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### Messages - Mark Kazakevich

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

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

$$r^6 - r^2 = 0$$

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:
$$y(t) = c_1 + c_2t + c_3\cos{t} + c_4\sin{t} + c_5e^{t} + c_6e^{-t}$$

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

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

$$r^3 - r^2 - r + 1 = 0$$

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:
$$y(t) = c_1e^{t} + c_2e^{t}t + c_3e^{-t}$$

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