Steady state value, a possible shortcut?

Question

I have this differential equation:

$\dot{\eta (t)}=\frac{-2\eta(t)}{C(\widehat{R}+\Delta R)}+\frac{K^2 L}{C E^2 (\widehat{R}+\Delta R)}+K$

which, my book says, correspond to a LTI system, asymptotically stable with a constant input:

$\frac{K^2 L}{C E^2 (\widehat{R}+\Delta R)}+K$

It says also that the steady state value of the variable $\eta (t)$ is

$\tilde{\eta (t)} = \frac{K C(\widehat{R}+\Delta R)}{2} + \frac{K^2 L}{2E^2}$

The book doesn't say how to get this, but I reached the same result solving the differential equation (after a change a variable it can be solved separating variables) and calculating the limit for $t\rightarrow\infty$. My question is: is it really necessary to solve the differential equation or is there any property I can use to avoid it? I'm asking this because looking at the expression of the steady state value you can see that is equivalent to the ratio of the input and the coefficient of $\eta(t)$. Is it random or am I missing something?

Answer 1

Think for a moment about what it means for $\eta(t)$ to have reached its steady state. It means precisely that $\dot{\eta}(t)=0$. If you plug that into your first equation and solve for $\eta(t)$ you get your solution as you've already observed.