112年暨南大學碩士班－工程數學詳解

 國立暨南國際大學112學年度碩士班入學考試

科目：工程數學(線性代數、微分方程)
編號：342 適用：電機系

解答：$$\mathbf{(a)}\; 2xyy'=y^2-x^2 \Rightarrow y'-{y\over 2x} =-{x\over 2y} 為\text{Bernoulli equation} \\取v=y^2 \Rightarrow v'=2yy'代回原式\Rightarrow xv'-v=-x^2 \Rightarrow v'-{1\over x} v=-x \\ \Rightarrow 積分因子u(x)=e^{\int -(1/x)dx}={1\over x} \Rightarrow {v'\over x}-{v\over x^2} =-1 \Rightarrow ({v\over x})'=-1\\ \Rightarrow  {v\over x}={y^2\over x}=-x+C \Rightarrow y^2= -x^2+Cx \Rightarrow \bbox[red, 2pt]{y=\pm \sqrt{-x^2+Cx},C為常數} \\\mathbf{(b)}\;y''-4y'+4y=0 \Rightarrow 特徵方程式：\lambda^2-4\lambda+4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow \lambda=2二重根\\ \Rightarrow \bbox[red, 2pt]{y=C_1e^{2x}+ C_2xe^{2x},C_1,C_2為常數} \\ \mathbf{(c)}\; y'''+12y''+48y'+64y=0 \Rightarrow 特徵方程式：\lambda^3+ 12\lambda^2 +48\lambda+64=0\\ \Rightarrow (\lambda+4)^3=0 \Rightarrow x=-4,三重根\Rightarrow \bbox[red, 2pt]{y=e^{-4x}(C_1+ C_2x+ C_3x^2),C_1,C_2,C_3為常數}$$

解答：$$\mathbf{(a)}\;\mathcal L\{f(t)\} \equiv F(s)=\int_0^\infty f(t)e^{-st}\,dt\\ \mathcal L^{-1}\{F(s)\}=f(t),\text{ if }\mathcal L\{f(t)\} = F(s) \\\mathbf{(b)}\;F(s)=\int_0^{\infty} f(t)e^{-st}\,dt \Rightarrow -{d\over ds}F(s) =-{d\over ds}\int_0^{\infty} f(t)e^{-st}\,dt =\int_0^{\infty} tf(t)e^{-st}\,dt =L\{tf(t)\},\\ \qquad \bbox[red, 2pt]{故得證}\\ \mathbf{(c)}\;L\{ \cos(kt)\} =\int_0^\infty \cos(kt)e^{-st}\,dt = \left.\left[ {1\over k^2+s^2}\cdot e^{-st}(k\sin(kt)-s\cos(kt))\right] \right|_0^\infty \\ ={s\over s^2+k^2} \Rightarrow L\{t\cos(kt)\} =-{d\over ds}\left({s\over s^2+k^2}\right) =-{1\over s^2+k^2}+{2s^2\over (s^2+k^2)^2} \\={2s^2-(s^2+k^2)\over (s^2+k^2)^2} =\bbox[red,2pt]{s^2-k^2\over (s^2+k^2)^2}\\ 同理，L\{ \sin(kt)\} =\int_0^\infty \sin(kt)e^{-st}\,dt = \left. \left[-{1\over s^2+k^2}\cdot e^{-st}(s \sin(kt)+k\cos(kt)) \right]\right|_0^\infty \\={k\over s^2+k^2} \Rightarrow L\{ t\sin(kt)\} =-{d\over ds}\left({k\over s^2+k^2}  \right) =\bbox[red,2pt]{2ks\over (s^2+k^2)^2}\\ \mathbf{(d)}\; L\{y''\}+ 16L\{y\} =L\{\cos(4t)\} \Rightarrow s^2Y(s)-sy(0)-y'(0)+ 16Y(s)= {s\over s^2+4^2}\\ \Rightarrow (s^2+16)Y(s)={s\over s^2+ 16}+1 \Rightarrow Y(s)={s\over (s^2+16)^2}+ {1\over s^2+16} \\ \Rightarrow L^{-1}\{Y(s)\}=y(t)=L^{-1}\{{s\over (s^2+16)^2}\} +L^{-1}\{{1\over s^2+16}\} ={1\over 8}\sin(4t)+{1\over 4}\sin(4t)\\ \Rightarrow \bbox[red,2pt]{y={1\over 8}\sin(4t)+{1\over 4}\sin(4t)}$$

解答：$$A=\begin{bmatrix}2 & 4& -4 & 2 \\9 & 18& -15& 12\\ 2 & 4& -2 & 4 \end{bmatrix} \xrightarrow{-r_1+r_3\to r_3,-(9/2)r_1+r_2\to r_2}\begin{bmatrix}2 & 4& -4 & 2 \\0 & 0& 3& 3\\ 0 & 0& 2 & 2 \end{bmatrix} \\ \xrightarrow{r_1/2,-(2/3)r_2+r_3\to r_3} \begin{bmatrix}1 & 2& -2 & 1 \\0 & 0& 3& 3\\ 0 & 0& 0 & 0 \end{bmatrix} \xrightarrow{r_2/3}\begin{bmatrix}1 & 2& -2 & 1 \\0 & 0& 1& 1\\ 0 & 0& 0 & 0 \end{bmatrix} \xrightarrow{2r_2+r_1 \to r_1}\begin{bmatrix}1 & 2& 0 & 3 \\0 & 0& 1& 1\\ 0 & 0& 0 & 0 \end{bmatrix}\\ \Rightarrow rref(A)=\begin{bmatrix}1 & 2& 0 & 3 \\0 & 0& 1& 1\\ 0 & 0& 0 & 0 \end{bmatrix} \Rightarrow \text{rank}(A)=\bbox[red, 2pt]2 \\\mathbf{(b)}\; B=rref(A) \Rightarrow Bx=0 \Rightarrow \cases{x_1+2x_2+3x_4=0\\ x_3+x_4=0} \Rightarrow \cases{x_1= -2x_2-3x_4\\ x_2=x_2\\ x_3=-x_4\\x_4=x_4}\\ \Rightarrow x=x_2\begin{bmatrix}-2 \\1\\0 \\0 \end{bmatrix}+ x_4 \begin{bmatrix}-3 \\0\\-1 \\1 \end{bmatrix} \Rightarrow \text{Null space of }A=\bbox[red, 2pt]{\left\{\begin{bmatrix}-2x_2-3x_4 \\x_2 \\-x_4 \\x_4 \end{bmatrix}: x_2,x_4\in \mathbb R \right\}}$$

解答：$$\mathbf{(a)}\;\mathbf x=18(2,1,1)-18(2,-1,1)-10(1,2,1)=\bbox[red, 2pt]{(-10,16,-10)}\\ \mathbf{(b)}\; 假設[\mathbf x]_{B'}=(a,b,c) \Rightarrow a(3,1,-5)+ b(1,1,-3)+c(-1,0,2)= (-10,16,-10)\\ \Rightarrow \cases{3a+b-c=-10\\ a+b=16\\ -5a-3b+2c= -10} \Rightarrow \cases{a=-7\\ b=23\\c=12} \Rightarrow [\mathbf x]_{B'}=\bbox[red,2pt]{(-7,23,12)}$$

解答：$$\mathbf{(a)}\;\det(A-\lambda I)=-(\lambda+4)(\lambda-8)^2=0 \Rightarrow \text{eighenvalues: }\bbox[red,2pt]{-4,8}\\ \mathbf{(b)}\; \lambda_1=-4 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}6 & 6&0 \\6 & 6&0 \\ 0 & 0 & 12\end{bmatrix}\begin{bmatrix} x_1 \\x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=-x_2\\ x_3=0},取v_1=\begin{bmatrix} -1 \\1\\ 0 \end{bmatrix}\\ \lambda_2=8 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-6 & 6&0 \\6 & -6&0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} x_1 \\x_2\\ x_3 \end{bmatrix}=0 \Rightarrow x_1=x_2,取v_2=\begin{bmatrix} 1 \\1\\ 0 \end{bmatrix}, v_3=\begin{bmatrix} 0 \\0\\ 1 \end{bmatrix} \\ \text{the corresponding eigenvectors: } \bbox[red,2pt]{\begin{bmatrix} -1 \\1\\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\1\\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\0\\ 1 \end{bmatrix}}\\ \mathbf{(c)}\; \begin{bmatrix}-1  \\1 \\0 \end{bmatrix} \text{ is a basis for the eigenspace corresponding to }\lambda=-4 \\  \begin{bmatrix}1  \\1 \\0 \end{bmatrix} \text{and} \begin{bmatrix}0  \\0 \\1 \end{bmatrix} \text{together constitute the basis for the eigenspace corresponding to }\lambda=8$$