Laplace transform,partial fraction problem

[ttsky]

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.
y''-2y'-2y=0 with initial conditions y(0) = 2 , y'(0) = 0

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

Homework Equations

Laplace Transforms Tables

The Attempt at a Solution

y''-2y'-2y=0

s^2\mathcal Y(s) - 2s - 2s\mathcal Y(s) -4 -2\mathcal Y(s) = 0
(s^2-2s-2)\mathcal Y(s) -2s-4=0
\mathcal Y(s) = (2s+4)/(s^2-2s-2)
and stuck here.. I can't figure out how to decompose last line.

 

[LCKurtz]

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.
y''-2y'-2y=0 with initial conditions y(0) = 2 , y'(0) = 0

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

Homework Equations

Laplace Transforms Tables

The Attempt at a Solution

y''-2y'-2y=0

s^2\mathcal Y(s) - 2s - 2s\mathcal Y(s) -4 -2\mathcal Y(s) = 0
(s^2-2s-2)\mathcal Y(s) -2s-4=0
\mathcal Y(s) = (2s+4)/(s^2-2s-2)
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?

 

[HallsofIvy]

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, r^2- 2r+ 2= r^2- 2r+ 1+ 1= 0 gives r= 1\pm i as characteristic solution and so y(t)= e^{t}(C_1cos(t)+ C_2 sin(t)) as general solution to the differential equation.

 

[LCKurtz]

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, r^2- 2r \color{red} {\bf +} 2

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.

 

[ttsky]

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?

 

[ttsky]

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, r^2- 2r+ 2= r^2- 2r+ 1+ 1= 0 gives r= 1\pm i as characteristic solution and so y(t)= e^{t}(C_1cos(t)+ C_2 sin(t)) 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.

 

[LCKurtz]

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.

 

[ttsky]

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!