107年台綜大轉學考-工程數學D09詳解

臺灣綜合大學系統107學年度學士班轉學生聯合招生考試

科目名稱：工程數學
類組代碼：D09

解答：$$先求齊次解:y_h''+y_h=0 \Rightarrow y_h= A\cos x+B\sin x\\ 再由r(x)=4x+10\sin x可令y=y_h+ y_p=ax+b+ A\cos x+B\sin x+ Cx\cos x+Dx\sin x\\ \Rightarrow y'=a+(D-A)\sin x+ (B+C)\cos x-Cx\sin x+Dx\cos x\\ \Rightarrow y''=(2D-A)\cos x-(B+2C)\sin x-Cx\cos x-Dx\sin x\\ \Rightarrow y''+y=ax+b+ 2D\cos x-2C\sin x =4x+10\sin x\\ \Rightarrow \cases{a=4\\ b=0\\ C=-5\\ D=0} \Rightarrow y=4x+ A\cos x+B\sin x-5x\cos x \Rightarrow y'= 4+(5-A)\sin x+ (B-5)\cos x  \\ 再由初始值\cases{y(\pi)=0 \\ y'(\pi)=2} \Rightarrow \cases{4\pi -A+5\pi =0\\ 4+5-B=2 } \Rightarrow \cases{A=9 \pi\\ B=7} \\\Rightarrow \bbox[red,2pt]{y=4x+9\pi \cos x+7\sin x-5x\cos x}$$

解答：$$\mathbf{(a)}\; \mathcal L\{\cos (5t)\} ={s \over s^2+5^2} =\bbox[red,2pt]{s\over s^2+25} \\\mathbf{(b)}\; \mathcal L^{-1}\{{1\over s-1}\} = \bbox[red, 2pt]{e^t} \\\mathbf{(c)}\;\mathcal L\{y'\} -\mathcal L\{ y\} = 2\mathcal L\{\cos (5t)\} \Rightarrow sY(s)-y(0)-Y(s)={2s\over s^2+25} \\ \quad \Rightarrow Y(s)={2s \over (s^2+25)(s-1)} = \cfrac{-{1\over 13}s+ {25\over 13}}{s^2+25}+ \cfrac{1/13}{ s-1} \\=-{1\over 13}\cdot {s\over s^2+25} +{25\over 13}\cdot {1\over s^2+25}+{1\over 13}\cdot {1\over s-1} \\ \quad\Rightarrow y(t)=-{1\over 13}\mathcal L^{-1}\{{s\over s^2+25} \}+{5\over 13}\mathcal L^{-1}\{ {5\over s^2+25}\}+{1\over 13} \mathcal L^{-1} \{ {1\over s-1} \} \\=-{1\over 13} \cos(5t)+{5\over 13}\sin(5t)+{1\over 13} e^t \\\quad \Rightarrow \bbox[red, 2pt]{y(t)= -{1\over 13} \cos(5t)+{5\over 13}\sin(5t)+{1\over 13} e^t}$$

解答：$$\mathbf{(a)}\;A=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow \begin{vmatrix} -\lambda & 1 & 0 \\ 1 & -\lambda & 0\\ 0 & 0 & 1-\lambda\end{vmatrix} = -(\lambda-1)^2(\lambda+1)=0\\ \quad\Rightarrow A的特徵值＝\bbox[red,2pt]{1,-1} \\\mathbf{(b)}\; \lambda_1=1 \Rightarrow (A-\lambda_1 I)X=0 \Rightarrow \begin{bmatrix} -1 & 1 & 0 \\ 1 & -1 & 0\\ 0 & 0 & 0\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} =0 \Rightarrow x_1=x_2 \\ \quad \Rightarrow 取v_1=\begin{bmatrix} 1/\sqrt 2 \\ 1/\sqrt 2 \\ 0\end{bmatrix},v_2=\begin{bmatrix} 0\\ 0 \\ 1\end{bmatrix} \\ \lambda_2=-1\Rightarrow (A-\lambda_2 I)X=0 \Rightarrow \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0\\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} =0  \Rightarrow \cases{x_1=-x_2\\ x_3=0} \\ \qquad \Rightarrow 取v_3=\begin{bmatrix} 1/\sqrt 2 \\ -1/\sqrt 2 \\ 0\end{bmatrix} \\ 因此單範正交特徵向量\text{(orthonormal eigen vector)}為\bbox[red,2pt]{\begin{bmatrix} 1/\sqrt 2 \\ 1/\sqrt 2 \\ 0\end{bmatrix}, \begin{bmatrix} 0\\ 0 \\ 1\end{bmatrix} ,\begin{bmatrix} 1/\sqrt 2 \\ -1/\sqrt 2 \\ 0\end{bmatrix}}\\ \mathbf{(c)}\;A=\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix}  \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & -1\end{bmatrix}  \begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix} ^{-1}=P\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & -1\end{bmatrix} P^{-1} \\ \Rightarrow \sin(A)= P\begin{bmatrix} \sin(1) & 0 & 0 \\ 0 & \sin(1) & 0\\ 0 & 0 & -\sin(1)\end{bmatrix}  P^{-1} \\ \Rightarrow \sin(A^3) = P\begin{bmatrix} \sin^3(1) & 0 & 0 \\ 0 & \sin^3(1) & 0\\ 0 & 0 & -\sin^3(1)\end{bmatrix}  P^{-1} \\ \Rightarrow \sin(3A)= 3\sin(A)-4\sin^3(A) \\=P \begin{bmatrix} 3\sin(1)-4\sin^3(1) & 0 & 0 \\ 0 & 3\sin(1)-4\sin^3(1) & 0\\ 0 & 0 & -3\sin(1)+4\sin^3(1) \end{bmatrix} P^{-1}\\=\begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 1\\ 0 & 1 & 0\end{bmatrix} \begin{bmatrix} 3\sin(1)-4\sin^3(1) & 0 & 0 \\ 0 & 3\sin(1)-4\sin^3(1) & 0\\ 0 & 0 & -3\sin(1)+4\sin^3(1) \end{bmatrix}  \begin{bmatrix} 1/2 & 1/2 & 0 \\ 0 & 0 & 1\\ -1/2 & 1/2 & 0\end{bmatrix} \\ =\begin{bmatrix} 3\sin(1)-4\sin^3(1) & 0 & 3\sin(1)-4\sin^3(1) \\ 3\sin(1)-4\sin^3(1) & 0 & -3\sin(1)+4\sin^3(1)\\ 0 & 3\sin(1)-4\sin^3(1) & 0 \end{bmatrix}  \begin{bmatrix} 1/2 & 1/2 & 0 \\ 0 & 0 & 1\\ -1/2 & 1/2 & 0\end{bmatrix} \\ = \bbox[red,2pt]{\begin{bmatrix} 0 & 6\sin(1)-8\sin^3(1) & 0 \\ 6\sin(1)-8\sin^3(1) & 0 & 0\\ 0 & 0 & 3\sin(1) -4\sin^3(1)\end{bmatrix}}$$

解答：$$\iint_R(2x-3x^2)dA = \int_{-1}^1 \int_0^1 (2x-3x^2)\,dydx = \int_{-1}^1 (2x-3x^2)\,dx = \left.\left[ x^2-x^3 \right]\right|_{-1}^1 =-2\\ \cases{C_1=\{(2t-1,0)\mid t\in[0,1]\} \\ C_2=\{(1,t)\mid t\in [0.1]\} \\ C_3=\{(-2t+1,1)\mid t\in[0,1] \} \\ C_3=\{(-1,1-t)\mid t\in[0,1]\}} \Rightarrow \oint_C 3x^2ydx +(x^2-5y)dy  =\oint_{C_1} 3x^2ydx +(x^2-5y)dy \\+ \oint_{C_2} 3x^2ydx +(x^2-5y)dy + \oint_{C_3} 3x^2ydx +(x^2-5y)dy+ \oint_{C_4} 3x^2ydx +(x^2-5y)dy \\=0 +\int_0^1(1-5t)dt +\int_0^1 -6(-2t+1)^2\,dt +\int_0^1 5(1-t)-1\,dt \\=\int_0^1 -24t^2+14t-1\,dt =\left.\left[ -8t^3+7t^2-t \right]\right|_0^1 =-2 \\ 因此左式＝右式＝－2，\bbox[red,2pt]{故得證}$$

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