Inverse Laplace transform of $F(s)=\frac{3s+7}{s^2-2s-3}$

Question

I have to calculate the inverse Laplace Transform of this image: $$F(s)=\frac{3s+7}{s^2-2s-3}$$ I try decomposing it in this way: $$F(s)=\frac{3s-3+10}{(s-1)^2-4}=3\frac{s-1}{(s-1)^2-4}+5\frac{2}{(s-1)^2-4}$$ where I can identify that the original function is $$f(t)=3e^t\cosh(2t)+5e^t\sinh(2t)$$

But the textbooks says that the result should be: $f(t)=-e^{-t}+4e^{3t}$ and I can't find where my mistake is.

Answer 1

$$ -e^{-t}+4e^{3t} = 3e^t\cosh(2t)+5e^t\sinh(2t) \text{.} $$

Their solutions was found by partial fraction decomposition: $$F(s) = \frac{4}{s-3} + \frac{-1}{s+1} \text{.} $$

Answer 2

Partial fraction decomposition leads to $$F(s) = \frac{4}{s-3} - \frac{1}{s+1}$$

Answer 3

You are correct! Indeed, we have that $$f(t)=3e^t\cosh(2t)+5e^t\sinh(2t)=3e^t\frac{e^{2t}+e^{-2t}}{2}+5e^t\frac{e^{2t}-e^{-2t}}{2}=4e^{3t}-e^{-t}.$$