111年台綜大轉學考-工程數學D36詳解

臺灣綜合大學系統111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：$$x^2y'+xy=\sin x \Rightarrow y'+{1\over x}y= {\sin x\over x^2} \Rightarrow 積因子I(x)= e^{\int (1/x)\,dx} =x\\ \Rightarrow I(x) y'+I(x)\cdot {1\over x}y=I(x)\cdot {\sin x\over x^2} \Rightarrow xy'+ y={\sin x\over x} \Rightarrow (xy)'={\sin x\over x}\\ \Rightarrow xy = \int {\sin x\over x}\,dx =Si(x)+C ，將y(1)=A代入左式\Rightarrow A= Si(1)+C \Rightarrow C= A-Si(1) \\ \Rightarrow \bbox[red, 2pt]{xy= Si(x)-Si(1)+A}$$

解答：$$先求齊次解，即y'''+y''=0 \Rightarrow \lambda^3+\lambda^2=0 \Rightarrow \lambda^2(\lambda+1)=0 \Rightarrow \lambda=0,-1\\ \Rightarrow y_h= C_1 +C_2x+ C_3e^{-x}; 接著由3e^x+ 4x^2，假設y_p = ae^x + bx^4+ cx^3+dx^2 \\ \Rightarrow y_p'=ae^x+ 4bx^3+ 3cx^2 + 2dx \Rightarrow y_p'' = ae^x +12bx^2 +6cx +2d\\ \Rightarrow y_p''' =ae^x +24bx+6c \Rightarrow y_p'''+y_p'' =2ae^x +12bx^2 +(24b+6c)x+ 6c+2d= 3e^x +4x^2 \\ \Rightarrow \cases{2a=3 \\ 12b=4\\ 24b+6c=0\\ 6c+2d=0} \Rightarrow \cases{a=3/2\\ b=1/3\\ c=-4/3 \\ d=4} \Rightarrow y_p = {3\over 2}e^x +{1\over 3}x^4 -{4\over 3}x^3 +4x^2 \\ \Rightarrow \bbox[red, 2pt]{y=C_1 +C_2x+ C_3e^{-x} +{3\over 2}e^x +{1\over 3}x^4 -{4\over 3}x^3 +4x^2}$$

解答：$$令A= \left[\begin{matrix}9 & 4 & 0\\-6 & -1 & 0\\6 & 4 & 3\end{matrix}\right]，依題意y'= A y;\\ \Rightarrow \det(A-\lambda I) = (\lambda-3)^2(\lambda-5) \Rightarrow \cases{\lambda_1= 3\\ \lambda_2= 5}\\ \text{Case I: } \lambda_1=3 \Rightarrow (A-\lambda_1 I){x} =\left[\begin{matrix}6 & 4 & 0\\-6 & -4 & 0\\6 & 4 & 0\end{matrix}\right]\left[\begin{matrix}x_1\\ x_2 \\ x_3\end{matrix}\right]=0 \Rightarrow 3x_1+2x_2=0\\ \Rightarrow 取v_1=(0,0,1)^T,v_2=(2,-3,0)^T\\ \text{Case II: } \lambda_2=5 \Rightarrow (A-\lambda_2 I){x} =\left[\begin{matrix}4 & 4 & 0\\-6 & -6 & 0\\6 & 4 & -2\end{matrix}\right]\left[\begin{matrix}x_1\\ x_2 \\ x_3\end{matrix}\right]=0 \Rightarrow \cases{x_1+x_2=0\\ 3x_2+2x_2=x_3}\\ \Rightarrow 取v_3=(1,-1,1)^T\\ \Rightarrow \bbox[red, 2pt]{y=C_1\begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}e^{3x} +C_2 \begin{bmatrix} 2\\ -3\\ 0\end{bmatrix}xe^{3x} +C_3\begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix} e^{5x}}$$

解答：$$先考慮f(t)=\begin{cases} 0,&\text{if }-\pi\lt t\le 0\\ t,& \text{if }0\le t\lt \pi\end{cases} \Rightarrow \mathcal{F}(t)= a_0 +\sum_{n=1}^\infty (a_n\cos nt +b_n \sin nt)\\其\cases{ a_0={1\over 2\pi} \int_{-\pi}^\pi f(t)\,dt ={1\over 2\pi} \int_0^{\pi} t\,dt = {\pi\over 4}\\[1ex] a_n= {1\over \pi}\int_{-\pi}^\pi f(t)\cos nt\,dt= {1\over \pi} \int_0^\pi t\cos nt\,dt= {1\over n^2\pi}((-1)^n-1)\\[1ex] b_n= {1\over \pi}\int_{-\pi}^\pi f(t)\sin nt\,dt= {1\over \pi} \int_0^\pi t\sin nt\,dt= -{1\over n}(-1)^n} \\再考慮\cases{\lim_{t\to \pi^-} f(t)=\pi\\ \lim_{t\to \pi^+} f(t)=0} \Rightarrow \mathcal{F}(\pi)={\pi\over 2} ，同理\mathcal{F}(n\pi)={\pi\over 2},\forall n\in \mathbb{N}\\ 因此\bbox[red, 2pt] {\mathcal{F}(t) =\begin{cases} {\pi\over 2}, &\text{if }t=n\pi \\{\pi\over 4}+ \sum_{n= 1}^\infty ({1\over n^2\pi}((-1)^n-1)\cos nt -{1\over n}(-1)^n \sin nt),&\text{otherwise}\end{cases}}$$

解答：$$先求f(t)的\text{Fourier series}，由於f(t)為奇函數，所以a_n=0，只需要求b_n;\\ b_n= \int f(t)\sin(nt)\;dt =\int_{-\pi/2}^{\pi/2} t\sin(nt)\,dt +\int_{\pi/2}^{3\pi/2}(\pi-t)\sin(nt)\,dt =\cases{4/n^2,n=1,5,9,\dots\\ -4/n^2,n=3,7,11,\dots \\0, n=2,4,6,8,\dots}\\ 令y=A_n\cos(nt)+ B_n\sin(nt) \Rightarrow y'=-nA_n\sin(nt) +nB_n\cos (nt) \\\Rightarrow y''=-n^2 A_n\cos(nt)-n^2B_n\sin(nt) \\\Rightarrow y''+ky'+y =(A_n +nkB_n-n^2A_n) \cos(nt) +(-n^2B_n -nkA_n +B_n)\sin (nt) =b_n \sin(nt)\\ \Rightarrow \cases{(1-n^2)A_n+nkB_n =0\\ (1-n^2)B_n -nkA_n =b_n} \Rightarrow \cases{A_n =-{nk\over (1-n^2)^2+n^2k^2}b_n\\[1ex] B_n= {1-n^2\over (1-n^2)^2+n^2k^2}b_n} \\\Rightarrow \bbox[red, 2pt]{y=A_n\cos(nt)+ B_n\sin(nt),n\in \mathbb{N}，其中\cases{A_n =-{nk\over (1-n^2)^2+n^2k^2}b_n\\[1ex] B_n= {1-n^2\over (1-n^2)^2+n^2k^2}b_n}及b_n=\cases{4/n^2,n=1,5,9,\dots\\ -4/n^2,n=3,7,11,\dots \\0, n=2,4,6,8,\dots}}$$
 

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