111年雲科大電子系碩士班-工程數學詳解

 

解答:$$\mathbf{(a)}\;(x^3y^4)dx +(x^4y^3+2y)dy =0 \Rightarrow \cases{P(x,y)=x^3y^4\\ Q(x,y)=x^4y^3+2y} \Rightarrow \frac{\partial P}{\partial y} =4x^3y^3 =\frac{\partial Q}{\partial x} \Rightarrow \text{Exact}\\ \Phi(x,y)=\int x^3y^4\,dx =\int x^4y^3+2y \,dy \Rightarrow \Phi(x,y)={1\over 4}x^4y^4 +\phi(y) ={1\over 4}x^4y^4+ y^2 +\rho(x) \\ \Rightarrow \Phi(x,y)= \bbox[red, 2pt]{{1\over 4}x^4y^4+ y^2 +C=0} \\ \mathbf{(b)}\;({1\over 4}x^4y^4+ y^2 +C)' = x^3y^4+ x^4y^3y'+2yy'=0 \Rightarrow x^3y^4+(x^4y^3+ 2y)\frac{d y}{dx}=0\bbox[red, 2pt]{與原式相同}$$

解答:$$\mathbf{(a)}\;y''+9y'+20y=0 \Rightarrow \lambda^2+9\lambda+20=0 \Rightarrow (\lambda+4)(\lambda+5) =0 \\ \Rightarrow \lambda=-4,-5 \Rightarrow \bbox[red, 2pt]{y_h=c_1e^{-4t} +c_2e^{-5t}} \\ \mathbf{(b)}\; y_p = Ae^{-t} \Rightarrow y_p'=-Ae^{-t} \Rightarrow y_p''=Ae^{-t} \Rightarrow y_p''+ 9y_p'+20y_p= 12Ae^{-t}=-36e^{-t} \\ \quad \Rightarrow A=-3 \Rightarrow \bbox[red, 2pt]{y_p=-3e^{-t}} \\ \mathbf{(c)}\;  y=y_h+ y_p = c_1e^{-4t} +c_2e^{-5t}-3e^{-t} \Rightarrow y'=-4c_1e^{-4t}-5c_2e^{-5t}+ 3e^{-t}  \\ 又\cases{y(0)= c_1+c_2 -3=0\\ y'(0)=-4c_1-5c_2+3=0} \Rightarrow \cases{c_1=12\\ c_2=-9} \Rightarrow \bbox[red, 2pt]{y=12e^{-4t}-9e^{-5t}-3e^{-t}}$$

解答:$$\mathbf{(a)} L\{e^{2t}\} =\int_0^\infty e^{2t}e^{-st}\,dt = \int_0^\infty e^{(2-s)t} \,dt = \left. \left[{1\over 2-s}e^{(2-s)t} \right] \right|_0^\infty = \bbox[red, 2pt]{1\over s-2} \\\mathbf{(b)}\; y'=\begin{bmatrix} 2 & -2\\ -1& 3\end{bmatrix}y+ \begin{bmatrix} 4\\ 2 \end{bmatrix} e^{2t},y(0)=0 \Rightarrow \cases{y_1'=2y_1-2y_2+4e^{2t}, y_1(0)=0\\ y_2' = -y_1+3y_2 +2e^{2t},y_2(0)=0} \\ \Rightarrow \cases{L\{y_1'\} =L\{2y_1-2y_2+4e^{2t}\} \\ L\{y_2'\} = L\{-y_1+3y_2 +2e^{2t}\} } \Rightarrow \cases{sY_1(s)= 2Y_1(s)-2Y_2(s)+{4\over s-2} \cdots(1)\\sY_2(s) =-Y_1(s) +3Y_2(s)+{ 2\over s-2} \cdots(2)} \\ 由(1) \Rightarrow Y_2(s) ={2-s\over 2}Y_1(s)+{2\over s-2} 代入(2) \Rightarrow Y_1(s)={4\over (s-1)(s-2)} \\ \Rightarrow Y_2={2\over s-2}-{2\over s-1} \Rightarrow y_1(t)=L^{-1}\{Y_1(s)\} =4L^{-1}\left\{{1\over s-2}-{1\over s-1}\right\} =4(e^{2t}-e^t)\\ y_2(t)=L^{-1}\{Y_2(s)\} =2L^{-1}\left\{{1\over s-2}-{1\over s-1}\right\} =2(e^{2t}-e^t)\\ \Rightarrow \bbox[red, 2pt]{\cases{y_1(t)= 4(e^{2t}-e^t)\\ y_2(t)= 2(e^{2t}-e^t)}} \\\mathbf{(c)}\; \cases{lim_{t\to \infty} y_1(t) >> y_1(0) \\ lim_{t\to \infty} y_2(t) >> y_2(0)} \Rightarrow \bbox[red,2pt] {unstable}$$

解答:$$\textbf{(a)}\; \cases{A=\begin{bmatrix}1 & 0 & -1 \\2 & 1 & 3\end{bmatrix} \\[1ex] B=\begin{bmatrix}2 & 1 & 0 \\0 & 3 & 4\end{bmatrix}} \Rightarrow A+B=\begin{bmatrix}3 & 1 & -1 \\2 & 4 & 7\end{bmatrix} \Rightarrow (A+B)^T = \bbox[red, 2pt]{\left[\begin{matrix}3 & 2\\1 & 4\\-1 & 7\end{matrix}\right]} \\ \mathbf{(b)}\; \cases{2A-B= \left[\begin{matrix}0 & -1 & -2\\4 & -1 & 2\end{matrix}\right] \\[1ex] C+D= \left[\begin{matrix}-1 & 3\\2 & 4\\2 & 2\end{matrix}\right] } \Rightarrow (2A-B)(C+D) = \bbox[red, 2pt]{\left[\begin{matrix}-6 & -8\\-2 & 12\end{matrix}\right]} \\ \mathbf{(c)}\; A-B=\left[\begin{matrix}-1 & -1 & -1\\2 & -2 & -1\end{matrix}\right] \Rightarrow X=-{3\over 2}(A-B) = \bbox[red, 2pt]{\left[\begin{matrix} 3/2 & 3/2 & 3/2\\-3 & 3 & 3/2\end{matrix}\right]}$$

解答:$$\cases{0x+4y-2z=2\\ 6x-2y+z=29 \\ 4x+8y -4z=24} \Rightarrow \left[\begin{array}{rrr|r}0 & 4 & -2 & 2\\ 6 & -2 & 1& 29\\ 4& 8 & -4 & 24 \end{array} \right] \xrightarrow{R_3/4} \left[\begin{array}{rrr|r}0 & 4 & -2 & 2\\ 6 & -2 & 1& 29\\ 1& 2 & -1 & 6 \end{array} \right] \\ \xrightarrow{-6R_3+R_2 \to R_2}  \left[\begin{array}{rrr|r}0 & 4 & -2 & 2\\ 0 & -14 & 7& -7\\ 1& 2 & -1 & 6 \end{array} \right] \xrightarrow{-R_2/14} \left[\begin{array}{rrr|r}0 & 4 & -2 & 2\\ 0 & 1 & -1/2& 1/2\\ 1& 2 & -1 & 6 \end{array} \right] \xrightarrow{-4R_2+R_1\to R_1,-2R_2+R_3\to R_3} \\ \left[\begin{array}{rrr|r}0 & 0 & 0 & 0\\ 0 & 1 & -1/2& 1/2\\ 1& 0 & 0 & 5 \end{array} \right] \Rightarrow \cases{x=5\\ y-z/2=1/2} \Rightarrow \bbox[red, 2pt]{(x,y,z)=(5,k,2k-1),k\in \mathbb R}$$

解答:$$\mathbf{(a)}\; \cases{\vec A=(1,-3,2)\\ \vec B=(3,-2,3)} \Rightarrow \vec A\cdot \vec B=1\times 3+(-3)\times (-2)+ 2\times 3= \bbox[red, 2pt]{15} \\\mathbf{(b)}\; \vec A\times \vec B=\begin{vmatrix} \vec i& \vec j& \vec k\\ 1 & -3 & 2\\ 3 & -2 & 3\end{vmatrix} =\bbox[red, 2pt]{-5\vec i+3\vec j+7\vec k} \\ \mathbf{(c)}\; \left(\vec A\cdot {\vec B\over \Vert \vec B\Vert} \right){\vec B\over \Vert \vec B\Vert} ={15\over \sqrt{22}}({3\over \sqrt{22}}, -{2\over \sqrt{22}}, {3\over \sqrt{22}}) =({45\over 22},-{30\over 22},{45\over   {22}}) = \bbox[red, 2pt]{{45\over 22}\vec i-{15\over 11}\vec j+{45\over 22}\vec k}$$

解答:$$A=\left[\begin{matrix}2 & 0 & 1\\0 & 1 & 0\\2 & 0 & 3\end{matrix}\right] \Rightarrow \det(A-\lambda I) = -(\lambda-1)^2(\lambda-4) =0 \Rightarrow \lambda=1,4\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I) v =0 \Rightarrow \left[\begin{matrix}1 & 0 & 1\\0 & 0 & 0\\2 & 0 & 2\end{matrix}\right] \left[\begin{matrix}x_1\\ x_2\\x_3\end{matrix}\right] =0 \Rightarrow x_1+x_3=0 \\\Rightarrow v=\left[\begin{matrix} -x_3\\ x_2\\x_3\end{matrix}\right] =x_3\left[\begin{matrix} -1\\ 0\\1 \end{matrix}\right]+x_2 \left[\begin{matrix} 0\\ 1\\0\end{matrix}\right],取v_1= \left[ \begin{matrix} -1\\ 0\\1 \end{matrix}\right], v_2=  \left[\begin{matrix} 0\\ 1\\0\end{matrix}\right] \\ \lambda_2=4  \Rightarrow (A-\lambda_2 I) v =0 \Rightarrow \left[\begin{matrix}-2 & 0 & 1\\0 & -3 & 0\\ 2 & 0 & -1 \end{matrix}\right] \left[\begin{matrix}x_1\\ x_2\\x_3\end{matrix}\right] =0 \Rightarrow \cases{2x_1=x_3\\ x_2=0} \\ \Rightarrow v=\left[\begin{matrix} x_3/2\\ 0\\ x_3 \end{matrix}\right] =x_3 \left[\begin{matrix} 1/2\\ 0\\1\end{matrix}\right],取v_3= \left[\begin{matrix} 1/2\\ 0\\1\end{matrix}\right] \\ \text{eigenvalues: }\bbox[red, 2pt]{1,4}, \text{eighenvectors: } \bbox[red, 2pt]{\left[\begin{matrix} -1\\ 0\\1 \end{matrix}\right],  \left[\begin{matrix} 0\\ 1\\0\end{matrix}\right]\left[\begin{matrix} 1/2\\ 0\\1\end{matrix}\right] }$$

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