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2022年9月2日 星期五

109年台綜大轉學考-工程數學D36詳解

 臺灣綜合大學系統109學年度學士班轉學生聯合招生考試

科目名稱:工程數學
類組代碼:D36

解答(1)(mx+ny)dx+(kx+ly)dy=0{M(x,y)=mx+nyN(x,y)=kx+ly{My=nNx=kthe equation is exact if n=kif n=kΨ(x,y)=Mdx=Ndy{mx+nydx=12mx2+nxy+f(y)nx+lydy=nxy+12ly2+g(x)Ψ(x,y)=12mx2+nxy+12ly2=C(2)my+ny+ky=0mλ2+nλ+k=0λ=n±n24mk2m=n2myh=C1enx/2m+C2xenx/2myp=Aexyp=yp=AexmAex+nAex+kAex=lexA(m+n+k)=lA=l/(m+n+k)yp=lm+n+kex=exy=yh+ypy=C1enx/2m+C2xenx/2m+ex

解答(1)A=[1000cos(θ)sin(θ)0sin(θ)cos(θ)]{adj(A)=[1000cos(θ)sin(θ)0sin(θ)cos(θ)]det

解答f(x,y)={x\over y} \Rightarrow \cases{f_x= 1/y\\ f_y= -x/y^2}\\(1) (\text{grad}f) \cdot (\text{grad}f) =(f_x,f_y)\cdot (f_x, f_y) = ({1\over y},-{x\over y^2}) \cdot ({1\over y},-{x\over y^2}) = \bbox[red, 2pt]{{1\over y^2}+{x^2\over y^4}}\\ (2)\nabla^2(f^2) =\nabla^2({x^2\over y^2}) ={\partial^2\over \partial x^2 } {x^2\over y^2}+{\partial^2\over \partial y^2 } {x^2\over y^2} ={\partial\over \partial x } {2x \over y^2}+{\partial\over \partial y } {-2x^2\over y^3} =\bbox[red, 2pt]{ {2\over y^2}+ {6x^2\over y^4}}

解答(1)\;\mathcal{L}\{f\} =\int_0^\infty f(t)e^{-st}\,dt = \int_a^b ke^{-st}\,dt = \left.\left[ -{1\over s}ke^{-st} \right] \right|_a^b =\bbox[red,2pt]{ {k\over s}\left(e^{-as}-e^{-bs} \right)} \\(2)\;\mathcal{L}\{f\} =\int_0^\infty f(t)e^{-st}\,dt = \int_0^k {t\over k}e^{-st}\,dt = {1\over k}\left. \left[-{t\over s}e^{-st}-{1\over s^2}e^{-st} \right] \right|_0^k = \bbox[red, 2pt]{{1\over k}\left( {1\over s^2}-{1\over s}e^{-sk}(k+{1\over s})\right)}

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