Laplace transform,partial fraction problem

In summary, the conversation is about a pre-calculus question that involves differential equations and Laplace transforms. The problem given is y''-2y'-2y=0 with initial conditions y(0) = 2 and y'(0) = 0. The person is stuck trying to decompose the line \mathcal Y(s) = (2s+4)/(s^2-2s-2) and is looking for help. They are required to use Laplace transforms and have been studying this topic for a few days. The conversation also includes a discussion on the usefulness of Laplace transforms and completing the square in a quadratic equation. In the end, the person thanks the other for their help and understanding.
  • #1
ttsky
20
0
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 can't figure out how to decompose last line.
 
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  • #2
ttsky said:
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 can't 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
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
HallsofIvy said:
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
LCKurtz said:
Complete the square in the denominator: ##(s-1)^2-3##. Then write the numerator as ##2(s-1)+6##. Does that help?

I don't see how this came about. can you elaborate?
 
  • #6
HallsofIvy said:
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.
 
  • #7
ttsky said:
I don't see how this came about. can you elaborate?

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.
 
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  • #8
LCKurtz said:
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!
 

What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function from the time domain to the frequency domain. It allows us to analyze the behavior of a system over time by examining its frequency response.

How is a Laplace transform calculated?

The Laplace transform is calculated by integrating a function with respect to time, multiplied by an exponential term. This results in a complex function in the frequency domain, which can then be further manipulated and analyzed.

What is the purpose of performing a Laplace transform?

The main purpose of performing a Laplace transform is to simplify the analysis of a system in the frequency domain. It allows for easier calculations of system response to different inputs and helps in solving differential equations.

What is a partial fraction problem?

A partial fraction problem is a mathematical problem that involves breaking down a rational function into simpler fractions. This is often done to make the function easier to integrate or manipulate.

How do you solve a partial fraction problem?

To solve a partial fraction problem, the fraction must first be decomposed into simpler fractions. This can be done by using the method of partial fraction decomposition, which involves finding the appropriate coefficients for each term in the decomposition.

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