國立成功大學112學年度碩士班招生考試
系所:水利及海洋工程學系
科目:工程數學
解答:y″+4πy′+4π2y=0⇒特徵多項式λ2+4πλ+4π2=0⇒(λ+2π)2=0⇒λ=−2π⇒y(x)=c1e−2πx+c2xe−2πx,其中c1與c2均為常數解答:A=[353046001]⇒det(A−λI)=(3−λ)(4−λ)(1−λ)=0⇒λ=1,3,4λ1=1⇒(A−λ1I)v=0⇒[253036000][x1x2x3]=0⇒{2x1=7x3x2+2x3=0⇒v=[7k/2−2kk],k∈R,取v1=[7/2−21]λ2=3⇒(A−λ2I)v=0⇒[05301600−2][x1x2x3]=0⇒{x2=0x3=0⇒v=[k00],k∈R,取v2=[100]λ3=4⇒(A−λ3I)v=0⇒[−15300600−3][x1x2x3]=0⇒{x1=5x2x3=0⇒v=[5kk0],k∈R,取v3=[510]因此特徵值為1,3,4,相對應的特徵向量為[7/2−21],[100],[510]
解答:L{y″}+5L{y′}+6L{y}=L{u(t−1)}+L{δ(t−2)}⇒s2Y(s)−sy(0)−y′(0)+5(sY(s)−y(0))+6Y(s)=e−ss+e−2s⇒Y(s)=1s2+5s+6(e−ss+e−2s+1)=e−s(16s−12(s+2)+13(s+3))+e−2s(1s+2−1s+3)+1s+2−1s+3⇒y(t)=L−1{Y(s)}⇒y(t)=u(t−1)(16−12e−2(t−1)+13e−3(t−1))+u(t−2)(e−2(t−2)−e−3(t−2))+e−2t−e−3t
解答:{y′+z′+z=0y′+2y+6∫t0z(t)dt=−2u(t)⇒{L{y′}+L{z′}+L{z}=0L{y′}+2L{y}+6L{∫t0z(t)dt}=−2L{u(t)}⇒{sY(s)+5+sZ(s)−6+Z(s)=0⋯(1)sY(s)+5+2Y(s)+6Z(s)/s=−2/s⋯(2)由(1)可得Y(s)=1s−s+1sZ(s)代入(2)⇒Z(s)=2s−1+4s+4⇒Y(s)=2s−4s−1−3s+4⇒{y(t)=L−1{Y(s)}z(t)=L−1{Z(s)}⇒{y(t)=−4et−3e−4t+2u(t)z(t)=2et+4e−4t
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解題僅供參考,其他歷年試題及詳解
第4,5兩題答案有誤,麻煩再勘誤一下,感謝~
回覆刪除我再算算, 有點複雜,謝謝提醒!
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刪除對了,感謝!
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