MATH 308 Lecture 30

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Lecture Notes

Review Day for exam on Wednesday

Exercise 11

Consider the system

$X^{\prime }(t)=\left[\begin{array}{cc}1& 1\\ 4& 1\end{array}\right]X(t)$

1. Show that the vectors $X_1(t)=\left[\begin{array}{c}1\\ 2\end{array}\right]{\mathrm{e}}^{3t}$ and $X_2(t)=\left[\begin{array}{c}1\\ -2\end{array}\right]{\mathrm{e}}^{-t}$ are solutions
2. Are they linearly independent? Describe all solutions to the system
3. Find the solution to the initial value problem $X(0)=\left[\begin{array}{c}3\\ 2\end{array}\right]$

Solution for 1:

$\begin{array}{rlrlr}{X_1}^{\prime }(t)& =\left[\begin{array}{c}3{\mathrm{e}}^{3t}\\ 6{\mathrm{e}}^{3t}\end{array}\right]\left[\begin{array}{cc}1& 1\\ 4& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{e}}^{3t}\\ 2{\mathrm{e}}^{3t}\end{array}\right]& =\left[\begin{array}{c}3{\mathrm{e}}^{3t}\\ 6{\mathrm{e}}^{3t}\end{array}\right]{X_2}^{\prime }(t)& =\left[\begin{array}{c}-{\mathrm{e}}^{-t}\\ 2{\mathrm{e}}^{-t}\end{array}\right]\left[\begin{array}{cc}1& 1\\ 4& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{e}}^{-t}\\ -2{\mathrm{e}}^{-t}\end{array}\right]& =\left[\begin{array}{c}{\mathrm{e}}^{-t}\\ 2{\mathrm{e}}^{-t}\end{array}\right]\end{array}$

$X_1$ and $X_2$ are solutions

Solution for 2:

Note that all solutions are of form $c_1{X}_1+c_2{X}_2=0$

$W(X_1,X_2)=\left|\begin{array}{cc}{\mathrm{e}}^{3t}& {\mathrm{e}}^{-t}\\ 2{\mathrm{e}}^{3t}& -2{\mathrm{e}}^{-t}\end{array}\right|=-4{\mathrm{e}}^{2t}\ne 0$

Therefore, they are linearly independent.

Solution for 3:

Solution for initial value problem satisfies $c_1{X}_1+c_2{X}_2=0$

Therefore

$\begin{array}{rl}X(t)& =c_1\left[\begin{array}{c}{\mathrm{e}}^{3t}\\ 2{\mathrm{e}}^{3t}\end{array}\right]+c_2\left[\begin{array}{c}{\mathrm{e}}^{-t}\\ -2{\mathrm{e}}^{-t}\end{array}\right]\\ X(0)=\left[\begin{array}{c}3\\ 2\end{array}\right]& =c_1\left[\begin{array}{c}1\\ 2\end{array}\right]+c_2\left[\begin{array}{c}1\\ -2\end{array}\right]\\ c_1& =2\\ c_2& =1\end{array}$

Exercise 10.2

Find the eigenvalues and eigenvectors of the given matrix $A=\left[\begin{array}{cc}3& -2\\ 4& -1\end{array}\right]$

$\begin{array}{rl}0=\left|\begin{array}{cc}3-\lambda & -2\\ 4& -1-\lambda \end{array}\right|& =(3-\lambda )(-1-\lambda )-(4)(-2)\\ 0& ={\lambda }^2-2\lambda +5\\ \lambda & =1\pm 2i\\ \end{array}$

Eigenvector for $1+2i$

• Solve $AX=(1+2i)X$, or
• Solve $(A-(1+2i))X=0$

$\begin{array}{rl}\left[\begin{array}{cc}2-2i& -2\\ 4& -2-2i\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}1& -\frac{1}{2}-\frac{1}{2}i\\ 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right]\\ \overrightarrow{x}& =⟨\alpha ,(1-i)\alpha ⟩\end{array}$