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!