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2024年3月4日 星期一

110年雲科大電子碩士班-工程數學詳解

國立雲林科技大學110學年度碩士班招生考試

系所: 電子系
科目:工程數學



解答: (a)y+2y+5=0e2xy+2e2xy=5e2x(e2xy)=5e2xe2xy=5e2xdx=52e2x+c1y=52+c1e2x(b)y6y+10y=0λ26λ+10=0λ=3±iy=e3x(c1cosx+c2sinx)(c)Let y=xm, then y=mxm1y=m(m1)xm2x2yxy2y=m(m1)xmmxm2xm=(m22m2)xm=0m22m2=0m=1±3y=c1x(1+3)+c2x(13)
解答: {P(x,y)=ex+y+yeyQ(x,y)=xey1{Py=ex+y+ey+yeyQx=eyPyQxP=1u=u integration factor u(y)=ey{uP=ex+yuQ=xey{(uP)y=1(uQ)x=1ExactΦ(x,y)=uPdx=uQdyΦ=(ex+y)dx=(xey)dyex+xy+ϕ(y)=xy+ey+ρ(x)Φ=ex+xy+ey+c1=0

解答: y6y+9y=0λ26λ+9=0(λ3)2=0λ=3yh=c1e3x+c2xe3xApplying variation of parameters, let {y1=e3xy2=xe3xW=|y1y2y1y2|=e6xyp=e3xxe3xe3xx2e6xdx+xe3xe3xe3xx2e6xdx=e3x1xdx+xe3x1x2dx=e3xlnxe3xy=yh+ypy=c3e3x+c2xe3xe3xlnx


解答: (a)L{5}3L{e4t}+2L{t3}=5s31s+4+26s4=5s3s+4+12s4(b)L1{1(s+2)(s+3)}=L1{1s+21s+3}=e2te3t

解答{x+2y+3z=ax+3y+8z=bx+2y+2z=c[123138122][xyz]=[abc][xyz]=[123138122]1[abc]=[1027615101][abc]=[10a2b7c6a+b+5cac]{x=10a2b7cy=6a+b+5cz=ac

解答: (a)v1=(1,1),v2=(1,1)(b)T=[1111](c)T(x,y)(x,y)xy0(d)A=[12121212]A2=[12121212][12121212]=[12121212]=AA2=Alim



解答: \textbf{(a)}\; \begin{array}{c}& X& Y &X^2 & XY &Y^2 \\\hline & 0 & 3 & 0 & 0 & 9\\ & 1& 3& 1& 3 & 9\\ & 1& 6 & 1& 6 & 36\\\hdashline \sum & 2 & 12& 2 & 9 & 54\end{array} \Rightarrow c_1={n\sum XY-(\sum X)(\sum Y)\over n\sum X^2-(\sum X)^2} ={3\cdot 9-2\cdot 12\over 3\cdot 2-2^2}={3\over 2}\\ \cases{\bar x =\sum X/n=2/3\\ \bar y=\sum Y/n=4} \Rightarrow \bar y=c_0+c_1\bar x \Rightarrow 4=c_0+ {3\over 2}\cdot {2\over 3} \Rightarrow c_0=3 \Rightarrow \bbox[red, 2pt]{f=3+{3\over 2}t} \\\textbf{(b)}\; \cases{f(0)=3\\ f(1)=9/2} \Rightarrow \text{least square error }=(3-3)^2+(3-{9\over 2})^2 +(6-{9\over 2})^2=\bbox[red, 2pt]{9\over 2}

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