Solve the Initial Value Problem using method of Laplace Transforms:

y''-y'-2y= -8cos(t) -2sin(t)
y(pi/2)=1 y'(pi/2)=0

Question

Solve the Initial Value Problem using method of Laplace Transforms:

y''-y'-2y= -8cos(t) -2sin(t)
y(pi/2)=1  y'(pi/2)=0

usukidoll · Answer

do you have the laplace table with you?

usukidoll · Answer

you have to apply laplace that that original equation first and then subsitute all of the derivatives with its value.. afterwards use partial fractions and take the laplace.. sorry i need to go and study.

anonymous · Answer

I get how to solve it but i'm used to y(0) not y( another value)

accessdenied · Answer

I think we should make a substitution of a new independent value in this case.
We currently have t = pi/2.  We want to change it to a variable that equals zero when t=pi/2.  That's easy enough to find:
   $ \tau = t - \pi/2 $      ==>  $ t = \pi / 2 + \tau $
If we make this substitution:
  $ y'' (t) - y'(t) - 2y(t) = -8\cos t - 2\sin t $
  $ y''(\tau + \pi/2) - y'(\tau + \pi/2) - 2y(\tau + \pi/2) = -8 \cos (\tau + \pi/2) - 2 \sin (\tau + \pi/2) $
Then we can just call a new dependent variable:  $  u(\tau) = y(\tau + \pi/2) $ and its derivatives
  $ u' ' ( \tau ) - u ' ( \tau ) - 2 u ( \tau ) =  -8 \cos (\tau + \pi/2) - 2 \sin (\tau + \pi/2) $
With $ u(0) = y(\pi/2) $  and $ u'(0) = y'(\pi/2) $

accessdenied · Answer

I don't think I emphasized properly that the purpose of this is to obtain initial bounds of u(0) and u'(0), but that is all we wanted from this substitution.

In the end once we solve for u(tau) we just have to take one more step and revert back to y(t).
  y(t) = y(tau + pi/2) = u(tau) = u(t - pi/2)

accessdenied · Answer

I have to go for a while, so if you still need assistance please feel free to bump this question up. I will also leave a resource I was viewing for another example if it is helpful: http://tutorial.math.lamar.edu/Classes/DE/IVPWithLaplace.aspx Good luck!

anonymous · Answer

Ah, this was complicated but thank you so much for explaining it! It took a while but I figured it out. Thanks again!

accessdenied · Answer

Yeah, it is quite a bit to take in although I tried to explain it as thoroughly as I could.  I know it hadn't come to me at first either so I had to see another example to get it down for this problem.  :P
Anyways, glad to help!