114年台大碩士班-常微分方程詳解

國立臺灣大學 114 學年度碩士班招生考試試題

科目: 常微分方程

解答:$$\textbf{(a) }   u=3-x+y \Rightarrow {du\over dx}=0-1+{dy\over dx} \Rightarrow y'=u'+1 =1+u^3 \Rightarrow u'=u^3 \\ \quad \Rightarrow \int {1\over u^3}\,du = \int 1\,dx \Rightarrow -{1\over 2u^2}=x+c_1 \Rightarrow u^2=(3-x+y)^2={1\over -2x+c_2} \\\quad \Rightarrow \bbox[red, 2pt]{y=x-3\pm {1\over \sqrt{-2x+c_2}}} \\\textbf{(b) }   e^x y^2{dy\over dx}+(1+y^2)=0 \Rightarrow \int {y^2\over 1+y^2}\,dy  =\int -e^{-x}\,dx \Rightarrow  \bbox[red, 2pt]{y-\tan^{-1}y=e^{-x}+c_1} \\\textbf{(c) }  y''+xy'+y=y''+(xy)'=0 \Rightarrow y'+xy =c_1 \Rightarrow  y'(0)+0\cdot y(0)=0=c_1 \\ \quad \Rightarrow y'+xy=0 \Rightarrow \int{1\over y}\,dy =\int -x\,dx \Rightarrow \ln|y|=-{1\over 2}x^2+c_2  \Rightarrow y=c_3e^{-x^2/2} \\\quad \Rightarrow y(0)=c_3=3\Rightarrow \bbox[red, 2pt]{y=3e^{-x^2/2}}$$

解答:$$\textbf{(a) } \det(A-\lambda) =(\lambda-3)^2=0 \Rightarrow \lambda=3 \Rightarrow (A-3I)v=0 \Rightarrow  \begin{bmatrix}-2& 1\\ -4& 2 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix}=0 \Rightarrow x_2=2x_1 \\ \quad \text{Choosing }v_1= \begin{bmatrix}1\\ 2 \end{bmatrix} \Rightarrow (A-3I)v=v_1 \Rightarrow \begin{bmatrix}-2&1\\ -4&2 \end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} = \begin{bmatrix}1\\ 2 \end{bmatrix} \Rightarrow x_2=1+2 x_1 \\\quad \text{Choosing }v_2= \begin{bmatrix}0\\ 1 \end{bmatrix} \Rightarrow e^{At} =e^{(3I+(A-3I))t} =e^{3It }e^{(A-3I)t} =e^{3t}(I+(A-3I)t) \\=e^{3t} \left( \begin{bmatrix}1&0\\0& 1 \end{bmatrix}+t \begin{bmatrix}-2& 1\\ -4& 2 \end{bmatrix} \right) = \bbox[red, 2pt]{e^{3t} \begin{bmatrix}1-2t& t\\ -4t& 1+2t \end{bmatrix}} \\\textbf{(b) } x(t) =e^{At}x(0)= {e^{3t} \begin{bmatrix}1-2t& t\\ -4t& 1+2t \end{bmatrix} \begin{bmatrix}1\\ 0 \end{bmatrix}} = \bbox[red, 2pt]{e^{3t} \begin{bmatrix}1-2t\\ -4t \end{bmatrix}} \\\textbf{(c) }x_p= -A^{-1} \begin{bmatrix}-2\\ 1 \end{bmatrix}  =-{1\over 9} \begin{bmatrix}5&-1\\ 4& 1 \end{bmatrix} \begin{bmatrix}-2\\ 1 \end{bmatrix} = \begin{bmatrix}11/9\\ 7/9 \end{bmatrix} \Rightarrow x=x_h +x_p = e^{At}c+ \begin{bmatrix}11/9\\ 7/9  \end{bmatrix} \\\quad \Rightarrow x(0)= \begin{bmatrix}1\\ 0 \end{bmatrix}=\begin{bmatrix}1& 0\\ 0& 1 \end{bmatrix} c+ \begin{bmatrix}11/9\\ 7/9 \end{bmatrix} \Rightarrow c= \begin{bmatrix}-2/9\\ -7/9 \end{bmatrix} \Rightarrow x(t) =e^{3t} \begin{bmatrix}1-2t& t\\-4t& 1+2t \end{bmatrix} \begin{bmatrix}-2/9\\ -7/9 \end{bmatrix} + \begin{bmatrix}11/9\\ 7/9 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{x(t) = \begin{bmatrix} \displaystyle {-(2+3t)\over 9}e^{3t}+{11\over 9}\\ \displaystyle {-(7+6t)\over 9}e^{3t}+{7\over 9} \end{bmatrix}}$$

解答:$$\textbf{(a) }\mathbf{x\in [\pi/2, \pi] }: y'+xy=0 \Rightarrow y=Ae^{-x^2/2} \Rightarrow y_1(\pi)= Ae^{-\pi^2/2} =2 \Rightarrow A=2e^{\pi^2/2} \\\quad \Rightarrow y_1=2e^{(\pi^2-x^2)/2} \Rightarrow \cases{y_1(\pi/2) =2e^{3\pi^2/8} \\ y_1'(\pi/2) =-\pi e^{3\pi^2/8}} \\\mathbf {x\in [0,\pi/2]}: y''+y=0 \Rightarrow y=c_1\cos x+ c_2\sin x \Rightarrow \cases{y_1(\pi/2)=c_2= 2e^{3\pi^2/8} \\ y_1'(\pi/2)=-c_1=-\pi e^{3\pi^2}/8} \\ \qquad \Rightarrow y_1=e^{3\pi^2/8}(\pi \cos x+ 2\sin x) \\ \Rightarrow \bbox[red, 2pt]{\begin{cases}y_1=e^{3\pi^2/8}(\pi \cos x+ 2\sin x)& x\in[0,\pi/2] \\y_1=2e^{(\pi^2-x^2)/2} & x\in [\pi/2, \pi] \end{cases}} \\\textbf{(b) }  \cases{\mathbf{ x\to \pi/2^-}: y''+y=0 \Rightarrow y''(\pi/2^-) =-y(\pi/2) \\ \mathbf{x\to \pi/2^+}: y'+xy=0 \Rightarrow y''+y+xy'=0 \Rightarrow y''(\pi/2^+)=-y(\pi/2)-{\pi\over 2}y'(\pi/2)} \\ \Rightarrow -y(\pi/2)=-y(\pi/2)-{\pi\over 2}y'(\pi/2) \Rightarrow y'(\pi/2)=0 \Rightarrow 0+{\pi\over 2}y(\pi/2)=0 \Rightarrow y(\pi/2)=0 \\ \Rightarrow y(x)=0 \text{ is the only solution} \Rightarrow y_2(0)\ne 1 \Rightarrow \bbox[red, 2pt]{\text{No, }y_2(0)\ne 1} \\\textbf{(c) }   \text{Trivially, }\cases{y(x)=0 \text{ satisfy (E1) } \\ u,v\in S \Rightarrow u+v\in S\\ y\in S,c\in \mathbb R \Rightarrow cy\in S}\Rightarrow \bbox[red, 2pt]{S\text{ is a vector space}} \\ S=\{0\} \Rightarrow \bbox[red, 2pt]{dim(S)=0}$$

解答:$$\textbf{(a) }y''+y^3=0 \Rightarrow 2y'y''+2y'y^3=0 \Rightarrow {d\over dt}(y')^2 +{d\over dt} \left( {1\over 2}y^4 \right)=0 \\\quad \Rightarrow (y')^2+ {1\over 2}y^4=\text{ constant} \;\bbox[red, 2pt]{QED.} \\\textbf{(b) }y(0) \ne 0 \Rightarrow (y')^2+ {1\over 2}y^4=E \gt 0 \Rightarrow (y')^2 =E-{1\over 2}y^4 \ge 0\Rightarrow y\in[-\sqrt[4]{2E}, \sqrt[4]{2E}] \\ \Rightarrow y'=\pm \sqrt{E-{1\over 2}y^4} \Rightarrow T= 4\int_0^A {dy\over \sqrt{E-{1\over 2}y^4}}, \text{ where }A=\sqrt[4]{2E} \\y=Au \Rightarrow dy =Adu \Rightarrow T=4 \int_0^1{Adu\over \sqrt{E-Eu^4}} ={4A\over \sqrt{E}} \int_0^1{1\over \sqrt{1-u^4}}\,du  \lt \infty \\ \qquad \Rightarrow y \text{ is a periodic function} \quad \bbox[red, 2pt]{QED.}$$

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解題僅供參考，碩士班歷年試題及詳解