# Toronto Math Forum

## MAT244-2013F => MAT244 Math--Tests => Quiz 3 => Topic started by: Razak Pirani on November 07, 2013, 12:35:09 PM

Title: Problem 1 (Day section)
Post by: Razak Pirani on November 07, 2013, 12:35:09 PM
4.2 #18 Find the general solution of the given differential equation.

y(6) - y'' = 0

r6 - r2 = 0
r2(r4 - 1) = 0
r2(r2 - 1)(r2 + 1) = 0
r2(r - 1)(r + 1)(r - i)(r + i) = 0

r1,2 = 0
r3 = 1
r4 = -1
r5 = i
r6 = -i

y(t) = c1 + c2t + c3et + c4e-t + c5cost + c6sint
Title: Re: Problem 1 (Day section)
Post by: Mark Kazakevich 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:
$$y(t) = c_1 + c_2t + c_3\cos{t} + c_4\sin{t} + c_5e^{t} + c_6e^{-t}$$