MATH 308 Lecture 27

« previous | Monday, April 1, 2013 | next »

Lecture Notes

April Fool's!

Section 7.1

Exercise 4

$y^{″}+3y^{\prime }-y=\tan xy(0)=1y^{\prime }(0)=-3$

Let $x_1=y$ and $x_2=y^{\prime }$, then

$\{\begin{array}{l}{x_1}^{\prime }=y^{\prime }=x_2\\ {x_2}^{\prime }=y-3y^{\prime }+\tan x=x_1-3x_2+\tan x\end{array}$

In matrix form,

${\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]}^{\prime }=\left[\begin{array}{cc}0& 1\\ 1& -3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left[\begin{array}{c}0\\ \tan x\end{array}\right]$

Initial Conditions

$\begin{array}{rl}{x}_1(0)=y(0)& =1\\ x_2(0)=y^{\prime }(0)& =-3\end{array}$

In matrix form,

$\left[\begin{array}{c}{x}_1(0)\\ x_2(0)\end{array}\right]=\left[\begin{array}{c}1\\ -3\end{array}\right]$

Matrix Basics

(See MATH 323 Lecture 3→)

Given $A=\left[\begin{array}{ccc}1& -1& -1\\ 2& 1& 0\\ 3& -2& 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}1& 1& -1\\ 2& -1& 1\\ 1& 1& 2\end{array}\right]$

• $A+B=\left[\begin{array}{ccc}2& 0& -2\\ 4& 0& 1\\ 4& -1& 3\end{array}\right]$
• $3A-2B=\left[\begin{array}{ccc}1& -5& -1\\ 2& 5& -2\\ 7& -8& -1\end{array}\right]$
• $AB=\left[\begin{array}{ccc}-2& 1& -4\\ 4& 1& -1\\ 0& 6& -3\end{array}\right]$
• $BA=\left[\begin{array}{ccc}0& 2& -2\\ 3& -5& -1\\ 9& -4& 1\end{array}\right]$

Just for kicks, let $\overrightarrow{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$...

• $A\overrightarrow{x}=\left[\begin{array}{c}{x}_1-x_2-x_3\\ 2x_1+x_2\\ 3x_1-2x_2+x_3\end{array}\right]$

Exercise 6

Verify that the given vector satisfies the differential equations

$x^{\prime }(t)=\left[\begin{array}{cc}2& -1\\ 3& -2\end{array}\right]x(t)+\left[\begin{array}{c}1\\ -1\end{array}\right]{\mathrm{e}}^{t}x(t)=\left[\begin{array}{c}1\\ 0\end{array}\right]{\mathrm{e}}^t+2\left[\begin{array}{c}1\\ 1\end{array}\right]t{\mathrm{e}}^t$

Calculate the derivative of $x(t)$:

$\begin{array}{rl}{x}^{\prime }(t)& =\left[\begin{array}{c}1\\ 0\end{array}\right]\frac{\mathrm{d}}{\mathrm{d}t}\left({\mathrm{e}}^t\right)+2\left[\begin{array}{c}1\\ 1\end{array}\right]\frac{\mathrm{d}}{\mathrm{d}t}\left(t{\mathrm{e}}^t\right)\\ & =\frac{\mathrm{d}}{\mathrm{d}t}\left[\begin{array}{c}{\mathrm{e}}^t+2t{\mathrm{e}}^t\\ 0+2t{\mathrm{e}}^t\end{array}\right]\\ & =\left[\begin{array}{c}3{\mathrm{e}}^t+2t{\mathrm{e}}^t\\ 2{\mathrm{e}}^t+2t{\mathrm{e}}^t\end{array}\right]\end{array}$

And we find that $x^{\prime }(t)=Ax(t)$.

Section 7.3

Lecture Notes

Given the system

$\begin{array}{rl}{x}_1+2x_2-x_3& =2\\ 2x_1+x_2+x_3& =1\\ x_1-x_2+2x_3& =-1\end{array}$

Solve it.

(The teacher is using the long way, I'll cut to the chase...)

$\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{f}\left(\begin{array}{cccc}1& 2& -1& 2\\ 2& 1& 1& 1\\ 1& -1& 2& -1\end{array}\right)=\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& -1& 1\\ 0& 0& 0& 0\end{array}\right]$

Therefore, the system is inconsisent with solutions

$\begin{array}{rl}{x}_1& =-x_3\\ x_2& =x_3+1\\ x_3& =x_3\end{array}$

Or simply $⟨-\alpha ,\alpha +1,\alpha ⟩$ for $\alpha \in \mathbb{ℝ}$