Inverting a Laplace Transform w/non-factorable denominator

In summary, the student is struggling with an equation invertible for a function with a quadratic in denominator. After unsuccessfully trying different methods, he finds a table that provides the answer.
  • #1
monet A
67
0

Homework Statement



A 2nd order DE with a Delta Dirac forcing function, I have been asked to solve in my DE course.
y'' + 2y' + 10y = [tex]\delta[/tex](t-1)

Also given are IV's y'(0)=0, y(0)=0


Homework Equations



At this stage in my course I am being asked to only use the method of partial fractions to rewrite my Laplace equations into a recognisable invertable form I am not supposed to be using the integral definition to invert my solutions.


The Attempt at a Solution



The Transforms of each side I have so far obtained:

[tex] Y(s).(s^2 + 2s + 10) = e^{-x} [/tex]

This gives me a solution for Y(s) -->
[tex] Y(s) = \frac{ e^{-x} }{s^2 + 2s + 10}[/tex]

So I have [tex] Y(s) = e^{-x} . F(s)[/tex]
[tex] e^{-x} [/tex] is no problem to invert but [tex] F(s)[/tex] has a quadratic in denominator which cannot be factored. I am at a loss as to how I can find any invertable form for it and for some reason I'm not aware of any other technique that is possible.
 
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  • #2
try completing the square for s2+2s+10
 
  • #3
rock.freak667 said:
try completing the square for s2+2s+10

Ok I feel stupid, I was thinking that completing the square would give me Y(s) terms on the RHS which I would have to divide out but I just tried it and now I don't know why I was thinking that.

I have factors but now I have a denominator with a dangling 6 that I don't know what to do with.

[tex] Y(s)s^2 + Y(s)2s + Y(s)4 = e^{-s} - Y(s)6[/tex]
____

[tex]Y(s)((s+2)(s+2) + 6) = e^{-s} [/tex]
So what now?
 
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  • #4
While (s+2)2+6 will give you back the original function, that is not completing the square.

Remember you want s2+2s+10 =(s+1)2/+A, so what is A?
 
  • #5
rock.freak667 said:
While (s+2)2+6 will give you back the original function, that is not completing the square.

Remember you want s2+2s+10 =(s+1)2/+A, so what is A?


Yeah, it's 9. Thanks for pointing out my mistake there, but also that's not the problem that I am having. THe thing is that the 9 has a factor of Y(s) so I have no choice but to return to the original equation as far as I can see. That's what is bothering me what can I do about it?

[tex] Y(s)s^2 + Y(s)2s + Y(s) = e^{-s} - Y(s)9[/tex]
____

[tex]Y(s)((s+1)^2 +9) = e^{-s} [/tex]
 
  • #6
Thanks for your help Rock. I have discovered a table which answers my question so just ignore the recent post, I'm okay now. :)
 

FAQ: Inverting a Laplace Transform w/non-factorable denominator

How do I know if a Laplace Transform has a non-factorable denominator?

A Laplace Transform has a non-factorable denominator if it contains terms that cannot be factored or simplified, such as exponential functions or trigonometric functions. It is important to identify this before attempting to invert the transform.

Can a non-factorable denominator be inverted?

Yes, a non-factorable denominator can still be inverted, but the process may be more complex and may require the use of partial fraction decomposition or other techniques.

What is partial fraction decomposition?

Partial fraction decomposition is a technique used to break down a rational function with a non-factorable denominator into simpler fractions that can be more easily inverted. This method involves finding the appropriate coefficients for the partial fractions and then using the inverse Laplace Transform to find the original function.

Are there any shortcuts or tricks for inverting a Laplace Transform with a non-factorable denominator?

Unfortunately, there are no shortcuts or quick tricks for inverting a Laplace Transform with a non-factorable denominator. It requires a thorough understanding of the properties of Laplace Transforms and various techniques for inverting them, such as partial fraction decomposition.

Are there any common mistakes to avoid when inverting a Laplace Transform with a non-factorable denominator?

One common mistake to avoid is overlooking the need for partial fraction decomposition. Attempting to invert the transform without this step can lead to incorrect results. It is also important to carefully check the algebraic manipulations and substitutions made during the inversion process to avoid any errors.

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