# Thread: Laplace transform for IVP

1. ## Laplace transform for IVP

Hey guys, I'm wondering if anyone can show me the steps for this problem. The question is below, I have be able to solved part a and b, but for part c, I know how to get the first two terms but I am not sure how to get the answer for the last term.
thanks

2. Originally Posted by hazeleyes
Hey guys, I'm wondering if anyone can show me the steps for this problem. The question is below, I have be able to solved part a and b, but for part c, I know how to get the first two terms but I am not sure how to get the answer for the last term.
thanks

If you take the Laplace transform we get

For the first part just evaluate this integral

$\int_{0}^{\infty}e^{at}\cdot e^{-st}dt = \int_{0}^{\infty}e^{-t(s-a)}dt$

$s^2X(s)-s(x(0)-x'(0)-a^2X(s)=F(s) \iff X(s)=\frac{s-1}{xs^2-a^2}+F(s)\cdot \frac{1}{x^2-a^2}$

Now use the convolutions theorem to finish the job remember that

$\mathcal{L}^{-1}\{F(s)G(s) \}=\int_{0}^{t}f(t-u)g(u)du$

Can you finish from here?

3. Thanks for your replied. I'm sorry, is it ok if you can show to the step by using this theorem.

4. Originally Posted by hazeleyes
Thanks for your replied. I'm sorry, is it ok if you can show to the step by using this theorem.
So in your problem you have

$\mathcal{L}^{-1}\{F(s)\}=f(t)$

and

$\mathcal{L}^{-1}\{G(s)\}=\mathcal{L}^{-1}\{\frac{1}{s^2-a^2}\}=\frac{1}{a}\sinh(at)$

Now just plug these into the convolution theorem.