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**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?

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Will the format of the exam be similar to past finals? Or will we be tested on much different course material?

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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}

\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}

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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}

\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

Is the quiz at the beginning of the lecture?

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