inverse Laplace of $\frac{\left(x^2-6x+7\right)}{\left(x^2-4x+5\right)^2}$

Question

inverse Laplace of $\frac{\left(s^2-6s+7\right)}{\left(s^2-4s+5\right)^2}$
i tried partial frictions but i get $\frac{2-2s}{\left(s^2-4s+5\right)^2}+\frac{1}{s^2-4s+5}$
how to find inverse laplace of the first term? For Second term I used complete square

Answer 1

Just another way:

$$\color{red}{\mathcal{L}_s^{-1}\left[\frac{s^2-6 s+7}{\left(s^2-4 s+5\right)^2}\right](t)}=\\-t \left(\mathcal{L}_s^{-1}\left[\int \frac{s^2-6 s+7}{\left(s^2-4 s+5\right)^2} \, ds\right](t)\right)=\\-t \left(\mathcal{L}_s^{-1}\left[\frac{3-s}{5-4 s+s^2}\right](t)\right)$$

Using Heaviside's expansion formula (See example 32 page 62) for: $ \mathcal{L}_s^{-1}\left[\frac{3-s}{5-4 s+s^2}\right](t)$ we have:

$P(s)=3-s$,

$Q(s)=5-4 s+s^2$,

$Q'(s)=-4+2 s$,

solve roots for:$Q(s)$

$\text{$\alpha $1}=2-i$ and $\text{$\alpha $2}=2-i$

$$\frac{P(\text{$\alpha $1}) \exp (t \text{$\alpha $1})}{Q'(\text{$\alpha $1})}+\frac{P(\text{$\alpha $2}) \exp (t \text{$\alpha $2})}{Q'(\text{$\alpha $2})}=\frac{P(2-i) \exp (t (2-i))}{Q'(2-i)}+\frac{P(2+i) \exp (t (2+i))}{Q'(2+i)}=\left(-\frac{1}{2}+\frac{i}{2}\right) \exp (t (2-i))+\left(-\frac{1}{2}-\frac{i}{2}\right) \exp (t (2+i))=e^{2 t} (-\cos (t)+\sin (t))$$ and then:

$$-t \left(\mathcal{L}_s^{-1}\left[\frac{3-s}{5-4 s+s^2}\right](t)\right)=-t e^{2 t} (-\cos (t)+\sin (t))=\color{red}{e^{2 t} t (\cos (t)-\sin (t))}$$