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=0⇒e2xy′+2e2xy=−5e2x⇒(e2xy)′=−5e2x⇒e2xy=∫−5e2xdx=−52e2x+c1⇒y=−52+c1e−2x(b)y″−6y′+10y=0⇒λ2−6λ+10=0⇒λ=3±i⇒y=e3x(c1cosx+c2sinx)(c)Let y=xm, then y′=mxm−1⇒y″=m(m−1)xm−2⇒x2y″−xy′−2y=m(m−1)xm−mxm−2xm=(m2−2m−2)xm=0⇒m2−2m−2=0⇒m=1±√3⇒y=c1x(1+√3)+c2x(1−√3)
解答: {P(x,y)=ex+y+yeyQ(x,y)=xey−1⇒{Py=ex+y+ey+yeyQx=ey⇒−Py−QxP=−1⇒u′=−u⇒ integration factor u(y)=e−y⇒{uP=ex+yuQ=x−e−y⇒{(uP)y=1(uQ)x=1⇒Exact⇒Φ(x,y)=∫uPdx=∫uQdy⇒Φ=∫(ex+y)dx=∫(x−e−y)dy⇒ex+xy+ϕ(y)=xy+e−y+ρ(x)⇒Φ=ex+xy+e−y+c1=0

解答: y″−6y′+9y=0⇒λ2−6λ+9=0⇒(λ−3)2=0⇒λ=3⇒yh=c1e3x+c2xe3xApplying variation of parameters, let {y1=e3xy2=xe3x⇒W=|y1y2y′1y′2|=e6x⇒yp=−e3x∫xe3x⋅e3xx2e6xdx+xe3x∫e3x⋅e3xx2e6xdx=−e3x∫1xdx+xe3x∫1x2dx=−e3xlnx−e3x⇒y=yh+yp⇒y=c3e3x+c2xe3x−e3xlnx
解答: (a)L{5}−3L{e−4t}+2L{t3}=5s−3⋅1s+4+2⋅6s4=5s−3s+4+12s4(b)L−1{1(s+2)(s+3)}=L−1{1s+2−1s+3}=e−2t−e−3t

解答: {x+2y+3z=ax+3y+8z=bx+2y+2z=c⇒[123138122][xyz]=[abc]⇒[xyz]=[123138122]−1[abc]=[10−2−7−61510−1][abc]=[10a−2b−7c−6a+b+5ca−c]⇒{x=10a−2b−7cy=−6a+b+5cz=a−c
解答: (a)v1=(1,1),v2=(1,−1)(b)T=[111−1](c)T(x,y)相當於將點(x,y)投影到直線x+y=0,也因此不可逆(d)A=[12−12−1212]⇒A2=[12−12−1212][12−12−1212]=[12−12−1212]=A⇒A2=A⇒lim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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