113年台大碩士班-工程數學A詳解

 國立臺灣大學 113 學年度碩士班招生考試

科目: 工程數學(A)

解答:$$\cases{x+y+z=k\\ 3x+4y+5z=3k\\ 4x+ 2y+(k+1) z=5k} \Rightarrow \begin{bmatrix}1& 1& 1\\ 3& 4& 5\\ 4& 2& k+1 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} = \begin{bmatrix} k\\ 3k\\ 5k \end{bmatrix} \equiv A\mathbf x=\mathbf b \\ \Rightarrow rref([A\mid b])=\left[\begin{array}{rrr|r} 1 & 1 & 1 & k\\0 & 1 & 2 & 0\\0 & 0 & k+1 & k \end{array}  \right] \Rightarrow \cases{x+y+ z=k\\y+2z=0\\ (k+1)z= k} \Rightarrow \cases{k=-1 \Rightarrow 無解 \\ k\ne-1\Rightarrow 唯一解} \\ \textbf{(a) }\bbox[red, 2pt]{k=-1}  \\\textbf{(b) }\bbox[red, 2pt]{k值無法決定無限多解}$$

解答:$$\cases{Q(x,y,z)=x^2 +y^2+2z^2-2xy+4yz+4zx\\ g(x,y,z) =x^2+ y^2+z^2-1} \Rightarrow \cases{Q_x= \lambda g_x \\Q_y= \lambda g_y \\Q_z= \lambda g_z \\g=0} \Rightarrow \cases{2x-2y+4z= \lambda(2x)  \cdots(1)\\ 2y-2x+4z=\lambda(2y)  \cdots(2)\\ 4z+4y+4x= \lambda(2z)  \cdots(3)\\ x^2+y^2+z^2=1 \cdots(4)} \\ \Rightarrow \cases{\displaystyle { (1)\over (2)} ={x-y+2z\over y-x+2z} ={x\over y} \\ \displaystyle {(1) \over (3)} ={x-y+2z\over 2z+2y+2x} ={x\over z}} \Rightarrow \cases{x^2-2xy=y^2-2yz \cdots(5) \\ 2x^2+2xy+ xz+yz-2z^2=0 \cdots(6)} \\ Eq.(5) \Rightarrow (x-z)^2=(y-z)^2 \Rightarrow \cases{x=y\\ x+y=2z} 代入(6) \Rightarrow \cases{z=2x\\ z=-x} 代入(4) \\ \Rightarrow \cases{x^2+x^2+4x^2=1\\ x^2+x^2+x^2=1} \Rightarrow \cases{x=\pm 1/\sqrt 6\\ x=\pm 1/\sqrt 3} \Rightarrow\cases{(x,y,z) =(\pm 1/\sqrt 6,\pm 1/\sqrt 6, \pm 2/\sqrt 6) \\(x,y,z) =(\pm 1/\sqrt 3 \pm 1/\sqrt 3, \mp 1/\sqrt 3)} \\ \textbf{(a ) } \cases{Q(\pm 1/\sqrt 6,\pm 1/\sqrt 6, \pm 2/\sqrt 6) =4\\ Q(\pm 1/\sqrt 3, \pm 1/\sqrt 3, \mp 1/\sqrt 3) =-2} \Rightarrow \bbox[red, 2pt]{\cases{極大值4\\ 極小值-2}} \\\textbf{(b) } \bbox[red, 2pt]{\cases{(x,y,z)=(1/\sqrt 6,1/\sqrt 6,2/\sqrt 6) 或(-1/\sqrt 6,-1/\sqrt 6, -2/\sqrt 6 )時有極大值\\ (x,y,z)= (1/\sqrt 3, 1/\sqrt 3, -1/\sqrt 3) 或(-1/\sqrt 3, -1/\sqrt 3, 1/\sqrt 3)時有極小值}}$$

解答:$$\begin{array}{c|cc|} & X&Y &X^2& XY&Y^2 \\\hline & -2&-3 &4& 6& 9\\& -1& 1& 1& -1& 1\\ & 0& 1& 0& 0& 1\\ & 1& 1& 1& 1& 1\\ & 2& 3&  4& 6&9 \\\hdashline \sum & 0& 3& 10& 12& 21 \end{array} \Rightarrow \cases{ \bar x=0/5=0\\ \bar y=3/5} \\\Rightarrow   a= {n \sum xy-\sum x\sum y \over n \sum x^2-(\sum x)^2} ={5\cdot 12-0\cdot 3\over 5\cdot 10-0^2} =1.2 \Rightarrow b={\sum y\over n} -a\cdot {\sum x\over n} ={3\over 5}-1.2\cdot {0=0.6} \\ \Rightarrow \bbox[red, 2pt]{\cases{a=1.2\\ b=0.6}}$$

解答:$$\textbf{(a) }f(x,y)=e^{xy} \cos(x+y) \Rightarrow \nabla f= (f_x, f_y) \\=(e^{xy} (y\cos(x+y)- \sin(x+y)), e^{xy}(x \cos(x+y)-\sin(x+y))) \\ \Rightarrow \nabla f(0,\pi) =(-\pi,0) \Rightarrow D_{\vec n}f(0,\pi) =(-\pi,0) \cdot (\cos \theta, \sin \theta) =-\pi\cos \theta \Rightarrow 最大值為 \bbox[red, 2pt]{\pi} \\\textbf{(b) } -\pi\cos \theta={\pi\over 2} \Rightarrow \cos\theta=-{1\over 2} \Rightarrow \sin \theta= \pm {\sqrt 3\over 2} \Rightarrow 沿著方向\bbox[red, 2pt]{(-{1\over 2}, {\sqrt 3\over 2}) 或(-{1\over 2}, -{\sqrt 3\over 2})}$$

解答:$$\textbf{(a) } y''+2y'+2y=0  \Rightarrow r^2+2r+2=0 \Rightarrow r=-1\pm i \Rightarrow y_h=e^{-x}(c_1 \cos x+ c_2\sin x) \\\quad \Rightarrow y= y_h+y_p \Rightarrow \bbox[red, 2pt]{y(x)= e^{-x}(c_1 \cos x+ c_2\sin x) + x^5e^{-4x}(\sin(3x)+ \cos(3x))} \\ \textbf{(b) }r(x)=1+x^2+e^{-2x}+ \sin x \Rightarrow \cases{y_{p_1} =Ax^2+Bx+C\\ y_{p_2}=De^{-2x} \\ y_{p_3} =E\cos x+F\sin x} \Rightarrow \cases{y_{p_1}' = 2Ax+B \\ y_{p_2}'=-2De^{-2x} \\ y_{p_3}' =-E\sin x+F\cos x} \\\quad \Rightarrow \cases{y_{p_1}'' = 2A  \\ y_{p_2}''= 4De^{-2x} \\ y_{p_3}'' =-E\cos x-F\sin x}  \Rightarrow \cases{y_{p_1}'' +2y_{p_1}'+ 2y_{p_1} = 2Ax^2+ (4A+2B)x +2A+2B+2C= 1+x^2 \\ y_{p_2}'' +2y_{p_2}'+ 2y_{p_2} =2De^{-2x} =e^{-2x} \\y_{p_3}'' +2y_{p_3}'+ 2y_{p_3} =(E+2F) \cos x+ (F-2E) \sin x =\sin x} \\ \Rightarrow \cases{A=1/2, B=-1, C=1\\ D=1/2\\E=-2/5, F=1/5} \Rightarrow y_p={1\over 2}x^2-x+1+{1\over 2}e^{-2x}-{2\over 5}\cos x+{1\over 5}\sin x \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y(x)= e^{-x}(c_1 \cos x+ c_2\sin x) +{1\over 2}x^2-x+1+{1\over 2}e^{-2x}-{2\over 5}\cos x+{1\over 5}\sin x}$$

解答:$$\textbf{(a) }\cases{x'=2x-y+t\\ y'=3x-2y-1\\ x(0)=y(0)=0} \Rightarrow \cases{L\{x'\} =2L\{x\} -L\{y\} +L\{t\} \\L\{ y'\} =3L\{x\} -2L\{y\}-L\{1\}} \\\Rightarrow \cases{sX(s)=2X(s)-Y(s)+1/s^2 \cdots(1)\\ sY(s)= 3X(s)-2Y(s)-1/s \cdots(2)} \\ Eq.(1) \Rightarrow Y(s)=(2-s)X(s)+{1\over s^2} 代入Eq.(2) \\\Rightarrow   s(2-s)X(s)+{1\over s}=3X(s)-2(2-s)X(s)-{2\over s^2}-{1\over s} \\ \Rightarrow \bbox[red, 2pt]{X(s)  ={2\over s^2(s-1)}} \Rightarrow \bbox[red, 2pt]{Y(s)={3-s\over s^2(s-1)} } \\\textbf{(b) } x(t)=L^{-1}\{X(s)\}  = L^{-1} \left\{{2\over s^2(s-1)} \right\} = L^{-1} \left\{-{2\over s}-{2\over s^2}+{2\over s-1} \right\}\\ \quad \Rightarrow \bbox[red, 2pt]{x(t)= -2-2t+2e^t} \\ y(t)=L^{-1} \{Y(s) \}= L^{-1} \left\{{3-s\over s^2(s-1)} \right\} =L^{-1} \left\{ -{2\over s}-{3\over s^2}+{2\over s-1} \right\} \\\quad \Rightarrow \bbox[red, 2pt]{y(t)=-2-3t+2e^t}$$

解答:$$\textbf{Case I }\alpha=0 \Rightarrow y''=0 \Rightarrow y=c_1x+c_2 \Rightarrow y'=c_1 \Rightarrow \cases{y(0)=c_2=0\\ y'(L)+ y(L)=c_1 +c_1L+c_2=0} \\ \qquad \Rightarrow \cases{c_1(1+L)=0\\ c_2=0} \Rightarrow c_1=c_2=0 \Rightarrow y=0 \Rightarrow \alpha=0不是特徵值\\ \textbf{Case II }\alpha=-k^2 \lt 0 \Rightarrow y''-k^2y=0 \Rightarrow y=c_1e^{kx}+ c_2e^{-kx} \Rightarrow y'=c_1ke^{kx} -c_2ke^{-kx} \\\qquad \Rightarrow \cases{y(0)=c_1+c_2=0 \\ y'(L)+ y(L)=c_1e^{kL}(1+k) +c_2e^{-kL}(1-k)=0} \\\qquad \Rightarrow c_1e^{kL}(1+k) -c_1e^{-kL}(1-k)=0 \Rightarrow c_1=0\Rightarrow c_2= 0 \Rightarrow y=0 \Rightarrow \alpha \lt 0不是特徵值\\ \textbf{Case III }\alpha=k^2 \gt 0 \Rightarrow y''+k^2y=0 \Rightarrow y=c_1 \cos(kx)+ c_2 \sin(kx) \Rightarrow y'=-c_1k \sin(kx)+ c_2k \cos(kx) \\ \qquad \Rightarrow  y(0)=c_1=0\Rightarrow  y'(L)+y(L)= c_2( \sin(kL)+ k\cos(kL))=0 \Rightarrow \sin(kL)+ k\cos(kL)=0\\\qquad \Rightarrow \tan(kL)=-k \Rightarrow 所有實數\bbox[red, 2pt]{特徵值\alpha_n=k_n^2, k_n 為滿足 \tan(k_nL)=-k_n的所有正實根 }$$

解答:$$a(x,y)u_x+ b(x,y)u_y=c(x,y) \Rightarrow \cases{a=1\\ b=2x\\ c= 8xy} \Rightarrow {dx\over 1} ={dy\over 2x} ={du \over 8xy} \\ \Rightarrow {dy\over dx}= 2x \Rightarrow y=x^2+c_1 \Rightarrow c_1=y-x^2, 又{du\over dx} =8xy =8x(x^2+c_1) \Rightarrow u=2x^4+4c_1x^2+c_2 \\=2x^4+4(y-x^2)x^2+ f(y-x^2 ) =4x^2y-2x^4+ f(y-x^2) \\ \Rightarrow u(0,y)=f(y)=\sin y \Rightarrow \bbox[red, 2pt]{u(x,y)= 4x^2y-2x^4+ \sin(y-x^2)}$$

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