Laplace of air resistance

Question

I have the next linear equation to get its laplace transform:

$$\frac{dv}{dt}+cv=g$$

and its values are:

$g = 32ft/sec^2$

$c = 0.25/sec$

My teacher tell me that I need to do a laplace transform to every operation, like this:

$$ Laplace (\frac{dv}{dt}) + Laplace (cv) = Laplace (g) $$

But it is correct do that? Because my result was:

$$ Y(s) [s+c] = v(0) + \frac{g}{s}$$ but I don't know

1. with this result can I already sustituite the c value?
2. Do I need make an Laplace transformation to this equation to get the final result?

Answer 1

You will get $Y(s)\to \frac{g+s v(0)}{s (c+s)}$.

Now inverse laplace transform delivers the velocity as general solution: $$v(t)\mapsto\mathcal{L}_s^{-1}[Y(s)](t)=\mathcal{L}_s^{-1}[\frac{g+s v(0)}{s (c+s)}](t)=\frac{g}{c}(1-e^{-c t})+\text{v(0)} e^{-c t}$$

$v(t)$ starts with $v(0)$ and then tends towards: $$\underset{t\to \infty }{\text{lim}}v(t)=\frac{g}{c}$$