### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Mark Kazakevich

Pages: 
1
##### MAT244 Misc / Exam Format/Coverage
« on: December 11, 2013, 05:44:38 PM »
Will the format of the exam be similar to past finals? Or will we be tested on much different course material?

2
##### Quiz 3 / Re: Problem 1 (Day section)
« on: November 07, 2013, 12:36:03 PM »
For the differential equation:
\begin{equation} y^{(6)} - y'' \end{equation}

We assume that $y = e^{rt}$.
Therefore, we must solve the characteristic equation:

\begin{equation} r^6 - r^2 = 0 \end{equation}

We find:
$r^6 - r^2 = 0 \implies r^2(r^4-1) \implies r^2(r^2+1)(r^2-1) = 0 \implies r^2(r^2+1)(r-1)(r+1) = 0$

This means the roots of this equation are:

$r_1 = 0, r_2=0, r_3=i, r_4=-i, r_5=1,r_6=-1$
(We have a repeated root at r = 0)

So the general solution to (1) is:
\begin{equation} y(t) = c_1 + c_2t + c_3\cos{t} + c_4\sin{t} + c_5e^{t} + c_6e^{-t} \end{equation}

3
##### Quiz 3 / Re: Problem 1 (night sections)
« on: November 06, 2013, 08:33:04 PM »
For the differential equation:
\begin{equation} y'''-y''-y'+y=0 \end{equation}

We assume that $y = e^{rt}$.
Therefore, we must solve the characteristic equation:

\begin{equation} r^3 - r^2 - r + 1 = 0 \end{equation}

We find:
$r^3 - r^2 - r + 1 = 0 \implies (r^2-1)(r-1) = 0 \implies (r+1)(r-1)^2 = 0$

This means the roots of this equation are:

$r_1 = 1, r_2=1, r_3=-1$
(We have a repeated root at r = 1)

So the general solution to (1) is:
\begin{equation} y(t) = c_1e^{t} + c_2e^{t}t + c_3e^{-t} \end{equation}

4
##### MAT244 Misc / Re: duration of quiz1
« on: October 01, 2013, 02:07:30 AM »
Is the quiz at the beginning of the lecture?

Pages: