115年台大碩士班-常微分方程詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 常微分方程

解答:$$u={1\over y^2 } \Rightarrow u'=-{2\over y^3}y' \Rightarrow y'= -{1\over 2}u'y^3 \Rightarrow -{1\over 2}u'y^3+y= y^3e^t \Rightarrow -{1\over 2}u'+u=e^t \\ \Rightarrow u'-2u=-2e^t \Rightarrow u'e^{-2t}-2ue^{-2t}=-2e^{-t} \Rightarrow (ue^{-2t})'=-2e^{-t} \\\Rightarrow ue^{-2t} =\int-2e^{-t}\,dt =2e^{-t}+c_1 \Rightarrow u={1\over y^2}=2e^t+c_1e^{2t} \Rightarrow y^2={1\over 2e^t+c_1e^{2t}} \\ \Rightarrow \bbox[red, 2pt]{y= \pm \sqrt{1\over 2e^t+c_1e^{2t}}}$$

解答:$$\cases{x'=2x+y \\ y'=x+2y} \Rightarrow \begin{bmatrix}x'\\ y' \end{bmatrix} = \begin{bmatrix}2& 1\\1& 2 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} \\A= \begin{bmatrix}2& 1\\1&2  \end{bmatrix} \Rightarrow \det(A-\lambda I) =(\lambda-1)(\lambda-3)=0 \Rightarrow \cases{\lambda_1 =1 \Rightarrow v_1= \begin{bmatrix}1\\ -1 \end{bmatrix} \\[1ex]\lambda_2=3 \Rightarrow v_2= \begin{bmatrix}1\\1 \end{bmatrix}} \\ \Rightarrow \begin{bmatrix}x\\ y \end{bmatrix} =c_1v_1 e^{\lambda_1 t} +c_2v_2e^{\lambda_2 t} =c_1 \begin{bmatrix}1\\-1 \end{bmatrix}e^t +c_2 \begin{bmatrix}1 \\1 \end{bmatrix}e^{3t} \Rightarrow \cases{x(0)=c_1+c_2=1\\ y(0)=-c_1 +c_2=0}  \\ \Rightarrow \cases{c_1=1/2\\ c_2 =1/2} \Rightarrow \begin{bmatrix}x \\ y \end{bmatrix} = {1\over 2} \begin{bmatrix}1\\-1 \end{bmatrix}e^t+ {1\over 2} \begin{bmatrix}1\\1 \end{bmatrix}e^{3t} \Rightarrow  \bbox[red, 2pt]{\cases{x(t)=\displaystyle {1\over 2}e^t+ {1\over 2}e^{3t} \\y(t) =-\displaystyle {1\over 2}e^t +{1\over 2}e^{3t}}}$$

解答:$$y''+4y'+4y=0 \Rightarrow r^2+4r+4=0 \Rightarrow (r+2)^2=0 \Rightarrow r=-2 \Rightarrow y(t) =c_1e^{-2t}+ c_2te^{-2t} \\ \Rightarrow y'(t)=(-2c_1 +c_2)e^{-2t}-2c_2te^{-2t} \Rightarrow \cases{y(0)=c_1 =1 \\ y'(0)=-2c_1+c_2 =0} \Rightarrow \cases{c_1= 1\\ c_2=2} \\ \Rightarrow \bbox[red, 2pt]{y(t)=e^{-2t} +2te^{-2t}}$$

解答:$$y''+y=0 \Rightarrow r^2+1=0 \Rightarrow r=\pm i \Rightarrow y_h =c_1\cos t+ c_2\sin t \\ \text{Let } \cases{y_1=\cos t\\ y_2=\sin t} \Rightarrow W(t)= \begin{vmatrix} y_1& y_2\\ y_1'& y_2' \end{vmatrix} = \begin{vmatrix} \cos t & \sin t\\ -\sin t& \cos t\end{vmatrix}=1 \\\Rightarrow y_p=-\cos t\int \sin t\cdot \sec t\,dt +\sin t\int \cos t\cdot \sec t\,dt =-\cos t \int \tan t\,dt +\sin t \int 1\,dt\\=\cos t \ln \cos t+ t\sin t \Rightarrow y=y_h +y_p \Rightarrow \bbox[red, 2pt]{y=c_1\cos t+ c_2\sin t+ t\sin t+ \cos t\ln \cos t}$$

解答:$$\cases{x'=y\\ y'=-x-y^3} \Rightarrow J(x,y) = \begin{bmatrix}\displaystyle {\partial x'\over \partial x} &\displaystyle {\partial x'\over \partial y} \\ \displaystyle {\partial y'\over \partial x} &\displaystyle {\partial y'\over \partial y} \end{bmatrix} = \begin{bmatrix}0& 1\\ -1 &-3y^2 \end{bmatrix} \Rightarrow J(0,0)= \begin{bmatrix}0&1\\-1& 0 \end{bmatrix} \\ \Rightarrow \det(J-\lambda I) =0 \Rightarrow \lambda=\pm i \Rightarrow \text{ the eigenvalues are purely imaginary (zero real part)}\\ \Rightarrow \text{the linear stability test is inconclusive.} \\\text{Using Lyapunov's Direct Method: }V(x,y)={1\over 2}x^2+ {1\over 2}y^2, \text{ where }\cases{V(0,0)=0 \\\text{V is positive definite}} \\ \Rightarrow \dot{V} ={\partial V\over \partial x}x'+{\partial V\over \partial y}y'=xy-xy-y^4=-y^4 \le 0 \Rightarrow \text{the origin is stable.} \\ \text{Applying LaSalle's Invariance Principle: }E=\{(x,y) \mid -y^4=0\} =\{(x,0) \mid x\in \mathbb R\} =x-\text{axis} \\ \Rightarrow \text{ the largest invariant set $M$ in $E$ is just the equilibrium point itself} \\ \Rightarrow \text{every trajectory in the plane will eventually converge to the origin} \\ \Rightarrow \bbox[red, 2pt]{\text{(0,0) is globally asymptotically stable}}$$

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解題僅供參考，碩士班歷年試題及詳解