Laplace transform,partial fraction problem

  • Thread starter ttsky
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  • #1
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This is more of a pre calc question but it dose however come from diff eqs, just in case I have made fundamental mistakes, i have posted it here. I have been studying this topic for few days by myself, never had any problems with algebra until here.

really appreciate all of your help.

Homework Statement


problem comes from here.
[itex]y''-2y'-2y=0[/itex] with initial conditions [itex]y(0) = 2 , y'(0) = 0[/itex]

I am stuck trying to decompose this line
[itex]\mathcal Y(s) = (2s+4)/(s^2-2s-2)[/itex]

Homework Equations


Laplace Transforms Tables

The Attempt at a Solution


[itex]y''-2y'-2y=0[/itex]

[itex]s^2\mathcal Y(s) - 2s - 2s\mathcal Y(s) -4 -2\mathcal Y(s) = 0[/itex]
[itex](s^2-2s-2)\mathcal Y(s) -2s-4=0[/itex]
[itex]\mathcal Y(s) = (2s+4)/(s^2-2s-2)[/itex]
and stuck here.. I cant figure out how to decompose last line.
 

Answers and Replies

  • #2
LCKurtz
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This is more of a pre calc question but it dose however come from diff eqs, just in case I have made fundamental mistakes, i have posted it here. I have been studying this topic for few days by myself, never had any problems with algebra until here.

really appreciate all of your help.

Homework Statement


problem comes from here.
[itex]y''-2y'-2y=0[/itex] with initial conditions [itex]y(0) = 2 , y'(0) = 0[/itex]

I am stuck trying to decompose this line
[itex]\mathcal Y(s) = (2s+4)/(s^2-2s-2)[/itex]

Homework Equations


Laplace Transforms Tables

The Attempt at a Solution


[itex]y''-2y'-2y=0[/itex]

[itex]s^2\mathcal Y(s) - 2s - 2s\mathcal Y(s) -4 -2\mathcal Y(s) = 0[/itex]
[itex](s^2-2s-2)\mathcal Y(s) -2s-4=0[/itex]
[itex]\mathcal Y(s) = (2s+4)/(s^2-2s-2)[/itex]
and stuck here.. I cant figure out how to decompose last line.

Complete the square in the denominator: ##(s-1)^2-3##. Then write the numerator as ##2(s-1)+6##. Does that help?
 
  • #3
HallsofIvy
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Are you required to use "Laplace Transform"? I have never quite understood why "Laplace Transform" methods are even taught for differential equations! Just writing out the characteristice equation for the given differential equation, [itex]r^2- 2r+ 2= r^2- 2r+ 1+ 1= 0[/itex] gives [itex]r= 1\pm i[/itex] as characteristic solution and so [itex]y(t)= e^{t}(C_1cos(t)+ C_2 sin(t))[/itex] as general solution to the differential equation.
 
  • #4
LCKurtz
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Are you required to use "Laplace Transform"? I have never quite understood why "Laplace Transform" methods are even taught for differential equations! Just writing out the characteristice equation for the given differential equation, [itex]r^2- 2r \color{red} {\bf +} 2[/itex]

That's ##r^2 -2r -2##, which changes the answer a bit. While I somewhat agree with your sentiments, the transforms are certainly handy for non-homogeneous terms which are piecewise defined, not to mention the usefulness of the transform space in EE applications.
 
  • #5
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Complete the square in the denominator: ##(s-1)^2-3##. Then write the numerator as ##2(s-1)+6##. Does that help?

I dont see how this came about. can you elaborate?
 
  • #6
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Are you required to use "Laplace Transform"? I have never quite understood why "Laplace Transform" methods are even taught for differential equations! Just writing out the characteristice equation for the given differential equation, [itex]r^2- 2r+ 2= r^2- 2r+ 1+ 1= 0[/itex] gives [itex]r= 1\pm i[/itex] as characteristic solution and so [itex]y(t)= e^{t}(C_1cos(t)+ C_2 sin(t))[/itex] as general solution to the differential equation.
we spent last semester doing just that, I am only studying ahead so I have yet to find out why myself. i have heard it is important for EE students, which is what I am.
 
  • #8
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Are you asking how to complete the square in a quadratic? If so, look in any algebra book or look here:

http://en.wikipedia.org/wiki/Completing_the_square

For the second one, just expand it out to see it's the same.

thank you for that, this topic opened a can of loop holes in my algebra! really appreciate your help. understand it now!
 

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