113年中正大學電機碩士班-線性代數與微分方程詳解

 國立中正大學113學年度碩士班招生考試

科目名稱:線性代數與微分方程
系所組別:電機工程學系

解答: $$\textbf{a. }\cases{a=1\\ b=0} \Rightarrow \cases{C=\begin{bmatrix} 1 & 0\\ 1& 1\end{bmatrix} \\D= \begin{bmatrix} 1 & 0\\ 0& 0\end{bmatrix} \\ E=\begin{bmatrix} 0 & 1\\ 1& 0\end{bmatrix} } \Rightarrow \cases{rref(C)= rref(E)=I_2 \Rightarrow rank(C)= rank(E)=2\\ rank(D)=1} \\\qquad \Rightarrow \bbox[red, 2pt]{\text{C and E are row equivalent} }\\ \textbf{b. }\cases{a=0\\ b=1} \Rightarrow \cases{C=\begin{bmatrix} 0 & 1\\ 1& 1\end{bmatrix} \Rightarrow rref(C)= I_2\\D= \begin{bmatrix} 0 & 0\\ 0& 1\end{bmatrix} \\ E=\begin{bmatrix} 0 & 0\\ 1& 1\end{bmatrix} } \Rightarrow \cases{Null(C)= \{\vec 0\} \\ Null(D)= span\{\begin{pmatrix} 1\\0\end{pmatrix}\} \\ Nul(E)= span\{\begin{pmatrix} 1\\ -1 \end{pmatrix}\}} \\\qquad \Rightarrow \bbox[red, 2pt]{\text{matrix C} } \\\textbf{c. }  \bbox[red, 2pt]{\text{C,D, E}} \text{ satisfy rank+nullity=2} \\\textbf{d. }\cases{a=1\\ b=1} \Rightarrow \cases{C=\begin{bmatrix} 1 & 1\\ 1& 1\end{bmatrix}  \\D= \begin{bmatrix} 1 & 0\\ 0& 1\end{bmatrix} \\ E=\begin{bmatrix} 0 & 1\\ 1& 1\end{bmatrix} } \Rightarrow \cases{rref(C)= \begin{bmatrix} 1 & 1\\ 0& 0\end{bmatrix} \\ rref(D)=I_2\\ rref(E)= I_2} \\ \Rightarrow \bbox[red, 2pt]{\text{C is not full-rank}}$$

解答: $$A \text{ is 2 x 3} \Rightarrow B \text{ is 3 x 2} \Rightarrow B=\begin{bmatrix} a& b\\ c& d\\ 0 & 0\end{bmatrix}   \Rightarrow \cases{AB= \begin{bmatrix} 0& 1& 1\\ 1& 0 & 1 \end{bmatrix} \begin{bmatrix} a& b\\ c& d\\ 0 & 0\end{bmatrix} =\begin{bmatrix} c& d\\ a& b \end{bmatrix} \\BA= \begin{bmatrix} a& b\\ c& d\\ 0 & 0\end{bmatrix} \begin{bmatrix} 0& 1& 1\\ 1& 0 & 1 \end{bmatrix} =\begin{bmatrix} b& a& a+b\\ d& c & c+d\\0 & 0& 0 \end{bmatrix}} \\ \Rightarrow \det(AB)=bc-ad =\det(BA)=0 \Rightarrow bc=ad \Rightarrow \bbox[red, 2pt]{\det(B)=0}$$

解答: $$\textbf{a. }\cases{B=\{1,2x\} \\B'=\{1,x,2x^2\}} \Rightarrow \cases{T_1(1)=x\\ T_1(2x)= 2x^2} \Rightarrow \cases{[T_1(1)]_{B'}= \begin{bmatrix}0\\ 1\\ 0 \end{bmatrix} \\ [T_1(2x)]_{B'}= \begin{bmatrix}0\\ 0\\ 1 \end{bmatrix}}  \Rightarrow [T_1]_{B\to B'} = \bbox[red, 2pt]{\begin{bmatrix}0 & 0\\ 1& 0\\ 0 & 1\end{bmatrix}} \\ \textbf{b. } [T_1]_{B\to B'} = \begin{bmatrix}0 & 0\\ 1& 0\\ 0 & 1\end{bmatrix} \Rightarrow T_1(\begin{bmatrix}a\\ b\end{bmatrix}) =\begin{bmatrix}  0\\ a\\ b\end{bmatrix} =ax+b\cdot(2x^2) \\ \Rightarrow T_2(T_1( \begin{bmatrix} a\\ b\end{bmatrix})) =T_2(ax+2bx^2) =a(x+1)+2b(x+1)^2=a+2b +(a+4b)x+ 2bx^2 =\begin{bmatrix}a+2b\\ a+4b \\ b\end{bmatrix}_{B'} \\ \Rightarrow [T_2\circ T_1]_{B\to B'} =\bbox[red, 2pt]{\begin{bmatrix}1 & 2\\ 1& 4\\ 0 & 1\end{bmatrix}}$$

解答: $$\cases{y'' \Rightarrow \text{second order} \\ yy' \Rightarrow \text{ nonlinear}} ,故選\bbox[red, 2pt]{(e)}$$

解答: $$\textbf{a. }\times: y^2y' \Rightarrow \text{ nonlinear} \\\textbf{b. }\bigcirc:(x^2+y^2)y'-xy=0\\ \textbf{c. }\times: (x^2+y^2)dy =xydx \\\textbf{d. }\times: (x^2+y^2)dy-xydx=0 \Rightarrow \cases{P(x,y)=x^2+y^2\\ Q(x,y)=-xy} \Rightarrow \cases{P_x=2x\\ Q_y=-x} \Rightarrow P_x\ne Q_y, \text{not exact} \\ \textbf{e. }\times: y'-{x\over x^2+y^2}y=0 \\故選\bbox[red, 2pt]{(b)}$$

解答: $$y''+y=0 \Rightarrow \lambda^2+1=0 \Rightarrow \lambda=\pm i \Rightarrow y_h= c_1 \cos x+c_2 \sin x \\ y_p=Ax\cos x+ Bx\sin x + C\cos (2x)+D \sin(2x) \\ \Rightarrow y_p'=A\cos x-Ax\sin x+B\sin x+ Bx\cos x-2C\sin(2x)+2D\cos(2x) \\ \Rightarrow y_p''=-2A\sin x -Ax\cos x+2B\cos x-Bx\sin x-4C \cos(2x)-4D\sin(2x) \\ \Rightarrow y_p''+y_p=-2A\sin x+2B\cos x-3C\cos(2x)-3D\sin(2x) =8\cos(2x)-4\sin x\\ \Rightarrow \cases{A=2\\B=0\\ C=-8/3\\ D=0} \Rightarrow y_p=2x\cos x-{8\over 3}\cos(2x) \Rightarrow y=y_h+ y_p \\ \Rightarrow y=c_1 \cos x+c_2 \sin x +2x\cos x-{8\over 3}\cos(2x) \\\Rightarrow y'=-c_1\sin x+ c_2\cos x+2\cos x-2x\sin x+{16\over 3}\sin(2x) \\ \Rightarrow \cases{y(\pi/2)= c_2+8/3=-1\\ y'(\pi/2)=-c_1-\pi =0} \Rightarrow \cases{c_1=-\pi\\ c_2= -11/3 }\\ \Rightarrow \bbox[red, 2pt]{y=-\pi\cos x-{11\over 3}\sin x+2x\cos x-{8\over 3}\cos(2x)}$$

解答: $$L\{y'\}+ L\{y\}= L\{e^{-3t} \cos(2t)\} \Rightarrow sY(s)-y(0)+Y(s)= {s+3 \over (s+3)^2+2^2} \\\Rightarrow Y(s)= {s+3 \over ((s+3)^2+4)(s+1)} \Rightarrow y(t)= L^{-1}\{Y(s)\} = L^{-1}\left\{ {s+3 \over ((s+3)^2+4)(s+1)} \right\}\\ =L^{-1} \left\{ {1\over 4(s+1)} -{1\over 4} \cdot {s+3\over (s+3)^2+4} +{1\over 4} \cdot {2\over (s+3)^2+4}\right\}\\ \Rightarrow \bbox[red, 2pt]{y(t)={1\over 4} e^{-t}-{1\over 4}e^{-3t} \cos(2t)+{1\over 4} e^{-3t}\sin(2t)}$$

解答: $$\cases{(D+5)x+y=0\\ -4x+(D+1)y=0} \Rightarrow \cases{(D+1)(D+5)x+(D+1)y=0\\ -4x+(D+1)y=0} \Rightarrow (D^2+6D+5)x+4x=0 \\ \Rightarrow (D+3)^2x=0 \Rightarrow x=c_1e^{-3t}+ c_2 te^{-3t} \Rightarrow Dx=(-3c_1+c_2)e^{-3t}  -3c_2te^{-3t}\\ \Rightarrow  (D+5)x =(2c_1+c_2)e^{-3t}  +2c_2te^{-3t} \Rightarrow y=-(D+5)x =-(2c_1+c_2)e^{-3t}  -2c_2te^{-3t}\\ \Rightarrow \bbox[red, 2pt]{\cases{x(t) =c_1e^{-3t}+ c_2 te^{-3t}\\ y(t)=-(2c_1+c_2)e^{-3t}  -2c_2te^{-3t}}}$$

解答: $$y'''+4y'=0 \Rightarrow \lambda^3+4\lambda=0 \Rightarrow \lambda(\lambda^2+4)=0 \Rightarrow \lambda=0, \pm   2 i \Rightarrow y_c=c_1\cos  2x+ c_2\sin 2x +c_3\\ \Rightarrow \cases{y_1=\cos 2x\\ y_2=\sin 2x\\ y_3=1} \Rightarrow W= \begin{vmatrix} y_1& y_2& y_3 \\ y_1' & y_2'& y_3' \\y_1'' & y_2'' & y_3'' \end{vmatrix} = \begin{vmatrix} \cos 2x& \sin 2x& 1 \\ -2\sin 2x & 2\cos 2x& 0 \\-4\cos 2x & -4\sin 2x & 0 \end{vmatrix}  = \begin{vmatrix}    -2\sin 2x & 2\cos 2x  \\-4\cos 2x & -4\sin 2x  \end{vmatrix}=8\\ \Rightarrow \cases{W_1= \begin{vmatrix} y_2& y_3\\ y_2'& y_3'\end{vmatrix} = \begin{vmatrix} \sin 2x& 1\\ 2\cos 2x& 0\end{vmatrix} =-2\cos 2x \\ W_2=\begin{vmatrix} y_1& y_3\\ y_1'& y_3'\end{vmatrix} = \begin{vmatrix} \cos 2x& 1\\ -2\sin 2x& 0\end{vmatrix} =2\sin 2x\\ W_3= \begin{vmatrix} y_1& y_2\\ y_1'& y_2'\end{vmatrix} = \begin{vmatrix} \cos 2x& \sin 2x\\ -2\sin 2x& 2\cos 2x\end{vmatrix} = 2} \Rightarrow \cases{u_1'=-2\cos 2x \sec 2x/8=-1/4 \\ u_2'=-2\sin 2x \sec 2x /8= -\tan 2x/4 \\ u_3'=2 \sec 2x/8= \sec 2x/4} \\ \Rightarrow \cases{u_1= -x/4\\ u_2 =\ln(\cos 2x)/8 \\u_3= \ln(\sec 2x+ \tan 2x)/8} \Rightarrow y_p=u_1y_1+ u_2y_2+u_3y_3\\ \Rightarrow y_p= -{x\over 4}\cos 2x+{\ln \cos(2x)\over 8}\sin 2x+ {1\over 8} \ln(\sec 2x+\tan 2x) \Rightarrow y=y_c+y_p \\ \Rightarrow \bbox[red, 2pt]{y= c_1\cos  2x+ c_2\sin 2x +c_3-{x\over 4}\cos 2x+{\ln \cos(2x)\over 8}\sin 2x+ {1\over 8} \ln(\sec 2x+\tan 2x) }$$

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