y(s)={s+3\over s^2+2s+10}+{1\over s^2(s^2+2s+10)} \\={s+1\over (s+1)^2+3^2} +{2\over (s+1)^2+3^2}+{1\over 50}\cdot {s+1\over (s+1)^2+3^2}-{2\over 25}\cdot {1\over (s+1)^2+3^2}-{1\over 50s}+{1\over 10s^2} \\ ={51\over 50}\cdot {s+1\over (s+1)^2+3^2}+ {48\over 25}\cdot {1\over (s+1)^2+3^2}-{1\over 50s}+{1\over 10s^2}\\\Rightarrow y(t)=L^{-1}\{Y(s)\}\Rightarrow \bbox[red, 2pt]{y(t)= {51\over 50}e^{-t}\cos(3t)+{16\over 25}e^{-t}\sin(3t)-{1\over 50}+{t\over 10}}$$

解答： $$\cases{x_1+x_2+ x_3=6\\ 3x_1-x_2+x_3=4\\ -2x_1+x_2-2x_3=-6}\Rightarrow A\mathbf x=\mathbf b,其中A=\left[\begin{matrix}1 & 1 & 1 \\3 & -1 & 1 \\-2 & 1 & -2\end{matrix}\right],\mathbf x=\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}, \mathbf b=\begin{bmatrix} 6\\ 4\\ -6 \end{bmatrix}\\ A^{-1}=\left[\begin{matrix}\frac{1}{6} & \frac{1}{2} & \frac{1}{3} \\\frac{2}{3} & 0 & \frac{1}{3} \\\frac{1}{6} & \frac{-1}{2} & \frac{-2}{3}\end{matrix}\right] \Rightarrow \mathbf x=A^{-1}\mathbf b =\left[\begin{matrix}\frac{1}{6} & \frac{1}{2} & \frac{1}{3} \\\frac{2}{3} & 0 & \frac{1}{3} \\\frac{1}{6} & \frac{-1}{2} & \frac{-2}{3}\end{matrix}\right] \begin{bmatrix} 6\\ 4\\ -6 \end{bmatrix} =\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\Rightarrow \bbox[red, 2pt]{\cases{x_1=1\\ x_2=2\\ x_3=3}}$$

解答： $$令y(x)=\sum_{n=0}^\infty a_nx^n \Rightarrow y'(x)=\sum_{n=1}^\infty na_nx^{n-1}  \Rightarrow y''(x)=\sum_{n=2}^\infty n(n-1)a_nx^{n-2} \\ 又e^x=\sum_{n=0}^\infty {1\over n!}x^n，因此y''-xy'+e^xy=4\\ \Rightarrow \sum_{n=2}^\infty n(n-1)a_nx^{n-2}-\sum_{n=1}^\infty na_nx^{n}+ \left(\sum_{n=0}^\infty {1\over n!}x^n \right) \left( \sum_{n=0}^\infty a_nx^n \right)=4\\ 考慮x^n的係數b_n= (n+2)(n+1)a_{n+2}-na_n+ (a_n+a_{n-1}+ {1\over 2}a_{n-2}+{1\over 3!}a_{n-3}+\cdots +{1\over n!}a_0) \\\qquad =(n+2)(n+1)a_{n+2}-na_n+\sum_{k=0}^n {1\over k!}a_{n-k} \Rightarrow \cases{b_0=4\\ b_n=0,n\gt 0}\\ 又初始值\cases{y(0)=2\\ y'(0)=3} \Rightarrow \cases{a_0=2\\ a_1=3}  \Rightarrow b_0=2a_2+a_0=4 \Rightarrow a_2=1 \\ n\gt 0\Rightarrow b_n=(n+2)(n+1)a_{n+2}-na_n+\sum_{k=0}^n {1\over k!}a_{n-k}=0\\ \Rightarrow a_{n+2}={1\over (n+2)(n+1)}\left(na_n-\sum_{k=0}^n {1\over k!}a_{n-k} \right) \Rightarrow \cases{a_3={1\over 6}(a_1-a_1-a_0)=-{1\over 3}\\ a_4={1\over 12}(2a_2-a_2-a_1-{1\over 2}a_0)=-{1\over 4}\\ \cdots}\\ \Rightarrow \bbox[red,2pt]{y(x)=2+3x+x^2-{1\over 3}x^3-{1\over 4}x^4+\cdots}\\$$

解答：

A. $$A=\left[\begin{matrix}1 & -2 & 1 \\0 & 1 & -1 \\0 & 2 & -1\end{matrix}\right] \Rightarrow \det(A-\lambda I)=0 \Rightarrow -(\lambda-1)(\lambda^2+1)=0 \Rightarrow \lambda=1,\pm i\\ \lambda_1= 1: (A-\lambda_1 I)\mathbf x=0 \Rightarrow \begin{bmatrix} 0&-2& 1\\ 0 & 0 &-1\\ 0 & 2& -2\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_2=0\\x_3=0} \\\qquad \Rightarrow E_1(A)=\{\mathbf x\mid \mathbf x=t\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix},t\in \mathbb R\}\\ \lambda_2=i:(A-\lambda_2 I)\mathbf x=0 \Rightarrow \begin{bmatrix} 1-i&-2& 1\\ 0 & 1-i &-1\\ 0 & 2& -1-i\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_1={-1+i\over 2}x_3\\ x_2={1+i\over 2}x_3}\\ \qquad \Rightarrow E_i(A)=\{\mathbf x\mid \mathbf x=t\begin{bmatrix} (-1+i)/2\\ (1+i)/2\\ 1 \end{bmatrix},t\in \mathbb R\} \\\lambda_3=-i:(A-\lambda_3 I)\mathbf x=0 \Rightarrow \begin{bmatrix} 1+i&-2& 1\\ 0 & 1+i &-1\\ 0 & 2& -1+i\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=0 \Rightarrow \cases{x_1={-1-i\over 2}x_3\\ x_2={1-i\over 2}x_3}\\ \qquad \Rightarrow E_{-i}(A)=\{\mathbf x\mid \mathbf x=t\begin{bmatrix} (-1-i)/2\\ (1-i)/2\\ 1 \end{bmatrix},t\in \mathbb R\}\\ 特徵值為\bbox[red,2pt]{1,i,-i}相對應的特徵向量分別為\bbox[red,2pt]{E_1(A)=\{\mathbf x\mid \mathbf x=t\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix},t\in \mathbb R\}}, \\\bbox[red,2pt]{E_i(A)=\{\mathbf x\mid \mathbf x=t\begin{bmatrix} (-1+i)/2\\ (1+i)/2\\ 1 \end{bmatrix},t\in \mathbb R\},及E_{-i}(A)=\{\mathbf x\mid \mathbf x=t\begin{bmatrix} (-1-i)/2\\ (1-i)/2\\ 1 \end{bmatrix},t\in \mathbb R\}}$$ B. $$A=\left[\begin{matrix}1 & -2 & 1 \\0 & 1 & -1 \\0 & 2 & -1\end{matrix}\right] =PDP^{-1} =\left[\begin{matrix}1 & \frac{-1+i}{2} & \frac{-1-i}{2} \\0 & \frac{1+i}{2} & \frac{1-i}{2} \\0 & 1 & 1\end{matrix}\right]\left[\begin{matrix}1 & 0 & 0 \\0 & i & 0 \\0 & 0 & -i \end{matrix}\right]\left[\begin{matrix}1 & -1 & 1 \\0 & -i & \frac{1+i}{2} \\0 & i & \frac{1-i}{2}\end{matrix}\right] \\ \Rightarrow A^{2000}=PD^{2000}P^{-1} =\left[ \begin{matrix}1 & \frac{-1+i}{2} & \frac{-1-i}{2} \\0 & \frac{1+i}{2} & \frac{1-i}{2} \\0 & 1 & 1\end{matrix}\right]\left[\begin{matrix}1^{2000} & 0 & 0 \\0 & i^{2000} & 0 \\0 & 0 & (-i)^{2000} \end{matrix}\right] \left[\begin{matrix}1 & -1 & 1 \\0 & -i  & \frac{1+i}{2} \\0 & i & \frac{1-i}{2}\end{matrix}\right]\\=\left[\begin{matrix}1 & \frac{-1+i}{2} & \frac{-1-i}{2} \\0 & \frac{1+i}{2} & \frac{1-i}{2} \\0 & 1 & 1\end{matrix} \right]\left[ \begin{matrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{matrix}\right]\left[\begin{matrix}1 & -1 & 1 \\0 & -i & \frac{1+i}{2} \\0 & i & \frac{1-i}{2}\end{matrix}\right] =PIP^{-1}= PP^{-1} \\=I =\bbox[red, 2pt]{\left[ \begin{matrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{matrix}\right]}$$

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