【习题】5.2 一阶常系数线性微分方程组

标签：2t 5.2 end sqrt3 frac begin pmatrix 一阶常 习题

[T050201] 求方程组 \(\begin{cases}x'+y'=y+z\\ y'+z'=z+x\\ z'+x'=x+y\end{cases}\) 的通解.

    解 将方程组写成矩阵形式:

\[\begin{aligned} &\begin{pmatrix} 1&1&0\\0&1&1\\1&0&1 \end{pmatrix} \begin{pmatrix} x'\\y'\\z' \end{pmatrix}= \begin{pmatrix} 0&1&1\\1&0&1\\1&1&0 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix}\\ \\ \Longrightarrow &\begin{pmatrix} x'\\y'\\z' \end{pmatrix}=\begin{pmatrix} 1&1&0\\0&1&1\\1&0&1 \end{pmatrix}^{-1}\begin{pmatrix} 0&1&1\\1&0&1\\1&1&0 \end{pmatrix}\begin{pmatrix} x\\y\\z \end{pmatrix}=\begin{pmatrix} 0&1&0\\0&0&1\\1&0&0 \end{pmatrix}\begin{pmatrix} x\\y\\z \end{pmatrix} :=A\begin{pmatrix} x\\y\\z \end{pmatrix} \end{aligned} \]

则 \(|\lambda E-A|=\begin{vmatrix}\lambda&-1&0\\0&\lambda&-1\\-1&0&\lambda\end{vmatrix}=\lambda^3-1=0\Longrightarrow \lambda_1=1,\lambda_2=-\frac{1}{2}+\frac{\sqrt3}{2}i,\lambda_3=-\frac{1}{2}-\frac{\sqrt3}{2}i\).
    对于特征值 \(\lambda_1=1\), 有

\[A-\lambda_1E= \begin{pmatrix} -1&1&0\\0&-1&1\\1&0&-1 \end{pmatrix}\xrightarrow{初等行变换} \begin{pmatrix} 1&0&-1\\0&1&-1\\0&0&0 \end{pmatrix} \]

于是可取特征向量 \(v_1=(1,1,1)^T\), 从而常系数齐次线性方程组有一个解 \(X_1(t)=e^tv_1\).
    对于特征值 \(\lambda_2=-\frac{1}{2}+\frac{\sqrt3}{2}i\), 有

\[A-\lambda_2E= \begin{pmatrix} \frac12-\frac{\sqrt{3}}{2}i&1&0\\0&\frac12-\frac{\sqrt3}{2}i&1\\1&0&\frac12-\frac{\sqrt{3}}{2}i \end{pmatrix} \xrightarrow{初等行变换} \begin{pmatrix} 1&0&\frac12-\frac{\sqrt3}{2}i\\0&1&\frac12+\frac{\sqrt3}{2}i\\0&0&0 \end{pmatrix} \]

于是可取特征向量 \(v_2=\left(-\frac12-\frac{\sqrt3}{2}i,-\frac12+\frac{\sqrt3}{2}i,1\right)^T\). 注意到

\[\begin{aligned} e^{(-\frac{1}{2}+\frac{\sqrt3}{2}i)t}v_2&=e^{-\frac12t} \begin{pmatrix} (-\frac{1}{2}+\frac{\sqrt3}{2}i)e^{\frac{\sqrt3}{2}it}\\ (-\frac{1}{2}-\frac{\sqrt3}{2}i)e^{\frac{\sqrt3}{2}it}\\ e^{\frac{\sqrt3}{2}it} \end{pmatrix}\\ &=e^{-\frac12t} \left\{ \begin{pmatrix} -\frac12\cos\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\ -\frac12\cos\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\ \cos\frac{\sqrt3}{2}t \end{pmatrix}+i \begin{pmatrix} -\frac12\sin\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\ -\frac12\sin\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\ \sin\frac{\sqrt3}{2}t \end{pmatrix} \right\} \end{aligned} \]

于是另两个线性无关的解为

\[X_2(t)=e^{-\frac12t}\begin{pmatrix} -\frac12\cos\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\ -\frac12\cos\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\ \cos\frac{\sqrt3}{2}t \end{pmatrix},\quad X_3(t)=e^{-\frac12t}\begin{pmatrix} -\frac12\sin\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\ -\frac12\sin\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\ \sin\frac{\sqrt3}{2}t \end{pmatrix} \]

注意到 \(X_1(t),X_2(t),X_3(t)\) 是常系数齐次线性方程组的三个线性无关解, 故通解为

\[\begin{aligned} X(t)&=c_1X_1(t)+c_2X_2(t)+c_3X_3(t)\\ &=c_1e^t\begin{pmatrix} 1\\1\\1 \end{pmatrix}+ c_2e^{-\frac12t}\begin{pmatrix} -\frac12\cos\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\ -\frac12\cos\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\sin\frac{\sqrt3}{2}t\\ \cos\frac{\sqrt3}{2}t \end{pmatrix}+ c_3e^{-\frac12t}\begin{pmatrix} -\frac12\sin\frac{\sqrt3}{2}t+\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\ -\frac12\sin\frac{\sqrt3}{2}t-\frac{\sqrt3}{2}\cos\frac{\sqrt3}{2}t\\ \sin\frac{\sqrt3}{2}t \end{pmatrix} \end{aligned} \]

其中 \(c_1,c_2,c_3\) 为任意常数. #

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[T050202] 求常系数非齐次线性方程组的通解:

\[X'=\begin{pmatrix} -1&2\\-2&3 \end{pmatrix}X+\begin{pmatrix} 1\\0 \end{pmatrix}. \]

    解 记 \(X'=AX+F\), 于是

\[|\lambda E-A|=\begin{vmatrix} \lambda+1&-2\\2&\lambda-3 \end{vmatrix}=(\lambda-1)^2\Longrightarrow \lambda=1 \ (二 \ 重) \]

从而对应的常系数齐次线性方程组有两个解形如 \(X(t)=(R_0+R_1t)e^t\), 其中 \(R_0,R_1\) 满足

\[\begin{cases} (A-E)R_0=R_1\\ (A-E)^2R_0=0 \end{cases} \]

注意到

\[(A-E)^2=\begin{pmatrix} -2&2\\-2&2 \end{pmatrix}^2= O \]

故可取 \(R_0=(1,0)^T\) 和 \((0,1)^T\), 从而 \(R_1=(-2,-2)^T\) 和 \((2,2)\), 从而对应的常系数齐次线性方程组有两个线性无关解

\[X_1(t)=\left[ \begin{pmatrix} 1\\0 \end{pmatrix}+ \begin{pmatrix} -2\\-2 \end{pmatrix}t \right]e^t=e^t\begin{pmatrix} 1-2t\\-2t \end{pmatrix}\\ \\ X_2(t)=\left[ \begin{pmatrix} 0\\1 \end{pmatrix}+ \begin{pmatrix} 2\\2 \end{pmatrix}t \right]e^t=e^t\begin{pmatrix} 2t\\1+2t \end{pmatrix} \]

    由此可得对应的常系数齐次线性方程组的基解矩阵为 \(\Phi(t)=e^t\begin{pmatrix} 1-2t&2t\\-2t&1+2t \end{pmatrix}\),\(\Phi^{-1}(t)=e^{-t}\begin{pmatrix} 1+2t&-2t\\2t&1-2t \end{pmatrix}\), 于是原常系数非齐次线性方程组有特解

\[\begin{aligned} \bar X(t)&=\Phi(t)\int_0^t\Phi^{-1}(s)F\mathrm{~d}s=e^t\begin{pmatrix} 1-2t&2t\\-2t&1+2t \end{pmatrix}\begin{pmatrix} -3e^{-t}-2te^{-t}+3\\-2e^{-t}-2te^{-t}+2 \end{pmatrix}\\ &=e^t\begin{pmatrix} -3e^{-t}+3-2t\\-2e^{-t}+2-2t \end{pmatrix}=\begin{pmatrix} -3+(1-2t)e^t+2e^t\\-2+(-2t)e^t+2e^t \end{pmatrix} \end{aligned} \]

故原方程组的通解

\[\begin{aligned} X(t)&=c_1X_1(t)+c_2X_2(t)+\bar X(t)\\ &=c_1e^t\begin{pmatrix} 1-2t\\-2t \end{pmatrix}+c_2e^t\begin{pmatrix} 2t\\1+2t \end{pmatrix}+\begin{pmatrix} -3+(1-2t)e^t+2e^t\\-2+(-2t)e^t+2e^t \end{pmatrix}\\ &=c_1e^t\begin{pmatrix} 1-2t\\-2t \end{pmatrix}+c_2e^t\begin{pmatrix} 2t\\1+2t \end{pmatrix}+\begin{pmatrix} -3+2e^t\\-2+2e^t \end{pmatrix}\\ &=c_1e^t\begin{pmatrix} 1-2t\\-2t \end{pmatrix}+c_2e^t\begin{pmatrix} 2t\\1+2t \end{pmatrix}+\begin{pmatrix} -3\\-2 \end{pmatrix}. \quad\quad\# \end{aligned} \]

标签：2t,5.2,end,sqrt3,frac,begin,pmatrix,一阶常,习题
From： https://www.cnblogs.com/hznudmh/p/18049417