# help with 2nd order linear ODE

• Aug 26th 2009, 06:49 AM
jenny
help with 2nd order linear ODE
Hi, I'm a little unfamiliar with modelling, and having trouble understanding this one in particular about the dampled oscillating system, turned on at time t = 1:

y" + 5y" + 6y = 0 if 0<=t<=1
AND
= 6t - 6 if t>1

Q: what is the displacement y=y(t) during the time interval 0<= t <=1; and evaluate y(1) and y'(1) to two decimal places.

From what I've read, displacement is found by integrating and finding y (?) but I don't quite know how.. Any help or clues would really be appreciated.

Thanks(Nod)

Edit: I know how to get the characteristic roots (-2, -3) and general solution (y = yh = c1e^(-3x) + c2e^(-2x)), but how can i use these to find the displacement?
• Aug 26th 2009, 10:53 AM
chisigma
If we set $y(0)=y'(0)=0$ the 'general integral of the incomplete equation' vanishes so that we can consider the 'particular integral of the complete equation' only. In term of Laplace transform the solution is...

$Y(s) = \frac {6 e^{-s}}{s^{2} (s+2) (s+3)} = 6 e^{-s} (\frac{a}{s^{2}} + \frac{b}{s} + \frac{c}{s+2} + \frac{d}{s+3})$ (1)

The solution in t is derived from (1) finding the values of a,b,c and d and then computing the inverse Laplace transform...

Kind regards

$\chi$ $\sigma$
• Aug 26th 2009, 02:10 PM
HallsofIvy
Quote:

Originally Posted by jenny
Hi, I'm a little unfamiliar with modelling, and having trouble understanding this one in particular about the dampled oscillating system, turned on at time t = 1:

y" + 5y" + 6y = 0 if 0<=t<=1
AND
= 6t - 6 if t>1

Q: what is the displacement y=y(t) during the time interval 0<= t <=1; and evaluate y(1) and y'(1) to two decimal places.

From what I've read, displacement is found by integrating and finding y (?) but I don't quite know how.. Any help or clues would really be appreciated.

Thanks(Nod)

Edit: I know how to get the characteristic roots (-2, -3) and general solution (y = yh = c1e^(-3x) + c2e^(-2x)), but how can i use these to find the displacement?

That is the displacement (for $t\le 1$). You need to determine what c1 and c2 are but you would need some additional information, like y(0) and y'(0). With only the information you have here, there is no way to determine c1 and c2 and so can say nothing more than $y(1)= c1e^{-3}+ c2e^{-2}$ and $y'(1)= -3c1e^{-3}- 2c2e^{-2}$.
• Aug 26th 2009, 05:35 PM
Calculus26
y" + 5y" + 6y = 0 if 0<=t<=1
AND
= 6t - 6 if t>1

If you use Laplace Transforms you can write

y " + 5y ' + 6y = 6u(t-1)(t-1) which gives the sol'n for all t>0

where u(t) is the unit step function

and solve the single eqn. But as HallsofIvy pointed out we need initial conditons
• Aug 26th 2009, 06:07 PM
jenny
ok great. yes i forgot to include the intiial conditions: y(0)=1 and y'(0)=0

Would this mean for y(0) and y'(0) you can substitute the t and y into the 6u(t-1)(t-1) equation, and find u(t).. or..?
• Aug 26th 2009, 06:26 PM
Calculus26
No you use it in the formula forthe laplace transf

L{y''} = s^2F - sy(0) -y'(0)

L{y'} = sF -y(0)

L{u(t-1)f(t-1)} = e^(-s)F

s^2F -s + 5sF -5 + 6F = 6e^(-s) /s^2

F(s^2+5s +6)= 5 + s + 6 e^(-s)/s^2

F = (5+s)/(s+3)(s+2) + 6e^(-s)/[s^2(s+3)s+2)]

F = 1/(s+2) + 2/(s+3)(s+2) + 6e^(-s)/[s^2(s+3)s+2)]

Now I'll let you show we get

y(t) = e^(-2t) +[ 2e^(-2t)-2e^(-3t)] +u(t-1) [3/2e^(-2(t-1)) -2/3 e^-3(t-1)+ (t-1) -5/6)]

which of course you can clean up

Hint 1/[s^2(s+3)(s+2)] = 1/6s^2 -5/36s +1/4(s+2) - 1/9(s+3)
• Aug 26th 2009, 06:37 PM
Calculus26
By the way if you are not familiar with the Laplace transform

solve y" +5y' +6y = 0 on the inteval 0 to 1 --then compute y(1) and
y '(1) which become the initial conditions for the IVP
y" +5y' +6y = 6(t-1) on the interval t > 1
• Aug 26th 2009, 06:40 PM
jenny
ok.. sorry i dont get it. if i'm finding the displacement over the interval 0<=t<=1, why are we finding the IVP for interval t>1?
• Aug 26th 2009, 06:47 PM
Calculus26
I was thinking you wanted the sol'n for all t.

you are right Looking at the problem you don't care about t>1

However what seems strange to me is what is the point of specifing

a forcing function for t > 1 when we only care about what happens when t< 1 ?

The only thing I can think of is we want to match up at t =1 the 2 solutions in which case you would want the solution for t > 1.