1st order linear DE with step function input the 1st order linear equation is:
$y'(t) + \frac D M y(t) = f(t)$
with constants:
$D = 100kg/s$
$M = 1000kg$
$f(t) = Fu(t)$  <-- that's Force x the unit step function
an initial condition:
$y(0) = 20.8m/s$
the input is a step function scaled by the Force $F$ ($Fu(t)$)
we need to solve the DE and then find the Force needed to make the final velocity $27.8m/s$.
also a block diagram  with the Laplace transform:
$f(t) \longrightarrow {\frac 1M \over (s + \frac DM)}$
thank you!
here's what i have so far...
first i integrated the linear function.
$y'(t) + .1 y(t) = .001f(t)$
using $mu$ in the linear DE and the initial condition y(0) = 20.8
$y(t) = .01 + 20.79 e^(-.1t)$  that's e to the power of -.1t
the problem is i can't figure out what to do with the right side of the equation.  the step function scaled by force.  i need help integrating the right side.  $Fu(t)$  
i need to solve the equation to a point where i can input a constant value for the force in order to aim for the target velocity of 27.8m/s.
 A: We are given:
$$\tag 1 y'(t) + \frac D M y(t) = \dfrac{1}{M}f(t)$$
where:


*

*$D = 100kg/s$

*$M = 1000kg$

*$f(t) = Fu(t)$, Force $\times$ Heaviside unit step function

*Initial Condition (IC): $y(0) = 20.8m/s$


Rewriting $(1)$ yields:
$$\tag 2 y'(t) + \dfrac{1}{10} y = \dfrac{F}{1000} u(t)$$
Taking the Laplace Transform of $(2)$ yields:
$$\mathcal{L}\left(y'(t) + \dfrac{1}{10} y = \dfrac{F}{1000} u(t)\right) = s y(s) - y(0) +  \dfrac{1}{10} y(s) = \dfrac{F}{1000 s}$$
We want to group the $y(s)$ term on the LHS side and everything else on the RHS, so we have:
$$y(s)\left(s + \dfrac{1}{10}\right) = y(0) + \dfrac{F}{1000 s} = 20.8 + \dfrac{F}{1000 s}$$
So we have (that last part is a partial fraction expansion):
$$\tag 3 y(s) = \dfrac{20.8 + \dfrac{F}{1000 s}}{s + \dfrac{1}{10}} = \dfrac{0.01 (F+20800 s)}{s (10 s+1)} = \left(\dfrac{20.8-0.01F}{s+0.1} + \dfrac{0.01 F}{s}\right)$$
Now, we need to find the Inverse Laplace Transform of $(3)$, so we have:
$$ \mathcal{L}^{-1}~(y(s)) = y(t) = \mathcal{L}^{-1}~\left(\dfrac{20.8-0.01F}{s+0.1} + \dfrac{0.01 F}{s}\right) = 0.01 \left(F-(F-2080) e^{-t/10}\right)$$
So, we have:
$$y(t) = 0.01 \left(F-(F-2080) e^{-t/10}\right)$$
Now, we need to find $F$ such that the final velocity is $27.8~m/s$. We are given a final time for this velocity at $t = 100$, so we would have:
$$y(100) = 0.01 \left(F-(F-2080) e^{-10}\right) = 27.8 \rightarrow F =  2780.03$$
Thus, we have:
$$y(t) = 27.8003-7.0003 e^{-t/10}$$
A plot of this is:

