解答:$$\mathbf{(1)}\; y'=5\sin 3x \Rightarrow y=\int 5\sin(3x)\,dx =-{5\over 3}\cos (3x)+c \Rightarrow \bbox[red,2pt]{y= -{5\over 3}\cos (3x)+c}\\ \mathbf{(2)}\; y'+y=e^{5x} \Rightarrow \text{integrating factor }I(x)=e^x \Rightarrow y'e^x+ye^x=e^{6x} \\ \quad \Rightarrow (ye^x)'=e^{6x} \Rightarrow ye^x = \int e^{6x}={1\over 6}e^{6x}+c \Rightarrow \bbox[red, 2pt]{y={1\over 6}e^{5x}+{c\over e^x} }\\ \mathbf{(3)}\; u=x+y+3 \Rightarrow u'=1+y' \Rightarrow u'-1=u^2 \Rightarrow {1\over u^2+1}du =dx\\ \quad \Rightarrow \tan^{-1}u =x+c \Rightarrow u=x+y+3=\tan(x+c) \Rightarrow \bbox[red, 2pt]{y= \tan(x+c)-x-3}$$
解答:$$\mathbf{(1)}\; L\{f(t)\} =L\{te^{4t}\}+ L\{e^{2t}\sin(t)\}= \bbox[red, 2pt]{{1\over (s-4)^2}+{1\over (s-2)^2+1} }\\\mathbf{(2)}\; f(t)=L^{-1}\left\{{e^{-2s}\over s(s-1)} \right\} =L^{-1}\left\{e^{-2s}\left({1\over s-1}-{1\over s}\right) \right\} =u(t-2)L^{-1}\left\{ {1\over s-1}-{1\over s} \right\}(t-2) \\\quad =u(t-2)\left( e^{t-2}-u(t-2)\right) =\bbox[red, 2pt]{u(t-2)(e^{t-2}-1)} \\\mathbf{(3)}\; F(s)= L\left\{\int_0^t \sin \tau \cos(t-\tau) \,d\tau\right\} =L\{\sin t\}\cdot L\{\cos t\}={1\over s^2+1}\cdot {s\over s^2+1} =\bbox[red, 2pt]{2\over (s^2+1)^2}$$
解答:$$L\{y''\}-L\{y'\}= L\{t^t\cos t\} \Rightarrow s^2Y(s)-sy(0)-y'(0)-(sY(s)-y(0)) ={s-1\over (s-1)^2+1}\\ \Rightarrow (s^2-s)Y(s)={s-1\over (s-1)^2+1} \Rightarrow Y(s)={1\over s((s-1)^2+1)} \\ \Rightarrow y(t)=L^{-1}\left\{ {1\over s((s-1)^2+1)} \right\}=L^{-1}\left\{ {1\over 2s}-{1\over 2}\cdot {s-1\over (s-1)^2+1 }+{1\over 2} \cdot{1\over (s-1)^2+1 } \right\} \\\Rightarrow \bbox[red, 2pt]{y(t)={1\over 2}-{1\over 2}e^t \cos t+{1\over 2}e^t\sin t}$$
解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''= m(m-1)x^{m-2} \Rightarrow 2x^2y''+5xy'+y= 2m(m-1)x^m+5mx^{m}+x^m \\=(2m^2+3m+1)x^m=0 \Rightarrow (2m+1)(m+1)x^m=0 \Rightarrow m=-1,-1/2 \Rightarrow y_h=c_1x^{-1}+c_2x^{-1/2}\\ y_p=ax^2+bx+c \Rightarrow y_p'=2ax+b \Rightarrow y_p''=2a \Rightarrow 2x^2y_p''+ 5xy_p'+y_p=15ax^2+ 6bx+c=x^2-x \\ \Rightarrow \cases{15a=1\\ 6b=-1\\c=0} \Rightarrow \cases{a=1/15\\ b=-1/6\\ c=0} \Rightarrow y_p={1\over 15}x^2-{1\over 6}x \Rightarrow y=y_h+y_p\\ \Rightarrow \bbox[red, 2pt]{y={c_1\over x}+{c_2\over \sqrt x}+ {1\over 15}x^2-{1\over 6}x}$$
解答:$$$$
解答:$$A=\begin{bmatrix}1 & 2 & 2 & -1 \\3 & 6 &5 &0 \\1 &2 &1 &2\end{bmatrix} \xrightarrow{-R_1+R_3\to R_3, -3R_1+R_2\to R_2} \begin{bmatrix}1 & 2 & 2 & -1 \\0 & 0 &-1 &3 \\0 &0 &-1 &3 \end{bmatrix} \xrightarrow{-R_2+R_3 \to R_3} \begin{bmatrix}1 & 2 & 2 & -1 \\0 & 0 &-1 &3 \\0 &0 &0 &0 \end{bmatrix} \\ \xrightarrow{2R_2+R_1\to R_1, -R_2\to R_2}\begin{bmatrix}1 & 2 & 0 & 5 \\0 & 0 &1 &-3 \\0 &0 &0 &0 \end{bmatrix} \Rightarrow \bbox[red,2pt]{rank(A)=2}\\ \Rightarrow \text{the bases for the row space of }A: \bbox[red, 2pt]{\{(1,2,0,5),(0,0,1,-3)\}},\\ \text{and for the column space of }A: \bbox[red, 2pt]{\{\begin{pmatrix}1\\3\\1 \end{pmatrix}, \begin{pmatrix}2 \\5 \\1 \end{pmatrix}\}}$$
解答:$$A=\begin{bmatrix}1 & 1 & -1 & -1 \\3 & 2 &0 &1 \\1 &0 &1 &0\end{bmatrix} \xrightarrow{-R_1+R_3\to R_3, -3R_1+R_2\to R_2} \begin{bmatrix}1 & 1 & -1 & -1 \\0 & -1 &3 &4 \\0 &-1 &2 &1 \end{bmatrix} \xrightarrow{-R_2+R_3\to R_3} \begin{bmatrix}1 & 1 & -1 & -1 \\0 & -1 &3 &4 \\0 &0 &-1 &-3 \end{bmatrix} \\ \xrightarrow{R_2+R_1\to R_1,3R_3+R_2\to R_2} \begin{bmatrix}1 & 0 & -2 & -3 \\0 & -1 &0 &-5 \\0 &0 &-1 &-3 \end{bmatrix} \xrightarrow{-2R_3+R_1\to R_1} \begin{bmatrix}1 & 0 & 0 & 3 \\0 & -1 &0 &-5 \\0 &0 &-1 &-3 \end{bmatrix} \\ \xrightarrow{-R_2\to R_2,-R_3\to R_3} \begin{bmatrix}1 & 0 & 0 & 3 \\0 & 1 &0 &5 \\0 &0 &1 &3 \end{bmatrix} \Rightarrow \text{basis for row space of }A:\{\vec a_1,\vec a_2,\vec a_3\},\text{ where }\cases{\vec a_1=(1,0,0,3)\\ \vec a_2 =(0,1,0,5)\\ \vec a_3=(0,0,1,3)}\\ \text{Gram-Schmidt process: }\\\vec u_1=\vec a_1 \Rightarrow \vec e_1={\vec u_1\over \Vert \vec u_1\Vert}={1\over \sqrt{10}}(1,0,0,3)\\ \vec u_2 = \vec a_2-(\vec a_2\cdot \vec e_1)\vec e_1 = (0,1,0,5)-{15\over \sqrt{10}}\cdot {1\over \sqrt{10}}(1,0,0,3) =(-{3\over 2},1,0,{1\over 2}) \\\qquad \Rightarrow \vec e_2={\vec u_2\over \Vert \vec u_2\Vert}={1\over \sqrt{14}}(-3,2,0,1) \\ \vec u_3=\vec a_3-(\vec a_3\cdot \vec e_1)\vec e_1 -(\vec a_3\cdot \vec e_2)\vec e_2 =(0,0,1,3)-{9\over \sqrt{10}}\cdot {1\over \sqrt{10}}(1,0,0,3)-{3\over \sqrt{14}}\cdot {1\over \sqrt{14}}(-3,2,0,1) \\ \qquad = (-{9\over 35},-{3\over 7},1,{3\over 35}) \Rightarrow \vec e_3= {\vec u_3\over \Vert \vec u_3\Vert } ={1\over 2\sqrt{385}}(-9,-15,35,5)\\ \Rightarrow \bbox[red, 2pt]{ \text{orthogonal basis: }\{\vec e_1,\vec e_2,\vec e_3\},\text{ where }\cases{\vec e_1={1\over \sqrt{10}}(1,0,0,3)\\\vec e_2= {1\over \sqrt{14}}(-3,2,0,1)\\ \vec e_3={1\over 2\sqrt{385}}(-9,-15,35,5)}}$$
解答:$$\cases{X=\begin{bmatrix}1& -1\\ -1& 1\end{bmatrix}\\[1ex] Y=\begin{bmatrix}1& -1\end{bmatrix}\\[1ex] Z=[-2]} \Rightarrow \cases{X^2=\begin{bmatrix}2 & -2\\ -2& 2 \end{bmatrix}\\[1ex]\\ YX+ZY=0\\ Z^2=[4]} \Rightarrow \cases{X^4= \begin{bmatrix}8 & -8\\ -8 & 8\end{bmatrix}\\[1ex] Z^4=[16]} \Rightarrow \cases{X^8= \begin{bmatrix}128 & -128\\ -128 & 128\end{bmatrix}\\[1ex] Z^4=[256]}\\因此 A=\begin{bmatrix}X& 0\\ Y& Z \end{bmatrix} \Rightarrow A^2= \begin{bmatrix}X^2& 0\\ 0& Z^2 \end{bmatrix} \Rightarrow A^4= \begin{bmatrix}X^4& 0\\ 0& Z^4 \end{bmatrix} \Rightarrow A^8= \begin{bmatrix}X^8& 0\\ 0& Z^8 \end{bmatrix} \\ =\bbox[red,2pt]{\begin{bmatrix}128& -128& 0\\ -128& 128& 0\\ 0 & 0 & 256\end{bmatrix}}$$
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