107年文化大學碩士班-工程數學詳解

中國文化大學107學年度碩士班考試入學

系組：化學及材料奈米碩士班
科目：工程數學

解答：$$\mathbf{(a)}\;y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} 代回原式\Rightarrow m(m-1)x^m -3mx^m+ 4x^m=0\\ \Rightarrow x^m(m^2-4m+4)=0 \Rightarrow m=2(重根）\Rightarrow y=C_1x^2 +C_2x^2\ln x \Rightarrow y'=(2C_1+ C_2)x +2C_2x\ln x \\ 將\cases{y(1)=4\\ y'(1)=5}代入\Rightarrow \cases{C_1=4\\ 2C_1+ C_2 = 5} \Rightarrow C_2=-3 \Rightarrow \bbox[red,2pt]{y=4x^2-3x^2\ln x}\\ \mathbf{(b)}\;先求齊次解，即y''-y'-2y=0 \Rightarrow \lambda^2-\lambda -2=0 \Rightarrow \lambda =2,-1 \Rightarrow y_h= C_1e^{2x}+C_2 e^{-x}\\ 再假設y_p= A\cos x+B\sin x \Rightarrow y'=-A\sin x+B\cos x \Rightarrow y''=-A\cos x-B\sin x\\ \Rightarrow -A\cos x-B\sin x+A\sin x-B\cos x-2A\cos x-2B\sin x\\ =(-3A-B)\cos x+(A-3B)\sin x =10 \cos x \Rightarrow \cases{-3A-B= 10\\ A-3B=0} \Rightarrow \cases{A=-3\\B=-1} \\ \Rightarrow y=y_h+y_p = C_1e^{2x}+C_2 e^{-x}-3\cos x-\sin x \Rightarrow y'=2C_1e^{2x}-C_2e^{-x}+3\sin x-\cos x\\ 再將\cases{y(0)=2 \\ y'(0)=1}代入\Rightarrow \cases{C_1+C_2-3=2\\ 2C_1-C_2-1=1} \Rightarrow \cases{C_1=7/3\\ C_2=8/3} \\ \Rightarrow \bbox[red,2pt]{y={7\over 3}e^{2x}+{8\over 3}e^{-x}-3\cos x-\sin x} \\\mathbf{(c)}\;令\cases{M(x,y)= 2xy^4+ \sin y\\ N(x,y)= 4x^2y^3 + x\cos y} \Rightarrow \cases{{\partial \over \partial y}M= 8xy^3+ \cos y\\ {\partial \over \partial x}N = 8xy^3+ \cos y} \Rightarrow {\partial \over \partial y}M= {\partial \over \partial x}N \\ 取\Psi(x,y) = \int M\,dx = \int N\,dy \Rightarrow \Psi= \int 2xy^4+ \sin y\,dx = \int 4x^2y^3 + x\cos y\,dy \\ \Rightarrow \Psi = x^2y^4+ x\sin y + \phi(y)= x^2y^4+ x\sin y + \rho(x) \Rightarrow \bbox[red,2pt]{x^2+y^4+ x\sin y=C} $$

解答：$$L\{ \sin kt\} =I=\int_0^\infty \sin(kt)e^{-st}\,dt = -{1\over s}e^{-st}\sin(kt) +{k\over s}\int \cos(kt)e^{-st}\,dt \\ =-{1\over s}e^{-st}\sin(kt) -{k\over s^2} \cos(kt)e^{-st}-{k^2\over s^2}I \Rightarrow (1+{k^2\over s^2}) I = \left. \left[-{1\over s}e^{-st}\sin(kt) -{k\over s^2} \cos(kt)e^{-st}\right] \right|_0^\infty \\ \Rightarrow {s^2+k^2\over s^2} I={k\over s^2} \Rightarrow I= {k\over s^2}\cdot {s^2\over s^2+k^2} ={k\over s^2+k^2} \Rightarrow L\{ \sin kt\} ={k\over s^2+k^2},\bbox[red, 2pt]{故得證}$$

解答：$$\det(M-\lambda I)=k(\lambda+1) (\lambda-4)(\lambda-9) \Rightarrow \begin{vmatrix}\alpha-\lambda & 0 & 3\beta\\ 0 & \alpha-\lambda & 4\beta\\ 3\beta & 4\beta & \alpha-\lambda \end{vmatrix} \\=(\alpha-\lambda)^3 -25\beta^2(\alpha -\lambda) =(\alpha-\lambda)(\alpha-\lambda+5\beta)(\alpha-\lambda-5\beta) \\ \Rightarrow -(\lambda-\alpha)(\lambda-\alpha-5\beta)(\lambda-\alpha+5\beta)=k(\lambda+1) (\lambda-4)(\lambda-9)\\ \Rightarrow \cases{\alpha=4\\ \alpha+5\beta =9\\ \alpha-5\beta=-1} \Rightarrow \bbox[red,2pt]{\cases{\alpha=4\\ \beta= 1}}$$

4. (20%) Find the particular solution of following differential equation:

\( \cases{{dx\over dt}=2x+y+ 4e^{2t}\\ {dy\over dt} = x+2y}\) which satisfies the initial conduction \( \cases{x(0)=4 \\ y(0)=1}\)

解答：$$\cases{x'= 2x+y+4e^{2t} \cdots(1)\\ y'=x+2y\cdots(2)},式(2) \Rightarrow x=y'-2y \Rightarrow x'=y''-2y'\\ 將\cases{x=y'-2y\\ x'=y''-2y'} 代入(1) \Rightarrow y''-2y'=2(y'-2y)+ y+ 4e^{2t} \Rightarrow y''-4y'+3y=4e^{2t} \cdots(3)\\ \Rightarrow \cases{y_h =C_1e^{3t}+ C_2e^t\\ y_p= ke^{2t}} 將y_p'=2ke^{2t},y_p'' =4ke^{2t} 代入(3) \Rightarrow k=-4 \Rightarrow y=C_1e^{3t}+ C_2e^t-4e^{2t}\\ \Rightarrow x=y'-2y= C_1e^{3t}-C_2e^t；最後將\cases{x(0)= 4\\ y(0)=1}代入\Rightarrow \cases{C_1-C_2= 4\\ C_1+C_2-4 = 1} \\ \Rightarrow \cases{C_1=9/2\\ C_2= 1/2} \Rightarrow \bbox[red, 2pt]{\cases{x= {9\over 2} e^{3t}-{1\over 2}e^t\\ y= {9\over 2} e^{3t}+{1\over 2}e^t -4e^{2t}}}$$

 

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