Using a vector notation, the system reads
$$\frac{d\mathbf x}{dt}=A\mathbf x+\mathbf b.$$
Take the Laplace transform, giving
$$s\mathbf X-\mathbf x_0=A\mathbf X+\frac{\mathbf b}s,$$ solved by
$$\mathbf X=(sI-A)^{-1}\left(\mathbf x_0+\frac{\mathbf b}s\right).$$
To get the response at infinity, multiply by $s$ ant take the limit at $0$,
$$\mathbf x_\infty=\lim_{s\to0}s\mathbf X=\lim_{s\to0}\,(sI-A)^{-1}\left(s\mathbf x_0+\mathbf b\right)=-A^{-1}\mathbf b.$$
This is obviously the solution of the steady-state system
$$0=A\mathbf x+\mathbf b.$$
Using the Laplace transform is overkill for this problem.
Hint:
$$\mathcal{L}(t^2*f(t))=+\dfrac{{d}^2F(s)}{{ds}^2}$$
And you can rewrite the equation to the following:
$${cos}^2t=\dfrac{1}{2}(cos(2t)+1)$$
Applying this gives:
$$\mathcal{L}(t^2*{cos}^2t)=\mathcal{L}(t^2*\dfrac{1}{2}(cos2t+1))$$
$$=\dfrac{1}{2}\mathcal{L}(t^2cos2t)+\dfrac{1}{2}\mathcal{L}(t^2)$$
This gives:
$$\dfrac{1}{2}*\dfrac{d^2(\frac{s}{s^2+4})}{{ds}^2}+\dfrac{1}{s^3}=\dfrac{s(s^2-12)}{(s^2+4)^3}+\dfrac{1}{s^3}$$
Best Answer
Let $f(t)$ denote the time-domain function, and $F(s)$ denote its Laplace transform. The final value theorem states that: $$ \lim_{t \to \infty} f(t) = \lim_{s\to 0} sF(s), $$ where the LHS is the steady state of $f(t).$ Since it is typically hard to solve for $f(t)$ directly, it is much easier to study the RHS where, for example, ODEs become polynomials or rational functions in $s.$
You can get more info in course notes like this one: PDF, or in control engineering/system theory books such as Kailath: Linear Systems or Ogata: Modern Control Engineering.
Also, as noted in the comments, there are Physics Stackexchange and Electrical Engineering Stackexchange