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
(a)
(b)f(x)={x+π,−π<x<0−x+π,0<x<π⇒f(x)為偶函數⇒bn=0,只需計算an;a0=12π∫π−πf(x)dx=12π(∫0−πx+πdx+∫π0−x+πdx)=12π×π2=π2an=1π∫π−πf(x)cos(nx)dx=1π(∫0−π(x+π)cos(nx),dx+∫π0(−x+π)cos(nx)dx)=1π([xnsin(nx)+1n2cos(nx)+πnsin(nx)]|0−π+[−xnsin(nx)−1n2cos(nx)+πnsin(nx)]|π0)=1π⋅2n2(1−(−1)n)={4n2π,n是奇數0,n是偶數⇒f(x)=a0+∞∑n=1ancos(nx)⇒f(x)=π2+2π∞∑n=11n2(1−(−1)n)cos(nx)
解答:先求齊次解,即y″+ω2y=0⇒特徵方程式:λ2+ω2=0⇒λ=±ωi⇒yh=Acos(ωt)+Bsin(ωt)由於r(t)=sint,因此需考慮ω是否等於1;由題意知ω=0.5,1.2,2,10,皆不等於1因此yp=Csint⇒y″p=−Csint⇒−Csint+ω2Csint=sint⇒C(ω2−1)=1⇒C=1/(ω2−1)⇒y=yh+yp⇒y=Acos(ωt)+Bsin(ωt)+1ω2−1sint,A與B皆為常數
解答:y′=2xy⇒∫ydy=∫2xdx⇒12y2=x2+Cy(1)=5⇒252=1+C⇒C=232⇒y2=2x2+23
解答:f=2exyz⇒{fx=2yzexyzfy=2xzexyzfz=2xyexyz⇒{fxx=2y2z2exyzfyy=2x2z2exyzfzz=2x2y2exyz⇒∇2f=fxx+fyy+fzz=2exyz(x2y2+y2z2+z2x2)
解答:假設{→u=[abc]→v=[def]⇒→u×→v=[bf−cecd−afae−bd]⇒(→u×→v)′=[(bf−ce)′(cd−af)′(ae−bd)′]=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(1)而{→u′×→v=[a′b′c′]×[def]=[b′f−c′ec′d−a′fa′e−b′d]→u×→v′=[abc]×[d′e′f′]=[bf′−ce′cd′−af′ae′−bd′]⇒→u′×→v+→u×→v′=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(2)(1)=(2)⇒→u×→v=→u′×→v+→u×→v′,故得證
解答:A=[−1−10−1−10001]⇒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}}
解答:y′=2xy⇒∫ydy=∫2xdx⇒12y2=x2+Cy(1)=5⇒252=1+C⇒C=232⇒y2=2x2+23
解答:f=2exyz⇒{fx=2yzexyzfy=2xzexyzfz=2xyexyz⇒{fxx=2y2z2exyzfyy=2x2z2exyzfzz=2x2y2exyz⇒∇2f=fxx+fyy+fzz=2exyz(x2y2+y2z2+z2x2)
解答:假設{→u=[abc]→v=[def]⇒→u×→v=[bf−cecd−afae−bd]⇒(→u×→v)′=[(bf−ce)′(cd−af)′(ae−bd)′]=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(1)而{→u′×→v=[a′b′c′]×[def]=[b′f−c′ec′d−a′fa′e−b′d]→u×→v′=[abc]×[d′e′f′]=[bf′−ce′cd′−af′ae′−bd′]⇒→u′×→v+→u×→v′=[b′f+bf′−c′e−ce′c′d+cd′−a′f−af′a′e+ae′−b′d−bd′]⋯(2)(1)=(2)⇒→u×→v=→u′×→v+→u×→v′,故得證
解答:A=[−1−10−1−10001]⇒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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