115年台綜大轉學考-工程數學D37詳解

臺灣綜合大學系統 115 學年度學士班轉學生聯合招生考試

科目名稱 工程數學D37

解答:$$u=e^y \Rightarrow {du\over dx}={dy\over dx}e^y \Rightarrow e^y \left( {dy\over dx}+1 \right) ={du\over dx}+u=e^x \Rightarrow {du\over dx}e^x +ue^x =e^{2x} \\ \Rightarrow \left( ue^x \right)'=e^{2x} \Rightarrow ue^x =\int e^{2x}\,dx= {1\over 2}e^{2x}+C \Rightarrow u=e^y={1\over 2}e^x+Ce^{-x}\\ \Rightarrow \bbox[red, 2pt]{y= \ln \left( {1\over 2}e^x+Ce^{-x} \right)}$$

解答:$$\textbf{(a) } \mathbf{F}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j} ,  \text{where }\cases{P(x,y) = e^x \sin(y) + e^{-y} \\Q(x,y) = e^x \cos(y) - xe^{-y}} \\ \Rightarrow  \cases{P_y=e^x\cos y-e^{-y} \\ Q_x =e^x \cos y-e^{-y}} \Rightarrow P_y= Q_x \Rightarrow \mathbf F \text{ is conservative} \quad \bbox[red, 2pt]{QED.} \\ \textbf{(b) } \phi(x,y) = \int P(x,y)\,dx = \int Q(x,y)\,dy \Rightarrow \phi= \int (e^x\sin y+ e^{-y})\,dx = \int (e^x\cos(y)-xe^{-y})\,dy \\ \Rightarrow e^x\sin y+xe^{-y } +\rho(y)=e^x \sin y+xe^{-y}+ \omega(x) \Rightarrow \bbox[red, 2pt]{\phi(x,y)=e^x\sin y+xe^{-y } +C} \\ \textbf{(c) } \int_{(0,0)}^{(1,\pi/2)} \mathbf F\cdot dr =\phi(1,\pi/2)-\phi(0,0)=e+e^{-\pi/2}-0 = \bbox[red, 2pt]{e+e^{-\pi/2}}$$

解答:$$\textbf{a) }P =[v_1 \; v_2\; v_2] \Rightarrow \cases{v_1=(a,b,2/3)^T\\ v_2=(0, 1/\sqrt 2, c)^T\\ v_3=(4/3\sqrt 2, 1/3\sqrt 2,-1/3\sqrt 2)} \Rightarrow v_2\cdot v_3=0 \Rightarrow 0+{1\over 6}-{c\over 3\sqrt 2}=0 \\ \Rightarrow c={\sqrt 2\over 2} \Rightarrow v_2=(0,1/\sqrt 2, 1/\sqrt 2)^T \Rightarrow v_1\cdot v_2=0 \Rightarrow 0+{b\over \sqrt 2}+{\sqrt 2\over 3}=0 \Rightarrow b=-{2\over 3} \\ \Rightarrow v_1\cdot v_3={4a\over 3\sqrt 2 }-{2\over 9\sqrt 2}-{2\over 9\sqrt 2}=0 \Rightarrow a={1\over 3} \Rightarrow \bbox[red, 2pt]{\cases{a=1/3\\ b=-2/3\\ c=1/\sqrt 2}} \\ \textbf{b) } \mathbf{P} = \begin{pmatrix} \frac{1}{3} & 0 & \frac{4}{3\sqrt{2}} \\ -\frac{2}{3} & \frac{1}{\sqrt{2}} & \frac{1}{3\sqrt{2}} \\ \frac{2}{3} & \frac{1}{\sqrt{2}} & -\frac{1}{3\sqrt{2}} \end{pmatrix} \Rightarrow \mathbf{P}^T = \begin{pmatrix} \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{4}{3\sqrt{2}} & \frac{1}{3\sqrt{2}} & -\frac{1}{3\sqrt{2}} \end{pmatrix}  \Rightarrow \mathbf P^T \mathbf P= \begin{pmatrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{pmatrix} \quad \bbox[red, 2pt]{QED.}$$

解答:$$\bbox[cyan,2pt]{題目有誤}! 以n=1為例: A \begin{pmatrix} 3\\9 \end{pmatrix} = \begin{pmatrix} -1&2\\-7& 8 \end{pmatrix}  \begin{pmatrix} 3\\9 \end{pmatrix} = \begin{pmatrix} 15\\51 \end{pmatrix}, 而6 \begin{pmatrix} 2\\7 \end{pmatrix}+ \begin{pmatrix} 1\\1 \end{pmatrix} = \begin{pmatrix} 13\\43 \end{pmatrix}\\  也就是說A^n \begin{pmatrix}3\\9 \end{pmatrix} \ne 6^n \begin{pmatrix} 2\\7 \end{pmatrix}+ \begin{pmatrix} 1\\1 \end{pmatrix}, \mbox{for }n=1 \\ 以下是將A^n \begin{pmatrix}3\\9 \end{pmatrix}改成A^n \begin{pmatrix}3\\8 \end{pmatrix}的解答\\ \det(A-\lambda I)=0 \Rightarrow \lambda^2-7\lambda+6=0 \Rightarrow (\lambda-6)(\lambda -1)=0 \Rightarrow \lambda_1=6,\lambda_2=1\\ \cases{\lambda_1=6 \Rightarrow  \text{ eigenvector }v_1= \begin{pmatrix} 2\\7 \end{pmatrix} \\ \lambda_2=1 \Rightarrow \mbox{eigenvector }v_2= \begin{pmatrix} 1\\1 \end{pmatrix} } \Rightarrow \begin{pmatrix} 3\\ 8 \end{pmatrix}=v_1+v_2 \Rightarrow A^n \begin{pmatrix} 3\\8 \end{pmatrix} =A^n \left( \begin{pmatrix} 2\\7 \end{pmatrix} + \begin{pmatrix} 1\\1 \end{pmatrix}\right) \\ =A^n \begin{pmatrix} 2\\7 \end{pmatrix}+ A^n \begin{pmatrix} 1\\1 \end{pmatrix} =6^n \begin{pmatrix} 2\\7 \end{pmatrix}+1^n \begin{pmatrix} 1\\1 \end{pmatrix} \Rightarrow A^n \begin{pmatrix} 3\\8 \end{pmatrix}= 6^n \begin{pmatrix} 2\\7 \end{pmatrix} + \begin{pmatrix} 1\\1 \end{pmatrix} \;\bbox[red, 2pt]{故得證}$$

解答:$$y''-2y'+y=0 \Rightarrow r^2-2r+1=0 \Rightarrow (r-1)^2=0 \Rightarrow r=1 \Rightarrow y_h =c_1e^x +c_2 xe^x\\ y_p=A x^2e^x \Rightarrow y_p'= Ae^x(x^2+2x) \Rightarrow y_p''= Ae^x(x^2+4x+2) \\ \Rightarrow y_p''-2y_p'+y_p =  Ae^x(x^2 + 4x + 2) - 2Ae^x(x^2 + 2x) + Ax^2e^x = e^x \Rightarrow 2A=1 \Rightarrow A={1\over 2} \\ \Rightarrow y_p={1\over 2}x^2e^x \Rightarrow y=y_h+y_p =c_1e^x +c_2 xe^x+{1\over 2}x^2e^x \Rightarrow y'=(c_1+c_2)e^x+   (c_2+1)xe^x +{1\over 2}x^2e^x \\ \Rightarrow \cases{y(0)=c_1=1\\ y'(0)=c_1+c_2= 0} \Rightarrow \cases{c_1=1\\ c_2=-1} \Rightarrow \bbox[red, 2pt]{y=e^x \left( 1-x+{1\over 2}x^2 \right)}$$

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