Use Laplace to solve x''+3x'+5x=sin(8t)

Initial conditions x(0)= 2 and x'(0)= -3

I have worked it down to (s^{2}+3s+5)X(s)= 2s+3+ [8/(s^{2}+64)] but stuck. Please help

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- Mar 24th 2012, 06:47 AMsim907Using Laplace to solve a differential equation
Use Laplace to solve x''+3x'+5x=sin(8t)

Initial conditions x(0)= 2 and x'(0)= -3

I have worked it down to (s^{2}+3s+5)X(s)= 2s+3+ [8/(s^{2}+64)] but stuck. Please help - Mar 24th 2012, 07:03 AMRidleyRe: Using Laplace to solve a differential equation
The Laplace transform of $\displaystyle f'(x)$ is $\displaystyle sF(s)-f(0)$, and $\displaystyle f''(x)$ has the transform $\displaystyle s^2F(s) - sf(0) - f'(0)$.

$\displaystyle \mathcal{L}\left \{\sin {8t} \right\}=\frac{8}{s^2+8^2}$ - Mar 24th 2012, 04:45 PMsim907Re: Using Laplace to solve a differential equation
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.

- Mar 24th 2012, 09:18 PMProve ItRe: Using Laplace to solve a differential equation
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. - Mar 24th 2012, 11:24 PMsim907Re: Using Laplace to solve a differential equation
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.

- Mar 24th 2012, 11:50 PMProve ItRe: Using Laplace to solve a differential equation
Try completing the square.

- Mar 25th 2012, 01:29 AMsim907Re: Using Laplace to solve a differential equation
Thanks but completing the square gives (s+1.5)^2 +2.75 which I am not sure if it can be used in this case.

- Mar 25th 2012, 02:11 AMProve ItRe: Using Laplace to solve a differential equation
It can. You need to turn the numerator into a function of (s + 1.5) and then you can use a shift theorem.

- Mar 25th 2012, 03:31 AMsim907Re: Using Laplace to solve a differential equation
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

- Mar 25th 2012, 03:38 AMProve ItRe: Using Laplace to solve a differential equation
Why not try splitting it up into partial fractions instead?

- Mar 25th 2012, 03:47 AMProve ItRe: Using Laplace to solve a differential equation
In fact, you've already completed the square on $\displaystyle \displaystyle \begin{align*} \frac{1}{s^2 + 3s + 5} = \frac{1}{(s + 1.5)^2 + 2.75} \end{align*}$, use the shift theorem :)

- Mar 25th 2012, 04:17 AMsim907Re: Using Laplace to solve a differential equation
My calculator TI-nspire CAS is giving the partial fraction as -24x/[4057(x^2+64)]+24x/[4057(x^2+3x+5)]-472/[4057(x^2+64)]+544/[4057(x^2+3x+5)]. Does this look right?

- Mar 25th 2012, 04:21 AMsim907Re: Using Laplace to solve a differential equation
Sorry I did not see your last post. Did not refresh the page. I will work on the shift theorem. Thanks

- Mar 25th 2012, 05:00 AMsim907Re: Using Laplace to solve a differential equation
My answer to the whole problem is sin(8t) + (2cos1.658t + 0.603sin1.658t).e^-1.5t . Does this look right?

- Mar 25th 2012, 08:05 PMsim907Re: Using Laplace to solve a differential equation
Thank you very much for your help and patience. I got it finally. It was, however long and messy. Here is how I did it