The Laplace transform of is , and has the transform .
Use Laplace to solve x''+3x'+5x=sin(8t)
Initial conditions x(0)= 2 and x'(0)= -3
I have worked it down to (s2+3s+5)X(s)= 2s+3+ [8/(s2+64)] but stuck. Please help
Thank you very much. That is what I did to get where I am stuck at the moment. If you check in the previous post I am stuck with this expression (s^2 +3s+5)X(s)= 2s+3+ [8/(s^2+64)] and I need to find the inverse Laplace transform for it.
There are two possible methods of solution.
The first is to put the RHS over a common denominator, before dividing by s^2 + 3s + 5. You then could solve that using Partial Fractions.
The second is to recognise that when you divide by s^2 + 3s + 5 [in other words, multiply by 1/(s^2 + 3s + 5)], the RHS will be the product of two Laplace Transforms. Its inverse will be a convolution integral.
Thank you very much. The equation becomes X(s) = [2s/(s^2+3s+5)] + [3/(s^2+3s+5)] + [8/(s^2+64)(s^2+3s+5)]. My problem is splitting the denominator s^2+3s+5 into an expression that will make it easy to find the inverse Laplace transform.
Thank you very much. It has worked for the term [(2s+3)/(s^2+3s+5)], where I got 2e^-1.5t .cos 1.658t. However I am struggling with the term 8/[(s^2+64)(s^2+3s+5)]. I have tried to split it to get a product of 8/(s^2+64)and 1/(s^2+3s+5) to get a convolution integral but the term 1/(s^2+3s+5) becomes a problem