Laplace transform for IVP

**hazeleyes** · May 21st 2011, 06:15 AM

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

**TheEmptySet** · May 21st 2011, 06:29 AM

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?

**hazeleyes** · May 21st 2011, 07:02 AM

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

**TheEmptySet** · May 21st 2011, 07:32 AM

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.

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