一，上一講的例題，如圖：

設$x=T_1$，$y=T_2$
方程組為·：$\{\begin{array}{c}{x}^{\prime }=?2x+2y\\ y^{\prime }=2x?5y\end{array}$
用消元法求出的通解為：$\{\begin{array}{c}x=c_1{e}^{?t}+c_2{e}^{?6t}\\ y=\frac{1}{2}{c}_1{e}^{?t}?2c_2{e}^{?6t}\end{array}$

二，用矩陣重新表示方程組：
$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{cc}?2& 2\\ 2& ?5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

三，用矩陣重新表示通解：
$\left[\begin{array}{c}x\\ y\end{array}\right]=c_1\left[\begin{array}{c}1\\ \frac{1}{2}\end{array}\right]e^{?t}+c_2\left[\begin{array}{c}1\\ ?2\end{array}\right]e^{?6t}$

四，設解的形式為：
$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]e^{\lambda t}$

五，將解帶入方程組：
$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\lambda \left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]e^{\lambda t}=\left[\begin{array}{cc}?2& 2\\ 2& ?5\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]e^{\lambda t}=\left[\begin{array}{cc}?2& 2\\ 2& ?5\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

六，化簡，求出特征值：

$\lambda \left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=\left[\begin{array}{cc}?2& 2\\ 2& ?5\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]$

$\left[\begin{array}{cc}?2& 2\\ 2& ?5\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]?\lambda \left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=0$

$\left[\begin{array}{cc}?2?\lambda & 2\\ 2& ?5?\lambda \end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=0$

要使等式有非0解，必須滿足：$\left|\begin{array}{cc}?2?\lambda & 2\\ 2& ?5?\lambda \end{array}\right|=0$

解得：${\lambda }_1=?1,{\lambda }_2=?6$，（答案跟上一講求的特征值一樣）

七，將${\lambda }_1$和${\lambda }_2$分別代入等式，求出特征向量：

將${\lambda }_1$代入：$\left[\begin{array}{cc}?1& 2\\ 2& ?4\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=0$

設自由變量$a_1=1$，則$a_2=\frac{1}{2}$，$\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=c_1\left[\begin{array}{c}1\\ \frac{1}{2}\end{array}\right]$，$c_1$為任意常數

$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]e^{\lambda t}=c_1\left[\begin{array}{c}1\\ \frac{1}{2}\end{array}\right]e^{?t}$

將${\lambda }_2$代入：$\left[\begin{array}{cc}4& 2\\ 2& 1\end{array}\right]\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=0$

設自由變量$a_1=1$，則$a_2=?2$，$\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=c_2\left[\begin{array}{c}1\\ ?2\end{array}\right]$，$c_2$為任意常數

$\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]e^{\lambda t}=c_2\left[\begin{array}{c}1\\ ?2\end{array}\right]e^{?6t}$

八，得解空間：
$\left[\begin{array}{c}x\\ y\end{array}\right]=c_1\left[\begin{array}{c}1\\ \frac{1}{2}\end{array}\right]e^{?t}+c_2\left[\begin{array}{c}1\\ ?2\end{array}\right]e^{?6t}$

九，二階矩陣的特征值是如下方程的解：
${\lambda }^2?trace(A)\lambda +detA=0$
trace(A)是A的跡，detA是A的行列式

總結

以上是生活随笔為你收集整理的第二十五讲 用线性代数解微分方程组的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。