112年高科大碩士班-工程數學詳解

 

解答： $$\cases{A(6,10,4)\\ B(4,6,2) \\C(-2,-2,8)} \Rightarrow \cases{\overrightarrow{AB} =(-2,-4,-2)\\ \overrightarrow{AC}= (-8,-12,4)} \Rightarrow \vec n=\overrightarrow{AB} \times \overrightarrow{AC} =(-40,24,-8 ) \\ \Rightarrow 平面方程式: -40(x-6)+24(y-10)-8(z-4)=0 \Rightarrow \bbox[red, 2pt]{5x-3y+z=4}\\ \Rightarrow \triangle ABC面積={1\over 2}\Vert \vec n\Vert ={1\over 2}\sqrt{40^2+24^2+8^2} = \bbox[red, 2pt]{4\sqrt{35}}$$

解答： $$y''+y=2u(t-\pi)-2u(t-2\pi) \Rightarrow L\{y''\} +L\{y\}=2L\{u(t-\pi)\}-2 L\{u(t-2\pi)\} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0)+Y(s)={2\over s}(e^{-\pi s }-e^{-2\pi s}) \Rightarrow (s^2+1)Y(s)={2\over s}(e^{-\pi s }-e^{-2\pi s})+1 \\ \Rightarrow Y(s)={2\over s(s^2+1)}(e^{-\pi s }-e^{-2\pi s})+{1\over s^2+1}    \Rightarrow y(t)=L^{-1}\{Y(s)\} \\ \Rightarrow \bbox[red, 2pt]{y(t)=2u(t-\pi)(1-\cos(t-\pi))-2u(t-2\pi)(1-\cos(t-2\pi))+\sin(t)}$$

解答： $$y''-4y'+4y=(24x^2-12x)e^{2x} =r(x)\\ 先求齊次解，y''-4y'+4y=0\Rightarrow y_h=c_1e^{2x}+c_2xe^{2x}\\ 令\cases{y_1=e^{2x}\\ y_2=xe^{2x}} \Rightarrow W(y_1,y_2)= \begin{vmatrix}y_1& y_2\\ y_1'& y_2' \end{vmatrix}=e^{4x} \Rightarrow y_p=-y_1 \int{y_2r\over w}dx +y_2\int {y_1r\over w}\,dx\\ = -e^{2x}\int 24x^3-12x^2\,dx +xe^{2x} \int 24x^2-12x\,dx =2x^4e^{2x}-2x^3e^{2x}  \\ y=y_h+y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^{2x} +c_2xe^{2x} +2x^4e^{2x}-2x^3e^{2x}}$$

解答： $$\cases{M(x,y)=(x+y)^2\\ N(x,y)=2xy+x^2-1} \Rightarrow \frac{\partial M}{\partial y} =2x+2y =\frac{\partial N}{\partial x} \Rightarrow 恰當(exact)方程式\\ \Phi(x,y)= \int (x+y)^2\,dx = \int (2xy+x^2-1)\,dy \Rightarrow {1\over 3}x^3+x^2y +xy^2+ \phi(y)= xy^2+x^2y -y +\rho(x) \\ \Rightarrow \cases{\phi(y)=-y\\ \rho(x)=x^3/3} \Rightarrow \Phi(x,y)={1\over 3}x^3+x^2y+ xy^2 -y+c_1=0\\ 再將y(1)=2代入\Rightarrow {1\over 3}+ 2+ 4-2+c_1=0 \Rightarrow c_1=-{13\over 3} \Rightarrow \bbox[red, 2pt]{{1\over 3}x^3+x^2y+ xy^2 -y={13\over 3}}$$

解答： $$\left[\begin{array}{rrr|rrr} 4 & 2& 3& 1& 0 & 0\\ 4 & 2& 0 & 0 & 1& 0\\ -1& -2& 0 & 0& 0& 1\end{array}\right] \xrightarrow{-R_1+R_2\to R_2,R_1/4+R_3\to R_3} \left[\begin{array}{rrr|rrr} 4 & 2& 3& 1& 0 & 0\\ 0 & 0& -3 & -1 & 1& 0\\ 0& -3/2& 3/4 & 1/4& 0& 1 \end{array} \right] \\ \xrightarrow{R_1/4, R_2/(-3),-2R_3/3}\left[\begin{array}{rrr|rrr} 1 & 1/2& 3/4& 1/4& 0 & 0\\ 0 & 0& 1 & 1/3 & -1/3& 0\\ 0& 1& -1/2 & -1/6& 0& -2/3 \end{array} \right] \xrightarrow{-R_3/2+R_1\to R_1, -R_2/2+R_3\to R_3} \\\left[\begin{array}{rrr|rrr} 1 & 0& 1& 1/3& 0 & 1/3\\ 0 & 0& 1 & 1/3 & -1/3& 0\\ 0& 1& 0 & 0& -1/6& -2/3 \end{array} \right] \xrightarrow {-R_2+R_1 \to R_1}\left[\begin{array}{rrr|rrr} 1 & 0& 0& 0& 1/3 & 1/3\\ 0 & 0& 1 & 1/3 & -1/3& 0\\ 0& 1& 0 & 0& -1/6& -2/3 \end{array} \right] \\ \xrightarrow{R_2 \leftrightarrow R_3} \left[\begin{array}{rrr|rrr} 1 & 0& 0& 0& 1/3 & 1/3\\ 0 & 1& 0 & 0 & -1/6& -2/3\\ 0& 0& 1 & 1/3& -1/3& 0 \end{array} \right] \Rightarrow A^{-1}= \bbox[red,2pt]{\begin{bmatrix} 0& 1/3 & 1/3\\ 0& -1/6& -2/3\\ 1/3& -1/3 & 0\end{bmatrix}}$$

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