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2023年12月28日 星期四

111年雲林科大電機系-工程數學詳解

 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:1.y=5sin(3x)y=5sin(3x)dx=53cos(3x)+c1y=53cos(3x)+c12.I(x)=e5x4dx=ex5yI(x)+5x4yI(x)=3x2ex5yex5+5x4yex5=3x2(yex5)=3x2yex5=x3+c1y=x3ex5+c1ex53.x2y=y2+xyyyx=y2x2v=1yv=yy2y=y2vy2vyx=y2x2v+1xy=1x2v+vx=1x2xv+v=1x(xv)=1xxv=lnx+c1xy=lnx+c1y=xlnx+c1
解答:,y+9y=0λ2+9=0λ=±3iyh=c1cos(3x)+c2sin(3x)yp=Acosx+Bsinxyp=Asinx+Bcosxyp=AcosxBsinxyp+9yp=8Acosx+8Bsinx=16sinx{A=0B=2yp=2sinxy=yh+yp=c1cos(3x)+c2sin(3x)+2sinxy=3c1sin(3x)+3c2cos(3x)+2cosx{y(π)=c1=0y(0)=3c2+2=0{c1=0c2=23y=23sin(3x)+2sinx
解答:(1)F(s)=L{t2}+L{e6t}+L{sin(6t)}=2s3+1s+6+6s2+36(2)f(t)=L1{8s(s2)2}=L1{2s2s2+4(s2)2}=22e2t+4e2tt(3)L{y}+L{t0y(τ)dτ}=2L{1}2L{sint}sY(s)+Y(s)=2s2s2+1Y(s)=2s(s+1)2(s2+1)(s+1)y(t)=L1{2s(s+1)}L1{2(s2+1)(s+1)}=2L1{1s1s+1}L1{1s2+1ss2+1+1s+1}=22et(sintcost+et)y(t)=23etsint+cost


解答:y3y=te3tsin(t)L{y}3L{y}=L{te3tsin(t)}sY(s)3Y(s)=ddsL{e3tsin(t)}=dds1(s3)2+1=2s6(s26s+10)2Y(s)=2(s26s+10)2y(t)=L1{2(s26s+10)2}=L1{2((s3)2+1)2}=e3tL1{2(s2+1)2}=e3t(sinttcost)y(t)=e3t(sinttcost)
解答:(a)A=[003021201]det\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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