113年台綜大轉學考-工程數學D36詳解

 臺灣綜合大學系統113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:D36

解答: $$y'-y=4 \Rightarrow y'e^{-x}-ye^{-x}=4e^{-x} \Rightarrow (ye^{-x})'=4e^{-x} \Rightarrow ye^{-x}= \int 4e^{-x}=-4e^{-x}+c_1\\ \Rightarrow \bbox[red, 2pt]{y=-4+c_1e^x,c_1為常數}$$

解答: $$xy'+4y=8x^4 \Rightarrow y'+{4\over x}y=8x^3 \Rightarrow x^4y'+4x^3y=8x^7 \Rightarrow (x^4y)'=8x^7\\ \Rightarrow x^4y= \int 8x^7\,dx =x^8+c_1 \Rightarrow y=x^4+{c_1\over x^4}\\ y(1)=2 \Rightarrow1+c_1=2 \Rightarrow c_1=1 \Rightarrow \bbox[red, 2pt]{y=x^4+{1\over x^4}}$$

解答: $$y''-3y'-4y=0 \Rightarrow \lambda^2-3\lambda-4=0 \Rightarrow (\lambda-4)(\lambda+1)=0 \Rightarrow \lambda=4,-1\\ \Rightarrow y=c_1e^{4x}+ c_2e^{-x} \Rightarrow y'=4c_1e^{4x}-c_2e^{-x} \Rightarrow \cases{y(0)=c_1+c_2=1\\ y'(0)= 4c_1-c_2 =2} \Rightarrow \cases{c_1=3/5\\ c_2=2/5} \\ \Rightarrow \bbox[red, 2pt]{ y={3\over 5}e^{4x}+{2\over 5}e^{-x}}$$

解答: $$\begin{bmatrix} B&C \\ D& E\end{bmatrix} \begin{bmatrix} X& Y \\ Z& U\end{bmatrix} =I=\begin{bmatrix} I_m& 0 \\ 0 & I_n\end{bmatrix} \Rightarrow \cases{BX+CZ=I_m \cdots(1)\\ DX+EZ=0 \cdots(2)\\ BY+CU= 0 \cdots(3)\\ DY+EU=I_n \cdots(4)} \\ 式(2) \Rightarrow Z=-E^{-1}DX 代入(1) \Rightarrow BX-CE^{-1}DX= I_m \Rightarrow (B-CE^{-1}D)X = I_m\\ \Rightarrow \bbox[red, 2pt] {X=(B-CE^{-1}D)^{-1}} \\ 式(3) \Rightarrow Y=-B^{-1}CU代入(4) \Rightarrow -DB^{-1}CU+EU=I_n \Rightarrow (-DB^{-1}C+E)U=I_n \\ \Rightarrow \bbox[red, 2pt]{U=(-DB^{-1}C+E)^{-1}}$$

解答: $$\cases{-2w+x-y=1\\ w-2x+z=-5\\ w-2y+z=-7\\ x+y-2z=7} \Rightarrow \left[ \begin{array}{rrrr|r}-2& 1 & -1 & 0 & 1\\ 1& -2 & 0 & 1& -5\\ 1& 0 & -2 & 1& -7\\ 0& 1& 1& -2 & 7\end{array} \right] \\ \Rightarrow \cases{\triangle =\begin{vmatrix} -2& 1 & -1 & 0  \\ 1& -2 & 0 & 1 \\ 1& 0 & -2 & 1 \\ 0& 1& 1& -2  \end{vmatrix} =8 \\\triangle_w =\begin{vmatrix} 1& 1 & -1 & 0  \\ -5& -2 & 0 & 1 \\ -7& 0 & -2 & 1 \\ 7& 1& 1& -2  \end{vmatrix} =-8 \\ \triangle_x= \begin{vmatrix} -2& 1 & -1 & 0  \\ 1& -5 & 0 & 1 \\ 1& -7 & -2 & 1 \\ 0& 7& 1& -2  \end{vmatrix} =8 \\ \triangle_y= \begin{vmatrix} -2& 1 & 1 & 0  \\ 1& -2 & -5 & 1 \\ 1& 0 & -7 & 1 \\ 0& 1& 7& -2  \end{vmatrix} =16\\ \triangle_z= \begin{vmatrix} -2& 1 & -1 & 1  \\ 1& -2 & 0 & -5 \\ 1& 0 & -2 & -7 \\ 0& 1& 1& 7  \end{vmatrix} =-16} \Rightarrow \cases{w=\triangle_w/ \triangle =-8/8=-1 \\ x=\triangle_x/\triangle =8/8=1\\ y=\triangle_y/\triangle =16/8=2\\ z=\triangle_z/\triangle = -16/8=-2} \\ \Rightarrow \bbox[red, 2pt]{(w,x,y,z) =(-1,1,2,-2)}$$

解答: $$假設f(x)週期為2\pi的函數,其傅利葉級數可以寫成f(x)\sim{a_0\over 2} +\sum_{n=1}^\infty \left(a_n\cos (nx)+b_n \sin(nx) \right) \\取t={Lx\over \pi},則g(t)的週期就是2L \Rightarrow g(t)\sim \bbox[red, 2pt]{{a_0\over 2}\sum_{n=1}^\infty \left(a_n\cos{n\pi t\over L} +b_n \sin{n\pi t\over L} \right)}\\ 現在要決定g(t)傅利葉級數的係數, 由於原來f(x)傅利葉係數a_n={1\over \pi}\int_0^{2\pi} f(x) \cos(nx)\,dx\\ t={Lx\over \pi} \Rightarrow x={\pi t\over L} \Rightarrow dx={\pi\over L}dt \Rightarrow \bbox[red, 2pt]{a_n= {1\over L}\int_0^{2L} g(t)\cos{n\pi t\over L}\,dt, n=0,1,2,\dots}\\同理\bbox[red, 2pt]{b_n={1\over L}\int_0^{2L} g(t)\sin{n\pi t\over L}\,dt,n=1,2,\dots}$$

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