Homework Help: Laplace transform

1. Apr 14, 2012

MEM33

1. The problem statement, all variables and given/known data

Solve the differential equation using laplace transforms. Assume zero intial conditions and that the forcing functions are zero prior to t=0

x''(t) + 6x'(t) +8x(t)=sin3t

2. Relevant equations

3. The attempt at a solution

I took the laplace transform which I found to be after simplyfing

X(s)=-8s^2-57/(s^2+3^2)(s^2+6s)

after partial fraction decompistion and comparing coeffcients I found these equations

0=a+b+c+d
-8=6a+6b+6c
0= 9c
-57=54c+9d

When I plug that into Ax=b it gives me free variables. i am not sure if my logic is wrong or I just made some mistakes along the way. Any help would be gratly appreciated.

MEM33

2. Apr 14, 2012

dikmikkel

I just made it. Try to write the two polynomials as a factor of the roots like (s-r1)(s-r2)(s-r3)(s-r4).
I also think you made the transform wrong.
Btw, if when you factor the poly in the denominator you can get the part.frac coefficients like this:
$\lim\limits_{s\rightarrow r_1}\{X(s)(s-r_1)\}$
Can you see why?

Last edited: Apr 14, 2012
3. Apr 14, 2012

MEM33

I found my mistake

Thanks!

4. Apr 14, 2012

Ray Vickson

Is your expression supposed to be $$X(s) = -8s^2 -\frac{57}{(s^2+3^2)(s^2+6s)} \text{ or } X(s) = \frac{-8s^2 - 57}{(s^2+3^2)(s^2+6s)}?$$
Using standard rules the expression, as written, means the first one.

RGV

5. Apr 14, 2012

MEM33

Sorry after re-working it my X(s) looks like 15/(s^2 + 3^2)(s +4)(s +2)

and my solution is X(t) = -.4cos(3t) +.5sin(3t) - .3exp(-4t) +.47exp(-2t)

Can anyone tell me if that is the correct solution?

6. Apr 15, 2012

dikmikkel

i get (evaluated as decimal numbers):
$-0.06\cos(3t)-0.003\sin(3t)+0.12e^{-2t}-0.06e^{-4t}$