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Laplace transform and algebra

  • Thread starter Yann
  • Start date
  • #1
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Homework Statement



I must find the solution of a differential equation, but I'm stuck with a problem of algebra;

Homework Equations



The problem is

[tex]
y''+2y'+2y = sin(at)
[/tex]

With y(0) = y(0)' = 0

y''+2y'+2y = sin(at)

[tex]
s^2L[y]+2sL[y]+2L[y] = \frac{a}{s^2+a^2}
[/tex]

[tex]
L[y](s^2+2s+2) = \frac{a}{s^2+a^2}
[/tex]

[tex]
L[y] = \frac{a}{(s^2+2s+2)(s^2+a^2)}
[/tex]

The Attempt at a Solution



I transform it;

[tex]
L[y] = \frac{a}{([s+1]^2+1)(s^2+a^2)}
[/tex]

[tex]
\frac{a}{([s+1]^2+1)(s^2+a^2)} = \frac{A(s+1)+B}{[s+1]^2+1}+\frac{C}{s^2+a^2}
[/tex]

[tex]
a = [A(s+1)+B](s^2+a^2) + C[(s+1)^2+1]
[/tex]

I just don't have a clue how to find A, B and C from here...
 
Last edited:

Answers and Replies

  • #2
1,860
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You might want to check your partial fractions again, though you might have just made a typo. Are you allowed to use Matlab or Mathematica to crunch through your partials (that is what I would do)? Also, are you not allowed to simply use superposition of the homogeneous and particular solution (you have a linear system with a very well known answer), do you have to use Laplace transforms?
 
Last edited:
  • #3
48
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Thx for the answer..

I really have to use Laplace transformes, it's in the question, and no, I tcan't use MatLab or any other programs, I need to do it manually.
 
  • #4
1,860
0
Okay, well this will require a lot of pointless brute force. Alright, so the right hand side is a function of s, which is equal to a for all values of s. This also means that it must be true for particular values of s, and you can choose certain values of s to find your variables, so what if you conveniently make s = -1? How about s = ia (where i is the imaginary unit)? So far you should have B and C. What about A?
 
  • #5
48
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There must be some other way, it's only a basic course, after all. Perhaps I'm making a mistake in the Laplace transforms...
 
  • #6
48
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Nobody has an idea ?
 
  • #7
learningphysics
Homework Helper
4,099
5
[tex]
\frac{a}{([s+1]^2+1)(s^2+a^2)} = \frac{A(s+1)+B}{[s+1]^2+1}+\frac{C}{s^2+a^2}
[/tex]

[tex]
a = [A(s+1)+B](s^2+a^2) + C[(s+1)^2+1]
[/tex]
You haven't considered that the numerator of the second fraction may be of the form Cs + D...

I'd write the numerator of the first fraction on the right side as As + B. And the second one as Cs +D. Then do what you did before... bring it all under the same denominator... then multiply out the numerator on the right side, and gather terms together... finally equate the left side with the right side... you're equating coefficients of s^a...

So all the coefficients on the right side should be 0 except for the constant term... so immediately you'll get C = -A using the s^3 term... continue with all the coefficients...
 
Last edited:
  • #8
1,860
0
I didn't check your Laplace transforms, and I missed the s^2 in the second one, but given that you do all that right, the method I gave you will work. You can also use the "cover up" method, which works on the same principle.
 
  • #9
48
0
I tried, but I know the solution (because of Maple) is;

[tex]
y(t) = \frac{e^{-t}sin(t)a^3}{4+a^4}+\frac{2e^{-t}cos(t)a}{4+a^4}+\frac{2\sin(at)-2a\cos(at)-a^2\sin(at)}{4+a^4}
[/tex]

And I'm not getting there.
 
  • #10
48
0
In fact I think we have to go with;

[tex]
\frac{a}{([s+1]^2+1)(s^2+a^2)} = \frac{A(s+1)+B}{[s+1]^2+1}+\frac{Cs+D+E}{s^2+a^2}
[/tex]
 

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