Laplace Transform Help

Use the Laplace transform to solve the following initial value problem:

y''-5y'-14y=sin(8t) y(0)=-3, y'(0)=1

First, using Y for the Laplace transform of y(t), i.e., Y=L{y(t)},
find the equation you get by taking the Laplace transform of the differential equation

? = 0

I'm confused as how to go about this so if anyone could walk me through this I'd really appreciate it. Also there is no variable "t" in the solution Thanks in advance for your time!

This is long, so I will just give you a way to start.

First you find do the laplace transform of the whole equation, that is

L{ y''-5y'-14y=sin(8t) }

from tables (or http://en.wikipedia.org/wiki/Laplace_tra… ), this turns into

s^2 Y - s y(0) - y'(0) - 5( sY - y(0) ) - 14 Y = 8 / (s^2 + 64)

which can be rearranged as

(s^2 - 5 s - 14) Y = 8 / (s^2 + 64) - 3 s + 16

(s - 7) (s + 2) Y = 8 / (s^2 + 64) - 3 s + 16

where I substituted the values of y'(0) and y(0) and id a factorization of the quadratic equation for s. Solving for Y we get

Y = 8 / ((s^2 + 64)(s-7)(s+2)) + (-3 s + 16)/((s - 7) (s + 2))

your job is now to separate these into partial fractions and apply the inverse of the laplace transform. The partial fractions will look like this

Y = A/(s - 7) + B/(s + 2) + Cs/(s^2 + 64) + D/(s^2 + 64)

whose inverse laplace transform, from the tables mentioned, is

y(t) = A exp (7t) + B exp (-2t) + C/8 cos(8t) + D/8 sin(8t)

your job now is to find the values of A, B, C, and D. This is a rather long algebra so I'll leave it to you. That is your solution.

use this theorem: L{dy/dt}= sY-y(0) and L{d2y/dt2}= s^2 Y -sy(0) -dy/dt( 0)
so: s^2 Y +3s -s -5(sY +3) = 64/(s^2 + 64) ==> Y(s^2 -5s) = 64/(s^2 + 64) +2s +15 ==>
Y = F(s)... all left to do is take the inverse laplace transform using P.F.E. I'm not gonna do that for u!!
I hope it helped.