# Math Help - Laplace Transforms

1. ## Laplace Transforms

see the attachment for the questions i am stuck on.
includes first and second order differential equations which need to be solved using Laplace Transforms.
your help would be much appreciated
thanks

2. Originally Posted by louboutinlover
see the attachment for the questions i am stuck on.
includes first and second order differential equations which need to be solved using Laplace Transforms.
your help would be much appreciated
thanks
(s-3)/ [(s+3)(s-1)] = A/(s+3) + B/(s-1)

There is a quick trick to solving for the coefficients of this partial fraction. To find A for example I multiply both sides of the equation by (s+3), now evaluate the left hand side at s = -3 since B is gone.

For A i get (s-3)/(s-1) evaluated at -3 gives 6/4 or 3/2. For B i do the same but multiply both sides by (s-1) and get -1/2

So now take the inverse laplace transform by using a look up table if you wish and I get 1.5e^(-3t) - 0.5e^(-t)

3. Actually I assumed your previous workings were corrrect but I spotted your denominator is incorrect. Your function in the Laplace domain should be F(s) = (5-s)/[(s-1)(s+3)]. By employing the same procedure as above I seperated the function into its partial fraction components, namely (5-s)/[(s-1)(s+3)] = A/(s-1) + B/(s+3) and found A = 1 and B = -2. The answer I finally got was
F(x) = e^x - 2e^(-3x).

4. HI I am extremely sorry but I made another small error. The question is very simple its just a small algabraic mistake both me and you have been making.

Notice sX(s) - 1 + 3X(s) = 4/(s-1)

X(s)(s+3) = 4/(s-1) +1

now when you solve for X(s) you should obtain 1/(s-1) becausethe (s+3)'s cancel.

Finally the inverse laplace transfrom of 1/(s-1) is indeed e^t.

Sorry about the previous post, I have been running low on sleep the pastfew days.