國立雲林科技大學111學年度碩士班招生考試
系所:電機系
科目:工程數學
解答:$$\mathbf{1.}\;y'=5\sin(3x) \Rightarrow y=\int 5\sin(3x)\,dx = -{5\over 3}\cos (3x)+c_1 \Rightarrow \bbox[red, 2pt]{y= -{5\over 3}\cos (3x)+c_1}\\ \mathbf{2.}\; 積分因子I(x)=e^{\int 5x^4\,dx} =e^{x^5} \Rightarrow y'I(x)+5x^4yI(x) = 3x^2e^{-x^5} \Rightarrow y'e^{x^5} +5x^4 ye^{x^5} = 3x^2\\ \Rightarrow (ye^{x^5})'=3x^2 \Rightarrow ye^{x^5}=x^3+c_1 \Rightarrow \bbox[red, 2pt]{y=x^3e^{-x^5}+ c_1e^{-x^5}} \\\mathbf{3.}\; x^2y'=y^2+xy \Rightarrow y'-{y\over x}={y^2\over x^2}為白努利方程式\\ 因此取v={1\over y} \Rightarrow v'=-{y'\over y^2} \Rightarrow y'=-y^2v'代入原式 \Rightarrow -y^2v'-{y\over x}={y^2\over x^2} \Rightarrow v'+{1\over xy}=-{1\over x^2} \\ \Rightarrow v'+{v\over x}=-{1\over x^2} \Rightarrow xv'+v=-{1\over x} \Rightarrow (xv)'=-{1\over x} \Rightarrow xv=-\ln x+c_1\\ \Rightarrow {x\over y}=-\ln x+c_1 \Rightarrow \bbox[red, 2pt]{y={x\over -\ln x+c_1}}$$
解答:$$先求齊次解,y''+9y=0 \Rightarrow \lambda^2+9=0 \Rightarrow \lambda=\pm 3i \Rightarrow y_h= c_1\cos(3x)+ c_2\sin(3x)\\ y_p= A\cos x+B\sin x \Rightarrow y_p'=-A\sin x+B\cos x \Rightarrow y_p''=-A\cos x-B\sin x\\ \Rightarrow y_p''+9y_p=8A\cos x+ 8B\sin x= 16\sin x \Rightarrow \cases{A=0\\ B= 2} \Rightarrow y_p=2\sin x \\ \Rightarrow y=y_h+y_p = c_1\cos(3x)+ c_2\sin(3x)+ 2\sin x \Rightarrow y'=-3c_1\sin(3x)+ 3c_2\cos(3x)+ 2\cos x\\ \Rightarrow \cases{y(\pi)= -c_1=0\\ y'(0)=3c_2+2=0} \Rightarrow \cases{c_1=0\\ c_2=-{2\over 3}} \Rightarrow \bbox[red, 2pt]{y=-{2\over 3} \sin(3x)+2\sin x}$$
解答:$$\mathbf{(1)}\; F(s)=L\{t^2\} +L\{ e^{-6t}\} +L\{\sin(6t)\} =\bbox[red, 2pt]{{2\over s^3}+{1\over s+6}+ {6\over s^2+36}} \\\mathbf{(2)}\; f(t)=L^{-1} \{ {8 \over s(s-2)^2}\} =L^{-1}\{{2\over s}-{2\over s-2}+ {4\over (s-2)^2} \} =\bbox[red, 2pt]{2-2e^{2t}+4e^{2t}t}\\ \mathbf{(3)}\; L\{y'\}+ L\{\int_0^t y(\tau)\,d\tau\} = 2L\{1\}- 2L\{\sin t\} \Rightarrow sY(s)+Y(s)={2\over s}-{2\over s^2+1} \\ \Rightarrow Y(s)={2\over s(s+1)}-{2\over (s^2+1)(s+1)} \Rightarrow y(t)= L^{-1}\{ {2\over s(s+1)}\}-L^{-1}\{ {2\over (s^2+1)(s+1)}\} \\= 2L^{-1}\{{1\over s}-{1\over s+1}\}-L^{-1}\{{1\over s^2+1}-{s\over s^2+1} +{1\over s+1}\} =2-2e^{-t}-(\sin t-\cos t+e^{-t}) \\ \Rightarrow \bbox[red, 2pt]{y(t)=2-3e^{-t}-\sin t+\cos t}$$
解答:$$y'-3y=te^{3t} \sin(t) \Rightarrow L\{y'\}-3L\{y\} =L\{te^{3t} \sin(t)\} \\ \Rightarrow sY(s)-3Y(s)=-\frac{\text{d}}{\text{d}s} L\{ e^{3t}\sin(t)\}= -\frac{\text{d}}{\text{d}s} {1\over (s-3)^2+1} ={2s-6\over (s^2-6s+10)^2} \\ \Rightarrow Y(s)={2\over (s^2-6s+10)^2} \Rightarrow y(t)=L^{-1}\{ {2\over (s^2-6s+10)^2}\} =L^{-1}\{ {2\over ((s-3)^2+1)^2}\} \\ =e^{3t}L^{-1}\{ {2\over (s^2+1)^2}\} =e^{3t}(\sin t-t\cos t) \Rightarrow \bbox[red, 2pt]{y(t)=e^{3t}(\sin t-t\cos t)}$$
解答:$$\mathbf{(a)}\;A=\begin{bmatrix}0 & 0 & 3\\ 0 & 2& 1\\ 2& 0 & 1 \end{bmatrix} \Rightarrow \det(A-\lambda I)= -(\lambda+2)(\lambda-2)(\lambda-3) =0 \Rightarrow \lambda=-2, 2,3\\ \lambda_1=-2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}2 & 0 & 3\\ 0 & 4& 1\\ 2& 0 & 3 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{2x_1+3x_3=0 \\ 4x_2+x_3=0}\\ \qquad \Rightarrow v= \begin{bmatrix}-3k/2 \\ -k/4\\ k \end{bmatrix},k\in \mathbb R, 取v_1= \begin{bmatrix}-3/2 \\ -/4\\ 1 \end{bmatrix}\\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-2 & 0 & 3\\ 0 & 0& 1\\ 2& 0 & -1 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=0\\ x_3=0}\\\qquad \Rightarrow v= \begin{bmatrix}0 \\ k\\ 0 \end{bmatrix},k\in \mathbb R, 取v_2= \begin{bmatrix}0 \\ 1\\ 0 \end{bmatrix}\\ \lambda_3=3 \Rightarrow (A-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix}-3 & 0 & 3\\ 0 & -1& 1\\ 2& 0 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ x_2=x_3}\\ \qquad \Rightarrow v= \begin{bmatrix}k \\ k\\ k \end{bmatrix},k\in \mathbb R, 取v_3= \begin{bmatrix}1 \\ 1\\ 1 \end{bmatrix}\\ \Rightarrow \text{eigenvalues: } \bbox[red,2pt]{ -2,2,3} \text{ and corresponding eigenvectors are } \bbox[red, 2pt]{\begin{bmatrix}-3/2 \\ -/4\\ 1 \end{bmatrix}, \begin{bmatrix}0 \\ 1\\ 0 \end{bmatrix}, \begin{bmatrix}1 \\ 1\\ 1 \end{bmatrix}}$$$$\mathbf{(b)}\; X=[v_1 v_2 v_3], D=\begin{bmatrix}\lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix} \\\Rightarrow \bbox[red, 2pt]{A=\begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}-2 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} -3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}^{-1}} \\ \Rightarrow A^5= \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}(-2)^5 & 0 & 0\\ 0 & 2^5 & 0 \\ 0 & 0 & 3^5 \end{bmatrix} \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}^{-1} \\= \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix}-32 & 0 & 0\\ 0 & 32 & 0 \\ 0 & 0 & 243 \end{bmatrix} \begin{bmatrix}-3/2 & 0 & 1\\ -1/4 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix}^{-1}\Rightarrow \bbox[red, 2pt]{A^5 = \left[ \begin{matrix} 78 & 0 & 165 \\78 & 32 & 133 \\110 & 0 & 133\end{matrix}\right]}$$
解答:$$\mathbf{(a)}\;A=\left[\begin{matrix}-4 & 3 & 4\\1 & -2 & 3\\6 & 0 & 0\end{matrix}\right] \Rightarrow rref(A)=\left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right] \Rightarrow \bbox[red,2pt]{\text{linearly independent}} \\ \mathbf{(b)}\; -\begin{bmatrix}6\\ -7\\ 8\\ 6 \end{bmatrix} +2\begin{bmatrix}4\\ 6\\ -4\\1 \end{bmatrix} =\begin{bmatrix} 2\\ 19\\ -16\\ -4\end{bmatrix} \Rightarrow \bbox[red, 2pt]{\text{linearly dependent}}$$
解答:$$A=\left[\begin{matrix}1 & 4 & 2 & 1\\0 & 1 & 1 & -1\\-2 & -8 & -4 & -2 \end{matrix}\right] \Rightarrow rref(A)= \left[ \begin{matrix}1 & 0 & -2 & 5\\0 & 1 & 1 & -1\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow \bbox[red, 2pt]{Rank(A)=2} \\ 又\left[ \begin{matrix}1 & 0 & -2 & 5\\0 & 1 & 1 & -1\\0 & 0 & 0 & 0\end{matrix}\right] \begin{bmatrix} x_1\\ x_2\\ x_3\\x_4\end{bmatrix} =0 \Rightarrow \cases{x_1-2x_3+5x_4=0\\ x_2+x_3-x_4=0} \\ \Rightarrow \begin{bmatrix} x_1\\ x_2\\ x_3\\x_4\end{bmatrix} = \begin{bmatrix} 2x_3-5x_4\\ -x_3+x_4\\ x_3\\x_4\end{bmatrix} = x_3 \begin{bmatrix} 2\\ -1\\ 1\\0\end{bmatrix}+ x_4 \begin{bmatrix} -5\\ 1\\ 0 \\1 \end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{ Null(A)=Span\left\{ \begin{bmatrix} 2\\ -1\\ 1\\0\end{bmatrix}, \begin{bmatrix} -5\\ 1\\ 0 \\1 \end{bmatrix} \right\}, Nulity(A)=2}$$
解答:$$A=\left[\begin{matrix}1 & 0 & 1\\7 & 7 & 8\\1 & 2 & 1\\7 & 7 & 6\end{matrix}\right] \Rightarrow rref(A)=\left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0 \\0 & 0 & 1\\0 & 0 & 0\end{matrix}\right] \Rightarrow \text{a basis for column space of }A=\{a_1,a_2, a_3\},\\ \text{where }a_1=\begin{bmatrix} 1\\ 7\\1\\7 \end{bmatrix}, a_2=\begin{bmatrix} 0 \\ 7 \\ 2\\7 \end{bmatrix}, a_3= \begin{bmatrix}1\\ 8\\ 1\\ 6\end{bmatrix} \\ b_1=a_1 \Rightarrow e_1={b_1\over \Vert b_1\Vert}=\begin{bmatrix} 1/10 \\ 7/10 \\ 1/10\\7/10 \end{bmatrix}\\ b_2=a_2-(a_2\cdot e_1)e_1 =\begin{bmatrix} 0 \\ 7 \\ 2\\7 \end{bmatrix}-\begin{bmatrix} 1\\ 7\\1\\7 \end{bmatrix}= \begin{bmatrix} -1\\ 0\\1\\0 \end{bmatrix} \Rightarrow e_2={b_2\over \Vert b_2 \Vert} = \begin{bmatrix} -1/\sqrt 2\\ 0\\1/\sqrt 2\\0 \end{bmatrix} \\ b_3=a_3-(a_3\cdot e_1)e_1-(a_3\cdot e_2)e_2 =\begin{bmatrix}1\\ 8\\ 1\\ 6\end{bmatrix}-\begin{bmatrix} 1\\ 7\\1\\7 \end{bmatrix} -0=\begin{bmatrix} 0\\ 1\\0\\-1 \end{bmatrix}\\ \qquad \Rightarrow e_3={b_3\over \Vert b_3 \Vert}=\begin{bmatrix} 0\\ 1/\sqrt 2\\0\\-1/\sqrt 2 \end{bmatrix}\\ \Rightarrow \text{an orthonomal basis for the column space of }A= \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1/10 \\ 7/10 \\ 1/10\\7/10 \end{bmatrix}, \begin{bmatrix} -1/\sqrt 2\\ 0\\1/\sqrt 2\\0 \end{bmatrix} , \begin{bmatrix} 0\\ 1/\sqrt 2\\0\\-1/\sqrt 2 \end{bmatrix} \right\}}$$
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解題僅供參考,其他歷年試題及詳解
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