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2022年9月24日 星期六

108年台綜大轉學考-工程數學D04詳解

臺灣綜合大學系統108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:D04

解答
(a)

(b)f(x)={x+π,π<x<0x+π,0<x<πf(x)bn=0an;a0=12πππf(x)dx=12π(0πx+πdx+π0x+πdx)=12π×π2=π2an=1πππf(x)cos(nx)dx=1π(0π(x+π)cos(nx),dx+π0(x+π)cos(nx)dx)=1π([xnsin(nx)+1n2cos(nx)+πnsin(nx)]|0π+[xnsin(nx)1n2cos(nx)+πnsin(nx)]|π0)=1π2n2(1(1)n)={4n2π,n0,nf(x)=a0+n=1ancos(nx)f(x)=π2+2πn=11n2(1(1)n)cos(nx)
解答y+ω2y=0:λ2+ω2=0λ=±ωiyh=Acos(ωt)+Bsin(ωt)r(t)=sintω1;ω=0.5,1.2,2,10,1yp=Csintyp=CsintCsint+ω2Csint=sintC(ω21)=1C=1/(ω21)y=yh+ypy=Acos(ωt)+Bsin(ωt)+1ω21sint,AB
解答y=2xyydy=2xdx12y2=x2+Cy(1)=5252=1+CC=232y2=2x2+23
解答f=2exyz{fx=2yzexyzfy=2xzexyzfz=2xyexyz{fxx=2y2z2exyzfyy=2x2z2exyzfzz=2x2y2exyz2f=fxx+fyy+fzz=2exyz(x2y2+y2z2+z2x2)
解答{u=[abc]v=[def]u×v=[bfcecdafaebd](u×v)=[(bfce)(cdaf)(aebd)]=[bf+bfcececd+cdafafae+aebdbd](1){u×v=[abc]×[def]=[bfcecdafaebd]u×v=[abc]×[def]=[bfcecdafaebd]u×v+u×v=[bf+bfcececd+cdafafae+aebdbd](2)(1)=(2)u×v=u×v+u×v
解答A=[110110001]det
解答\cases{y_1'= y_2+2-u(t-1)\\ y_2'= -y_1+1-u(t-1)} \Rightarrow \begin{bmatrix} y_1'\\ y_2'\end{bmatrix} = \begin{bmatrix} 0 & 1\\ -1& 0\end{bmatrix}  \begin{bmatrix} y_1\\ y_2\end{bmatrix} + \begin{bmatrix} 2-u(t-1)\\ 1-u(t-1) \end{bmatrix} \equiv y'=Ay+ \mathbf b\\ A= \begin{bmatrix} 0 & 1\\ -1& 0\end{bmatrix} \Rightarrow \det(A-\lambda I)=\lambda^2+1=0 \Rightarrow \lambda=\pm i\\  \lambda_1=i \Rightarrow (A-\lambda_1 I)\mathbf x=\begin{bmatrix} -i & 1\\ -1& -i\end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow x_1=-ix_2,取v_1=\begin{bmatrix} -i\\ 1\end{bmatrix}\\ \lambda_2=-i \Rightarrow (A-\lambda_2 I)\mathbf x=\begin{bmatrix} i & 1\\ -1& i\end{bmatrix}\begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow x_1= ix_2,取v_2=\begin{bmatrix} i\\ 1\end{bmatrix} \\ 因此y'=Ay 的齊次解為\Rightarrow \begin{bmatrix} y_1\\ y_2\end{bmatrix} =C_1e^{it}\begin{bmatrix} -i\\ 1\end{bmatrix} +C_2e^{-it} \begin{bmatrix} i\\ 1\end{bmatrix} \Rightarrow \cases{y_1=-C_1ie^{it} +C_2ie^{-it}\\ y_2= C_1e^{it}+ C_2e^{-it}} \\ \Rightarrow \cases{y_1'=  C_1e^{it}+ C_2e^{-it}+2-u(t-1)\\ y_2'=C_1 ie^{it} -C_2ie^{-it} +1-u(t-1)} \Rightarrow   \cases{\mathcal L\{y_1'\}= \mathcal L\{ C_1e^{it}+ C_2e^{-it}+2-u(t-1) \}\\ \mathcal L\{y_2'\}= \mathcal L\{C_1 ie^{it} -C_2ie^{-it} +1-u(t-1)\} } \\ \Rightarrow \cases{sY_1(s)-y_1(0) ={C_1\over s-i} +{C_2\over s+i} +{2\over s}-e^{-s}/s \\ sY_2(s)-y_2(0)= {C_1i\over s-i}-{C_2 i\over s+i}+ {1\over s}-e^{-s}/s} \Rightarrow \cases{Y_1(s) = {C_1\over s(s-i)} +{C_2\over s(s+i)} +{2\over s^2}+{2\over s}-e^{-s}/s^2 \\ Y_2(s)= {C_1i\over s(s-i)}-{C_2 i\over s(s+i)}+ {1\over s^2}+ {1\over s}-e^{-s}/s^2} \\ \Rightarrow  \cases{Y_1(s) = {C_1\over i}({1\over s-i}-{1\over s}) +{C_2\over i}({1\over s }-{1\over s+i}) +{2\over s^2}+{2\over s}-e^{-s}/s^2 \\ Y_2(s)= C_1({1\over s-i}-{1\over s})-C_2 ({1\over s}-{1\over s+i})+ {1\over s^2}+ {1\over s}-e^{-s}/s^2}  \\ \Rightarrow \cases{y_1= {C_1\over i}(e^{it}-1 ) +{C_2\over i}(1-e^{-it}) +2t+ 2-(t-1)u(t-1) \\ y_2= C_1(e^{it}-1)-C_2 (1-e^{-it})+ t+ 1-(t-1)u(t-1)}
解答A= \left[\begin{matrix}1 & 0 & 3\\1 & 1 & 1\\0 & 1 & 0\end{matrix}\right] \Rightarrow rref(A)=\left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right] \Rightarrow Rank(A)=3 \Rightarrow \bbox[red,2pt]{線性獨立}
解答(2,0, a_1)\cdot (1,0,8) = 2+8a_1=0 \Rightarrow a_1= \bbox[red,2pt]{-{1\over 4}}
解答\cases{3x+y+ 2z=1\\ x-y+3z=-3 \\ y-2z=1} \Rightarrow \begin{bmatrix}3 & 1 &2 \\ 1 & -1 & 3\\ 0 & 1 &-2 \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 1\\ -3\\1\end{bmatrix} \equiv A\mathbf x=\mathbf b \\ \Rightarrow \cases{A_x= \begin{bmatrix}1 & 1 &2 \\ -3 & -1 & 3\\ 1 & 1 &-2 \end{bmatrix} \\[1ex] A_y= \begin{bmatrix}3 & 1 &2 \\ 1 & -3 & 3\\ 0 & 1 &-2 \end{bmatrix} \\[1ex] A_z =\begin{bmatrix}3 & 1 &1 \\ 1 & -1 & -3\\ 0 & 1 &1 \end{bmatrix}} \Rightarrow \cases{\det(A_x)= -8 \\ \det(A_y) =13 \\\det(A_z) =6 \\\det(A)=1} \Rightarrow \cases{x=\det(A_x)/\det(A)= -8\\ y=\det(A_y) /\det(A) = 13\\ z= \det(A_z) / \det(A) = 6}\\ \Rightarrow \bbox[red,2pt]{\cases{x=  -8\\ y=  13\\ z=   6}}

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