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： $$(1)\;(mx+ny)dx +(kx+ly)dy=0 \Rightarrow \cases{M(x,y)=mx+ny\\ N(x,y)=kx+ly} \Rightarrow \cases{M_y =n\\ N_x =k}\\ \Rightarrow \text{the equation is exact if }\bbox[red, 2pt]{ {n=k}} \\\text{if }n=k \Rightarrow \Psi(x,y)=\int M\,dx = \int N\,dy \Rightarrow \cases{\int mx+ny\,dx ={1\over 2}mx^2+nxy +f(y)\\ \int nx+ly\,dy = nxy +{1\over 2}ly^2 +g(x)}\\ \Rightarrow \Psi(x,y)=\bbox[red, 2pt]{{1\over 2}mx^2 +nxy+ {1\over 2}ly^2=C}\\(2)\;my''+ ny'+ky=0 \Rightarrow m\lambda^2+ n\lambda+k=0 \Rightarrow \lambda ={-n \pm \sqrt{n^2-4mk}\over 2m} ={-n\over 2m}\\ \Rightarrow y_h= C_1e^{-nx/2m} +C_2xe^{-nx/2m} \\ 令y_p= Ae^x \Rightarrow y_p'=y_p''=Ae^x \Rightarrow mAe^x+ nAe^x+kAe^x = le^x \Rightarrow A(m+n+k)= l\\ \Rightarrow A=l/(m+ n+k) \Rightarrow y_p = {l\over m+n+k}e^x =e^x\Rightarrow y=y_h +y_p\\ \Rightarrow \bbox[red, 2pt]{y=C_1e^{-nx/2m} +C_2xe^{-nx/2m} + e^x }$$

解答： $$(1)\;A=\left[\begin{matrix}1 & 0 & 0 \\0 & \cos\left(\theta\right) & -\sin\left(\theta\right) \\0 & \sin\left(\theta\right) & \cos\left(\theta \right)\end{matrix}\right] \Rightarrow \cases{\text{adj}(A)=\begin{bmatrix} 1 & 0 & 0\\ 0 & \cos(\theta) & \sin(\theta)\\ 0 & -\sin (\theta) & \cos(\theta)\end{bmatrix} \\ \det(A)=1} \\ \Rightarrow A^{-1}={1\over \det(A)} \text{adj}(A) = \bbox[red, 2pt]{\begin{bmatrix} 1 & 0 & 0\\ 0 & \cos(\theta) & \sin(\theta)\\ 0 & -\sin (\theta) & \cos(\theta)\end{bmatrix} }\\(2)\;A=\left[\begin{matrix}1 & 0 & 0 \\0 & \cos\left(\theta\right) & -\sin\left(\theta\right) \\0 & \sin\left(\theta\right) & \cos\left(\theta \right)\end{matrix}\right] \Rightarrow \det(A- \lambda I)= -(\lambda -1)(\lambda^2 -2\cos\theta \lambda+1)=0\\ \Rightarrow \text{eigenvalues of } A \text{ are: } \bbox[red,2pt]{1, \cos\theta\pm i\sin \theta }$$

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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