Laplace Transforms...again....

[PhillDubya]

Equation:
d³/dt³ (y) + d²/dt² (y) + dy/dt + y = d²/dt² (u) + du/dt + u
Initial Conditions all = 0, and u(t) = unit step, and solve for y(t)


I used the Laplace for solving it and got:

Y(s) = [(s^2 + s + 1) / (s^3 + s^2 + s +1)] U(s)

U(s) = unit step = 1/s

Y(s) = (s^2 + s + 1) / s(s^3 + s^2 + s +1) = s^2 + s +1/ s(s²+1)(s+1)

I am having problems converting this back (the inverse Laplace) in order to get a y(t) function , or back to the time domain.

1/s = 1, 1/s²+1 =sint, 1/s+1 = e^at However, I am assuming I cannot convert like this because of the numerator [(s^2 + s + 1)] which will not factor.

So, is partial fraction expansion the only option with this?

Would it be easier to just solver the 3rd order differential?


Thanks for your time.

 

[user_88]

Is the initial equation correct as written?
It looks like you have an extra +1 when taking the transform of the left hand side.
If so, you get some cancellation and simplification....

 

[PhillDubya]

Yes, you are exactly right, I forgot the + y.

Equation should be: d³/dt³ (y) + d²/dt² (y) + dy/dt + y = d²/dt² (u) + du/dt + u

Corrected in the orginal post, thanks User 88.

 

[user_88]

This result is from Mathcad 13 Laplace Transform Solver.

 

[PhillDubya]

Ok thanks for the help!

I am doing it now with partial fractions to confirm.

I have never used MathCAD, is there a way to put the original differential equation: d³/dt³ (y) + d²/dt² (y) + dy/dt + y = d²/dt² (u) + du/dt + u, into it and have it solve that way?

Is it a lot of trouble to do this?

Thanks for your time, this is a great help.


Edit:
Ok, partial fractions definitely does not work, because of the combination of an (s) and (s^2 +1) in the denominator.

 

[user_88]

One form of the Mathcad 13 differential equation solver yields a graphical output, and is probably not what you are looking for.
I am not absolutely sure about the symbolic/closed form solution. You may be able to do it, but I haven't worked through the procedure yet.

If you can correctly take the Laplace Transform of the differential equation, it is very simple to take the partial fraction of that result, and also simple to take the inverse of the partial fraction expansion....The procedure is basically a matter of just typing the transform of the DE into the Mathcad window...
selecting it with a dotted box, clicking on any one of the algebraic letters, and then clicking the appropriate symbolic command .... inverse laplace, or partial fraction... from the symbolic menu.

 

[MrAl]

Ok thanks for the help!

I am doing it now with partial fractions to confirm.

I have never used MathCAD, is there a way to put the original differential equation: d³/dt³ (y) + d²/dt² (y) + dy/dt + y = d²/dt² (u) + du/dt + u, into it and have it solve that way?

Is it a lot of trouble to do this?

Thanks for your time, this is a great help.


Edit:
Ok, partial fractions definitely does not work, because of the combination of an (s) and (s^2 +1) in the denominator.

Hi there,

Partial fractions should work one way or another. Perhaps you could
try this again, by hand, just to get it down.

Another trick is to think about one of the properties of 1/s, and that
is that 1/s in the frequency domain is an integration in the time domain.
This allows you to simplify slightly (if you want to do so of course) a
form like this:
H(s)=(1/s)*H1(s)
to this:
H1(t)=L(-1)H1(s)
and then once you find the inverse Laplace of this you then integrate
that result to obtain the final result
H(s)=integrate(H1(t),dt).