Inverting a Laplace Transform w/non-factorable denominator

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SUMMARY

The discussion focuses on solving a second-order differential equation (DE) with a Dirac Delta forcing function using the method of partial fractions. The equation presented is y'' + 2y' + 10y = δ(t-1), with initial conditions y'(0)=0 and y(0)=0. The participant successfully derived the Laplace transform Y(s) = e^{-s} / (s^2 + 2s + 10) but struggled with inverting the non-factorable denominator. Ultimately, the participant resolved their confusion by utilizing a reference table for Laplace transforms.

PREREQUISITES
  • Understanding of second-order differential equations
  • Familiarity with Laplace transforms
  • Knowledge of the method of partial fractions
  • Ability to complete the square for quadratic expressions
NEXT STEPS
  • Study the method of completing the square for quadratic expressions in Laplace transforms
  • Explore Laplace transform tables for common functions and their inverses
  • Practice solving second-order differential equations with various forcing functions
  • Learn about the properties of the Dirac Delta function in the context of differential equations
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Students in differential equations courses, mathematicians dealing with Laplace transforms, and anyone seeking to understand the inversion of transforms with non-factorable denominators.

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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 = \delta(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:

Y(s).(s^2 + 2s + 10) = e^{-x}

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

So I have Y(s) = e^{-x} . F(s)
e^{-x} is no problem to invert but F(s) 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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try completing the square for s2+2s+10
 
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.

Y(s)s^2 + Y(s)2s + Y(s)4 = e^{-s} - Y(s)6
____

Y(s)((s+2)(s+2) + 6) = e^{-s}
So what now?
 
Last edited:
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?
 
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?

Y(s)s^2 + Y(s)2s + Y(s) = e^{-s} - Y(s)9
____

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

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