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2022年9月25日 星期日

111年台綜大轉學考-工程數學D04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:工程數學
類組代碼:D04

解答{f1(x)=ex{f1(x)=exf1(x)=ex{f1(x)f1(x)f1(x)f1(x)exf2(x)=ex2{f2(x)=ex2f2(x)=ex2f2(x)=f2(x)ex2f3(x)=tanx{f3(x)=tanxf3(x)=tanxf3(x)=f3(x)tanxf4(x)=sinhx{f4(x)=sinhxf4(x)=sinhxf4(x)=f4(x)sinhx{exex2tanxsinhx
解答{f=x+yz{fx=1fy=1fz=1g=xyz{gx=yzgy=xzgz=xy{xfg=fxg+fgx=xyz+(x+yz)yzyfg=fyg+fgy=xyz+(x+yz)xzzfg=fzg+fgz=xyz+(x+yz)xygrad(fg)=(xfg,yfg,zfg)=(yz(2x+yz),xz(x+2yz),xy(x+y2z))div(grad(fg))=xyz(2x+yz)+yxz(x+2yz)+zxy(x+y2z)=2yz+2xz2xy=2(yz+xzxy)
解答A=(110110002)det(AλI)=0λ(λ2)(λ+2)=0{λ1=0λ2=2λ3=2λ1=0(Aλ1I)x=0(110110002)(x1x2x3)=0{x1=x2x3=0,v1=(110)λ2=2(Aλ2I)x=0(310130000)(x1x2x3)=0{x1=0x2=0,v2=(001)λ3=2(Aλ3I)x=0(110110004)(x1x2x3)=0{x1=x2x3=0,v3=(110)P=(v1v2v3)D=(λ1000λ2000λ3)A=PDP1eigenbasis={(110),(001),(110)},A=(101101010)(000020002)(1212000112120)
解答{y12y1+3y2=0y2y1+2y2=0{L{y12y1+3y2}=0L{y2y1+2y2}=0{sY1(s)y1(0)2Y1(s)+3Y2(s)=0sY2(s)y2(0)Y1(s)+2Y2(s)=0{(s2)Y1(s)+3Y2(s)=1(1)Y1(s)+(s+2)Y2(s)=0(2),(2)Y1(s)=(s+2)Y2(s)(1)(s24)Y2(s)+3Y2(s)=1Y1(s)=1s21y1(t)=L1{1s21}=12(etet)Y1(s)=1s21(2)Y2(s)=1(s+2)(s21)y2(t)=L1{1(s+2)(s21)}=13e2t12et+16ety1(t)=12(etet),y2(t)=13e2t12et+16et
解答y=a0+a1x+a2x2+a3x3+a4x4+y=a1+2a2x+3a3x2+4a4x3+3xy=3a1x+6a2x2+9a3x3+12a4x4+(3x1)y=a1+(3a12a2)x+(6a23a3)x2+(9a34a4)x3+y
解答\mathbf{(a)}\;令\cases{M(x,y)=e^{x+y}+ ye^y\\ N(x,y)= xe^y-1},則(e^{x+y}+ ye^y)dx +(xe^y-1)dy=0 相當於Mdx+Ndy=0\\ 由於\cases{M_y= e^{x+y}+e^y+ ye^y\\ N_x=e^y} \Rightarrow M_y\ne N_x \Rightarrow 非恰當(Not\; Exact)\\ 令積分因子\mu(y) \Rightarrow \cases{{\partial \over \partial y}M\mu =M_y\mu +M\mu_y= (e^{x+y}+e^y+ ye^y)\mu +(e^{x+y}+ ye^y)\mu_y\\ {\partial \over \partial x}N\mu = N_x\mu =e^y\mu} \\ \Rightarrow (e^{x+y}+e^y+ ye^y)\mu +(e^{x+y}+ ye^y)\mu_y=e^y\mu \Rightarrow \mu+\mu_y=0 \Rightarrow \mu=e^{-y}\\ \Rightarrow \cases{M\mu= e^x+y\\ N\mu = x-e^{-y}}\Rightarrow \Psi(x,y)=\int (e^x+y)\,dx = \int x-e^{-y}\,dy \Rightarrow \Psi =e^x+xy+ \phi(y)=xy+e^{-y}+ \varphi(x)\\ \Rightarrow \Psi=e^x+xy+ e^{-y}=C,再將初始值y(0)=-3代入\Rightarrow 1+e^{3} =C\\ \Rightarrow \bbox[red,2pt]{e^x+xy+ e^{-y}= 1+e^{3}}
解答\lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n\to \infty} \left|{x^{2n+3}\over (2n+3)!} \cdot { (2n+1)!\over x^{2n+1}} \right| = x^2\lim_{n\to \infty} {1\over (2n+3)(2n+2)} =0\\ \Rightarrow 收斂半徑= \bbox[red, 2pt]{\infty}

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