Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
thank you for helping with the edit Mahdi. i'll put in where i'm at so far. – Eli Jul 20 '13 at 17:48
$f(t) = Fu(t)$ yes. the Laplace of the step being the integral of $u(t)e^(-st)$ ? i don't know why the -st won't superscript. – Eli Jul 20 '13 at 18:05
$u(t)e^{-st}$ Yes, the input is a step function scaled by the force F. $f(t) = Fu(t)$. Once solved i need to find a Force that will yield a new given final velocity. – Eli Jul 20 '13 at 18:09
F is constant insofar as once it is applied to the object it will remain the same, in order to achieve the final velocity. Also, thank you for helping me Azmoti. – Eli Jul 20 '13 at 18:23
It just says "choose F such that the final velocity is 27.8m/s". The only time given is when the velocity is 20.8 at t=0. But the next part of the problem says "plot the velocity y(t) versus time. Your axis should go from 0 to 100 sec." So I guess we will use 100s as the final time. – Eli Jul 20 '13 at 18:33
up vote 3 down vote accepted

We are given:

$$\tag 1 y'(t) + \frac D M y(t) = \dfrac{1}{M}f(t)$$


  • $D = 100kg/s$
  • $M = 1000kg$
  • $f(t) = Fu(t)$, 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:

enter image description here

share|cite|improve this answer
Azmoti thank you so much. Right before you posted that I was figuring out for myself about Laplacing the differentials on the left side but I was still stuck. THANK YOU!!! – Eli Jul 20 '13 at 19:05
on the original equation it is $y'(t) + (D/M) y(t) = (1/M) f(t)$ – Eli Jul 20 '13 at 19:08
so would we then divide $F/s$ by .001? – Eli Jul 20 '13 at 19:10
Oops yes I did I apologize. Little mistakes add up though, don't they. – Eli Jul 20 '13 at 19:14
I'm working it through with your solution by just multiplying $1/1000$ on the RHS. Is it as simple as that? – Eli Jul 20 '13 at 19:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.