臺灣綜合大學系統112學年度學士班轉學生聯合招生
類組代碼：D36
科目名稱：工程數學
解答：$$\mathbf{(a)}\; 先求齊次解,y'''+y'=0 \Rightarrow \lambda^3+\lambda=0 \Rightarrow \lambda(\lambda^2+1)=0 \Rightarrow \lambda=0,\pm i\\ \Rightarrow y_h=c_1+ c_2\cos x+c_3\sin x\\又r(x)=\sin x \Rightarrow 非齊次解y_p=Ax\sin x \Rightarrow y_p'=A\sin x+Ax\cos x \\\Rightarrow y_p''=2A\cos x-Ax\sin x \Rightarrow y_p'''=-3A\sin x-Ax\cos x\\ \Rightarrow y_p'''+y_p'= -2A\sin x=\sin x \Rightarrow A=-{1\over 2} \Rightarrow y_p=-{1\over 2}x\sin x \\ \Rightarrow y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y=c_1+c_2\cos x+c_3\sin x-{1\over 2}x\sin x}\\ \mathbf{(b)}\; L\{ y''-4y\} =L\{ -7e^{2x}+x\} \Rightarrow s^2Y(s)-sy(0)-y'(0)-4Y(s)=-{7\over s-2}+{1\over s^2} \\ \Rightarrow (s^2-4)Y(s)=s+3-{7\over s-2}+{1\over s^2} \Rightarrow Y(s)= {s+3\over s^2-4}-{7\over (s-2)^2(s+2)}+ {1\over s^2(s^2-4)} \\ \Rightarrow y(x)=L^{-1} \left\{-{1\over 4(s+2)}+{5\over 4(s-2)} -{7\over 16(s-2)}- {7\over 4(s-2)^2} -{7\over 16(s+2)}\\ \qquad - {1\over 4s^2} -{1\over 16(s+2)} +{1\over 16(s-2)} \right\} \\=-{1\over 4}e^{-2x} +{5\over 4}e^{2x}+ {7\over 16}e^{2x}- {7\over 4}xe^{2x}-{7\over 16}e^{-2x}-{x\over 4}-{1\over 16}e^{-2x}+{1\over 16}e^{2x}\\ \Rightarrow \bbox[red, 2pt]{y={{1\over 4}(-3e^{-2x}+ 7e^{2x}-7xe^{2x}-x)}}$$
解答：$$P=A(A^T A)^{-1}A^T \Rightarrow P^2 =A(A^T A)^{-1}A^TA(A^T A)^{-1}A^T =A(A^T A)^{-1}A^T =P \\\Rightarrow P^3=P =\bbox[red,2pt]{{1\over 6} \begin{bmatrix}5 & 2& -1 \\2 & 2& 2\\ -1& 2& 5 \end{bmatrix}}$$
解答：$$若A=\begin{bmatrix}a & b \\c & d \end{bmatrix} \Rightarrow A^{-1}=\begin{bmatrix}{d\over ad-bc} & -{b\over ad-bc} \\-{c \over ad-bc} & {a \over ad-bc} \end{bmatrix} \Rightarrow \det(A^{-1})={ad-bc \over (ad-bc)^2}=\bbox[red, 2pt]{1\over ad-bc} \ne 1$$
解答：$$b_n={2\over \pi}\int_0^\pi 1\cdot \sin(nx)\,dx = {2\over \pi} \left. \left[-{1\over n} \cos(nx) \right] \right|_0^\pi ={2\over n\pi}(1-(-1)^n)= {4\over n\pi}, n為奇數\\ \Rightarrow f(x)=1= \sum_{n=1}^\infty b_n\sin(nx) = \bbox[red, 2pt]{{4\over \pi}(\sin(x)+{1\over 3}\sin(3x)+ {1\over 5}\sin(5x)+\cdots)}\\ \Rightarrow f(1)=1={4 \over \pi}(\sin 1+{1\over 3}\sin 3+{1\over 5}\sin 5+\cdots)\\ \Rightarrow \sin 1+{1\over 3}\sin 3+{1\over 5}\sin 5+\cdots={4\over \pi}, \bbox[red,2pt]{故得證}$$
解答：$$假設u(x,t)= X(x)T(t),則u_t=ku_{xx} \Rightarrow ＸT'=kＸ''Ｔ \Rightarrow {T'\over kT} ={X''\over X}=\lambda 為一常數\\ \text{case I }{\lambda \gt 0:} X''-\lambda X=0 \Rightarrow X= c_1e^{\sqrt \lambda x} +c_2e^{-\sqrt \lambda x} \Rightarrow X'=c_1\sqrt \lambda e^{\sqrt \lambda x}- c_2\sqrt \lambda e^{-\sqrt \lambda x}\\ \qquad 初始值\cases{X'(0)=0 \\ X'(L)=0} \Rightarrow \cases{c_1\sqrt \lambda-c_2\sqrt \lambda=0 \cdots(1)\\ c_1\sqrt \lambda e^{\sqrt \lambda L}-c_2\sqrt \lambda e^{-\sqrt \lambda L}=0 \cdots(2)}\\ \qquad 由(1)得c_1=c_2 代入(2) \Rightarrow c_1\sqrt \lambda(e^{2\sqrt \lambda L}-1)=0 \Rightarrow c_1=0 \Rightarrow X=0 明顯解不討論\\ \text{Case II }\lambda=0: X''=0 \Rightarrow X=c_1x+c_2
\Rightarrow X'=c_1\\ \qquad 初始值\cases{X'(0)=0 \\ X'(L)=0} \Rightarrow c_1=0 \Rightarrow X=c_2,又T'=0 \Rightarrow T=c_3 \\ \qquad u(x,t)=c_2c_3,此時我們可假設u(x,t)=C常數解\\ \text{Case III }\lambda=-\rho^2 \lt 0: X''+\rho^2 X=0 \Rightarrow X=A\cos \rho x+ B\sin \rho x \Rightarrow X'=-A\rho \sin \rho x+B \rho \cos \rho x\\ \qquad 初始值\cases{X'(0)=0\\ X'(L)=0} \Rightarrow \cases{B\rho =0 \Rightarrow B=0\\ -A\rho \sin \rho L +B\rho \cos \rho L=0} \Rightarrow \cases{A=B=0 \Rightarrow X=0明顯解,不討論\\ \sin \rho L=0} \\ \qquad \Rightarrow \rho L=n\pi \Rightarrow \rho ={n\pi\over L} \Rightarrow X=\cos {n\pi \over L}x ,n\in \mathbb N,此時T'+\rho^2k T=0 \Rightarrow T=Ce^{-\rho^2kt}\\ \qquad \Rightarrow u_n(x,t)=Ce^{-\rho^2kt} \cos n\pi x/L\\ 將\lambda=0,\lambda=-\rho^2的結果合併,可得 u(x,t)=a_0+ \sum_{n=1}^\infty a_n e^{-kn^2\pi^2t/L^2} \cos(n\pi x/L) \\ \Rightarrow \lim_{t\to \infty}u(x,t)=a_0 ={1\over L}\int_0^L f(x)\,dx 也就是f(x)在區間(0,L)與x軸所圍的面積=\bbox[red,2pt]{平均溫度}$$
解答：$$\mathbf{(a)}\;f(x,y)=\ln(x^2+y^2+1)+e^{2xy} \Rightarrow \cases{f_x=2x/(x^2+y^2+1) +2ye^{2xy} \\ f_y=2y/(x^2+y^2+1) +2xe^{2xy}} \\ \Rightarrow \text{gradient of }f= \nabla f=(f_x,f_y) \Rightarrow \nabla f(0,-2)=\bbox[red, 2pt]{(-4,-4/5)} \\\mathbf{(b)}\; \nabla f(0,-2)\cdot {\mathbf v\over \Vert \mathbf v\Vert} =(-4,-{4\over 5}) \cdot {(7,-24) \over \sqrt{7^2+24^2}} =(-4,-{4\over 5}) \cdot ({7\over 25},-{24\over 25})= \bbox[red, 2pt]{4\over 5} \\ \mathbf{(c)}\; \text{minimum value of the directional derivative} =-\nabla f(0,-2)\cdot {\nabla f(0,-2) \over \Vert \nabla f(0,-2)\Vert} \\=-\sqrt{(-4)^2+(-{4\over 5})^2}=-\sqrt{416\over 25} =\bbox[red, 2pt]{-{４\sqrt{26}}\over 5}$$ ==================== END ====================
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