115年台北科技大學製造所碩士班-微分方程詳解

國立臺北科技大學115學年度碩士班招生考試

系所組別:製造科技研究所 科目:微分方程

解答:$$\cases{P(x,y)= 2x+e^y\\ Q(x,y)=xe^y-1} \Rightarrow P_y=e^y =Q_x \Rightarrow \text{ Exact} \\ \Rightarrow \Psi(x,y)= \int (2x+e^y)\,dx =\int (xe^y-1)\,dy \Rightarrow \Psi(x,y)=x^2+xe^y +\rho(y)=xe^y-y+ \phi(x) \\ \Rightarrow \Psi(x,y)= \bbox[red, 2pt]{x^2+xe^y-y=C}$$

解答:$$y'=2xe^{-y} \Rightarrow e^y y'=2x\Rightarrow \int e^y\,d y= \int 2x\,dx \Rightarrow e^y=x^2+C \Rightarrow y=\ln(x^2+C)\\ \Rightarrow y(0)=\ln C=0 \Rightarrow C=1 \Rightarrow \bbox[red, 2pt]{y=\ln(x^2+1)}$$

解答:$$y''+y=0 \Rightarrow r^2+1=0 \Rightarrow r=\pm i \Rightarrow y_h= c_1\cos x+ c_2\sin x\\ \text{Let }\cases{y_1=\cos x\\ y_2= \sin x} \Rightarrow W= \begin{vmatrix} \cos x& \sin x\\ -\sin x& \cos x \end{vmatrix} =1 \\ \text{By variation of parameters, }y_p=-y_1 \int{y_2\cdot r(x)\over W}\,dx +y_2\int {y_1\cdot r(x)\over W} \,dx , \text{ where }r(x)=\sec x \\ \Rightarrow y_p=-\cos x\int \tan x\,dx +\sin x \int1\,dx=-\cos x \ln|\sec x|+\sin x\cdot x \\ \Rightarrow y=y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1\cos x+c_2\sin x-\cos x\ln|\sec x|+x\sin x}$$

解答:$$y''+2y'+y = 2e^{-t} u(t-2) \Rightarrow  \mathcal{L}\{y''\} + 2\mathcal{L}\{y'\} + \mathcal{L}\{y\} =  \mathcal{L}\{2e^{-t}u(t-2)\}\\ \Rightarrow  [s^2 Y(s) - sy(0) - y'(0)] + 2[s Y(s) - y(0)] + Y(s) = 2e^{-2}  \mathcal{L}\{ e^{-(t-2)} u(t-2)\}  \\ \Rightarrow (s+1)^2 Y(s) =2e^{-2} \left( e^{-2s} \cdot {1\over s+1} \right) \Rightarrow  Y(s) = e^{-2s} \frac{2e^{-2}}{(s + 1)^3} \\  \mathcal{L}^{-1}\left\{\frac{1}{(s + 1)^3}\right\} = \frac{t^2}{2!} e^{-t} = \frac{1}{2}t^2 e^{-t}  \Rightarrow  \mathcal{L}^{-1}\left\{\frac{2e^{-2}}{(s + 1)^3}\right\} =2e^{-2} \left( {1\over 2}t^2e^{-t} \right) =e^{-2}t^2e^{-t} \\ \Rightarrow y(t) =\mathcal{L}^{-1}\left\{e^{-2s} \frac{2e^{-2}}{(s + 1)^3}\right\} = u(t-2)\cdot e^{-2}(t-2)^2 e^{-(t-2) } \Rightarrow \bbox[red, 2pt]{y(t) = (t - 2)^2 e^{-t} u(t - 2)}$$

解答:$$\cases{x'=4x-y\\ y'=2x+y} \Rightarrow \begin{bmatrix}x'\\ y' \end{bmatrix} = \begin{bmatrix}4& -1\\2& 1 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix}\\ A=\begin{bmatrix}4& -1\\2& 1 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2-5\lambda+6 =(\lambda-2)(\lambda-3)=0 \Rightarrow \lambda_1=2, \lambda_2=3 \\ \cases{\lambda_1=2 \Rightarrow (A-\lambda_1 I)v_1= 0 \Rightarrow v_1= \begin{bmatrix}1\\2 \end{bmatrix} \\ \lambda_2=3 \Rightarrow(A-\lambda_2I )v_2=0 \Rightarrow v_2= \begin{bmatrix}1\\1 \end{bmatrix}} \Rightarrow X(t) =c_1v_1e^{\lambda_1 t}+ c_2v_2 e^{\lambda_2 t} \\ \Rightarrow \begin{bmatrix} x(t)\\ y(t) \end{bmatrix} =c_1 \begin{bmatrix}1\\2 \end{bmatrix}e^{2t}+c_2 \begin{bmatrix}1\\1 \end{bmatrix} e^{3t} \Rightarrow \begin{bmatrix}x(0) \\y(0) \end{bmatrix} = \begin{bmatrix}1\\ 0 \end{bmatrix} =c_1 \begin{bmatrix}1\\2 \end{bmatrix}+c_2 \begin{bmatrix}1\\1 \end{bmatrix} = \begin{bmatrix}c_1+c_2\\ 2c_1+c_2 \end{bmatrix} \\ \Rightarrow \cases{c_1+c_2=1\\ 2c_1+c_2=0} \Rightarrow \cases{c_1=-1\\ c_2=2} \Rightarrow \bbox[red, 2pt]{\cases{x(t)=-e^{2t}+2e^{3t} \\ y(t)=-2e^{2t}+2e^{3t}}}$$

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解題僅供參考，碩士班歷年試題及詳解