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)=e−x−f1(x)=−ex⇒{f1(−x)≠f1(x)f1(−x)≠−f1(x)⇒ex不是偶函數也不是奇函數f2(x)=ex2⇒{f2(−x)=ex2−f2(x)=−ex2⇒f2(x)=f2(−x)⇒ex2為偶函數f3(x)=tanx⇒{f3(−x)=−tanx−f3(x)=−tanx⇒f3(−x)=−f3(x)⇒tanx為奇函數f4(x)=sinhx⇒{f4(−x)=−sinhx−f4(x)=−sinhx⇒f4(−x)=−f4(x)⇒sinhx為奇函數⇒{ex不是偶函數也不是奇函數ex2為偶函數tanx為奇函數sinhx為奇函數解答:{f=x+y−z⇒{fx=1fy=1fz=−1g=xyz⇒{gx=yzgy=xzgz=xy⇒{∂∂xfg=fxg+fgx=xyz+(x+y−z)yz∂∂yfg=fyg+fgy=xyz+(x+y−z)xz∂∂zfg=fzg+fgz=−xyz+(x+y−z)xy⇒grad(fg)=(∂∂xfg,∂∂yfg,∂∂zfg)=(yz(2x+y−z),xz(x+2y−z),xy(x+y−2z))⇒div(grad(fg))=∂∂xyz(2x+y−z)+∂∂yxz(x+2y−z)+∂∂zxy(x+y−2z)=2yz+2xz−2xy=2(yz+xz−xy)
解答:A=(−1−10−1−10002)⇒det(A−λI)=0⇒−λ(λ−2)(λ+2)=0⇒{λ1=0λ2=2λ3=−2λ1=0⇒(A−λ1I)x=0⇒(−1−10−1−10002)(x1x2x3)=0⇒{x1=−x2x3=0,取v1=(−110)λ2=2⇒(A−λ2I)x=0⇒(−3−10−1−30000)(x1x2x3)=0⇒{x1=0x2=0,取v2=(001)λ3=−2⇒(A−λ3I)x=0⇒(1−10−110004)(x1x2x3)=0⇒{x1=x2x3=0,取v3=(110)取P=(v1v2v3)及D=(λ1000λ2000λ3)⇒A=PDP−1因此eigenbasis={(−110),(001),(110)},對角化A=(−101101010)(00002000−2)(−1212000112120)
解答:{y′1−2y1+3y2=0y′2−y1+2y2=0⇒{L{y′1−2y1+3y2}=0L{y′2−y1+2y2}=0⇒{sY1(s)−y1(0)−2Y1(s)+3Y2(s)=0sY2(s)−y2(0)−Y1(s)+2Y2(s)=0⇒{(s−2)Y1(s)+3Y2(s)=1⋯(1)−Y1(s)+(s+2)Y2(s)=0⋯(2),由(2)得Y1(s)=(s+2)Y2(s)代入(1)⇒(s2−4)Y2(s)+3Y2(s)=1⇒Y1(s)=1s2−1⇒y1(t)=L−1{1s2−1}=12(et−e−t)再將Y1(s)=1s2−1代入(2)⇒Y2(s)=1(s+2)(s2−1)⇒y2(t)=L−1{1(s+2)(s2−1)}=13e−2t−12e−t+16et⇒y1(t)=12(et−e−t),y2(t)=13e−2t−12e−t+16et
解答:y=a0+a1x+a2x2+a3x3+a4x4+⋯⇒y′=a1+2a2x+3a3x2+4a4x3+⋯⇒3xy′=3a1x+6a2x2+9a3x3+12a4x4+⋯⇒(3x−1)y′=−a1+(3a1−2a2)x+(6a2−3a3)x2+(9a3−4a4)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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