I am just beginning my study on Laplace transform and I am stuck by a problem asking me to find the inverse Laplace transform of:
$$\dfrac{12(s-2)}{(s^2-4s+85)^2} \tag{$\ast$}$$
Here the definition of inverse Laplace transform is: If $L(f(t)) = F(s) = \int_{0}^{\infty} e^{-st} f(t)$ defines a Laplace transform, then the inverse Laplace transform is $L^{-1}(F(s)) = f(t)$.
My attempt: I wrote $s^2-4s+85$ as $(s-2)^2+ 9^2$ and tried to write $(*)$ as a sum of something, but I failed.
Can somebody give me some hint for this question? Thanks!
Best Answer
Note that $$ \mathcal{L}^{-1} \left[ \dfrac{12(s-2)}{(s^2-4s+85)^2} \right](t) = 12 \cdot \mathcal{L}^{-1} \left[ \frac{s-2}{(s-2)^2 + 9^2} \cdot \frac{1}{(s-2)^2 + 9^2}\right](t) $$ by your algebraic manipulations and linearity.
Now you just need to know the following:
The Laplace transforms of sine and cosine: $$ \mathcal{L} \left[ \sin \alpha t \right](s) = \frac{\alpha}{s^2 + \alpha^2} \qquad \mathcal{L} \left[ \cos \alpha t \right](s) = \frac{s}{s^2 + \alpha^2} $$
The shifting property:
$$ \mathcal{L}[e^{\alpha t} f(t)](s) = \mathcal{L}[f](s-\alpha) $$
The convolution theorem: defining $$ (f\ast g)(t) := \int_0^t f(t-\tau) g(\tau) \, \mathrm{d} \tau $$ we have $$ \mathcal{L}[f \ast g](s) = \mathcal{L}[f](s) \cdot \mathcal{L}[g](s) $$
These should be enough to conclude.