1. ## laplace question

Hi can anyone help me with this...?

The inverse Laplace transform, L^-1{s+3/s^2+16}

2. Break into 2 elementary inverse transforms

L^-1{s/(s^2+16} = cos(4t)

For the second multiply and divide by 4:

L^-1 {3/(s^2+16} = 3/4 L^(-1){4/s^2+16} = 3/4 sin(4t)

3. can you explain why you multiply and divide by 4 please?

4. L { sin(at)} = a/(s^2 +a^2)

So L^(-1){a/(s^2 +a^2)} = sin(at)

For L^(-1){3/(s^2+16}

needs to be put in the form 4/(s^2+16)

So we use the linear property to write 3/(s^2+16) as 3/4 (4/[s^2+16)])

5. thank you i understand now

6. Originally Posted by Ep!taph
Hi can anyone help me with this...?

The inverse Laplace transform, L^-1{s+3/s^2+16}
It would help if your expression was posted less ambiguously.

By s+3/s^2+16 do you mean s + (3/(s^2 + 16)), s + (3/s^2) + 16 or (s + 3)(s^2 + 16)??

That is, by s+3/s^2+16 do you mean $s + \frac{3}{s^2 + 16}$, $s + \frac{3}{s^2} + 16$ or $\frac{s+3}{s^2 + 16}$ ? (As you can see, learning some basic latex would also help ..... see http://www.mathhelpforum.com/math-he...-tutorial.html)