# Stuck on a Laplace Transform Question

• Apr 30th 2009, 06:46 PM
BlazingFire
Stuck on a Laplace Transform Question
Here's what I'm working on:

$\displaystyle y'' + 6y' + 5y = t; y(0) = a, y'(0) = b$

So far, I've done this. Taking the Laplace transform of both sides of the equation gives me:

$\displaystyle \mathfrak{L}\{y''\} + 6\mathfrak{L}\{y'\} + 5\mathfrak{L}\{y\} = \mathfrak{L}\{t\}$

Simplifying the above expression gets me to:

$\displaystyle s^2 Y(s) - sy(0) - y'(0) + 6sY(s) - 6y(0) + 5Y(s) = \frac{1}{s^2}$

Factoring out the Y(s) terms on the left and applying the initial conditions leads me to:

$\displaystyle Y(s)(s^2 + 6s + 5) - as - b - 6a = \frac{1}{s^2}$

$\displaystyle Y(s)(s^2 + 6s + 5) = \frac{1}{s^2} + as + b + 6a$

Multiplying both sides by $\displaystyle \frac{1}{s^2 + 6s + 5}$ gives me:

$\displaystyle Y(s) = (\frac{1}{s^2} + as + b + 6a)(\frac{1}{s^2 + 6s + 5})$

And this is where I'm stuck and need major help.
• Apr 30th 2009, 06:57 PM
Jhevon
Quote:

Originally Posted by BlazingFire
Here's what I'm working on:

$\displaystyle y'' + 6y' + 5y = t; y(0) = a, y'(0) = b$

So far, I've done this. Taking the Laplace transform of both sides of the equation gives me:

$\displaystyle \mathfrak{L}\{y''\} + 6\mathfrak{L}\{y'\} + 5\mathfrak{L}\{y\} = \mathfrak{L}\{t\}$

Simplifying the above expression gets me to:

$\displaystyle s^2 Y(s) - sy(0) - y'(0) + 6sY(s) - 6y(0) = \frac{1}{s^2}$

• Apr 30th 2009, 07:00 PM
BlazingFire
Quote:

Originally Posted by Jhevon

I edited my original post to put it back. Sorry about that. Still doesn't change my original question, though.
• Apr 30th 2009, 07:04 PM
Jhevon
Quote:

Originally Posted by BlazingFire

$\displaystyle Y(s) = (\frac{1}{s^2} + as + b + 6a)(\frac{1}{s^2 + 6s + 5})$

And this is where I'm stuck and need major help.

the next step is to combine the fractions. note that you have

$\displaystyle Y = \frac {as^3 + (6a + b)s^2 + 1}{s^2(s + 5)(s + 1)}$

now, do partial fractions decomposition