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:先求齊次解,即y‴+y″=0⇒λ3+λ2=0⇒λ2(λ+1)=0⇒λ=0,−1⇒yh=C1+C2x+C3e−x;接著由3ex+4x2,假設yp=aex+bx4+cx3+dx2⇒y′p=aex+4bx3+3cx2+2dx⇒y″p=aex+12bx2+6cx+2d⇒y‴p=aex+24bx+6c⇒y‴p+y″p=2aex+12bx2+(24b+6c)x+6c+2d=3ex+4x2⇒{2a=312b=424b+6c=06c+2d=0⇒{a=3/2b=1/3c=−4/3d=4⇒yp=32ex+13x4−43x3+4x2⇒y=C1+C2x+C3e−x+32ex+13x4−43x3+4x2
解答:令A=[940−6−10643],依題意y′=Ay;⇒det(A−λI)=(λ−3)2(λ−5)⇒{λ1=3λ2=5Case I: λ1=3⇒(A−λ1I)x=[640−6−40640][x1x2x3]=0⇒3x1+2x2=0⇒取v1=(0,0,1)T,v2=(2,−3,0)TCase II: λ2=5⇒(A−λ2I)x=[440−6−6064−2][x1x2x3]=0⇒{x1+x2=03x2+2x2=x3⇒取v3=(1,−1,1)T⇒y=C1[001]e3x+C2[2−30]xe3x+C3[1−11]e5x
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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