MATH 308 Lecture 27

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Lecture Notes

April Fool's!

Section 7.1

Exercise 4

$y''+3y'-y=\tan {x}\quad y(0)=1\quad y'(0)=-3$

Let $x_{1}=y$ and $x_{2}=y'$, then

${\begin{cases}x_{1}'=y'=x_{2}\\x_{2}'=y-3y'+\tan {x}=x_{1}-3x_{2}+\tan {x}\end{cases}}$

In matrix form,

${\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}'={\begin{bmatrix}0&1\\1&-3\end{bmatrix}}\,{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}+{\begin{bmatrix}0\\\tan {x}\end{bmatrix}}$

Initial Conditions

${\begin{aligned}x_{1}(0)=y(0)&=1\\x_{2}(0)=y'(0)&=-3\end{aligned}}$

In matrix form,

${\begin{bmatrix}x_{1}(0)\\x_{2}(0)\end{bmatrix}}={\begin{bmatrix}1\\-3\end{bmatrix}}$

Matrix Basics

(See MATH 323 Lecture 3→)

Given $A={\begin{bmatrix}1&-1&-1\\2&1&0\\3&-2&1\end{bmatrix}}$ and $B={\begin{bmatrix}1&1&-1\\2&-1&1\\1&1&2\end{bmatrix}}$

• $A+B={\begin{bmatrix}2&0&-2\\4&0&1\\4&-1&3\end{bmatrix}}$
• $3A-2B={\begin{bmatrix}1&-5&-1\\2&5&-2\\7&-8&-1\end{bmatrix}}$
• $A\,B={\begin{bmatrix}-2&1&-4\\4&1&-1\\0&6&-3\end{bmatrix}}$
• $B\,A={\begin{bmatrix}0&2&-2\\3&-5&-1\\9&-4&1\end{bmatrix}}$

Just for kicks, let ${\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}$...

• $A\,{\vec {x}}={\begin{bmatrix}x_{1}-x_{2}-x_{3}\\2x_{1}+x_{2}\\3x_{1}-2x_{2}+x_{3}\end{bmatrix}}$

Exercise 6

Verify that the given vector satisfies the differential equations

$x'(t)={\begin{bmatrix}2&-1\\3&-2\end{bmatrix}}\,x(t)+{\begin{bmatrix}1\\-1\end{bmatrix}}\,\mathrm {e} ^{t}\quad \quad x(t)={\begin{bmatrix}1\\0\end{bmatrix}}\,\mathrm {e} ^{t}+2\,{\begin{bmatrix}1\\1\end{bmatrix}}\,t\,\mathrm {e} ^{t}$

Calculate the derivative of $x(t)$:

${\begin{aligned}x'(t)&={\begin{bmatrix}1\\0\end{bmatrix}}\,{\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathrm {e} ^{t}\right)+2\,{\begin{bmatrix}1\\1\end{bmatrix}}\,{\frac {\mathrm {d} }{\mathrm {d} t}}\left(t\,\mathrm {e} ^{t}\right)\\&={\frac {\mathrm {d} }{\mathrm {d} t}}{\begin{bmatrix}\mathrm {e} ^{t}+2t\,\mathrm {e} ^{t}\\0+2t\,\mathrm {e} ^{t}\end{bmatrix}}\\&={\begin{bmatrix}3\mathrm {e} ^{t}+2t\,\mathrm {e} ^{t}\\2\mathrm {e} ^{t}+2t\,\mathrm {e} ^{t}\end{bmatrix}}\end{aligned}}$

And we find that $x'(t)=A\,x(t)$.

Section 7.3

Lecture Notes

Given the system

${\begin{aligned}x_{1}+2x_{2}-x_{3}&=2\\2x_{1}+x_{2}+x_{3}&=1\\x_{1}-x_{2}+2x_{3}&=-1\end{aligned}}$

Solve it.

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

$\mathrm {rref} \left({\begin{array}{ccc|c}1&2&-1&2\\2&1&1&1\\1&-1&2&-1\end{array}}\right)=\left[{\begin{array}{ccc|c}1&0&1&0\\0&1&-1&1\\0&0&0&0\end{array}}\right]$

Therefore, the system is inconsisent with solutions

${\begin{aligned}x_{1}&=-x_{3}\\x_{2}&=x_{3}+1\\x_{3}&=x_{3}\end{aligned}}$

Or simply $\left\langle -\alpha ,\alpha +1,\alpha \right\rangle$ for $\alpha \in \mathbb {R}$