Stuck on a Laplace Transform Question

**BlazingFire** · April 30th 2009, 06:46 PM

Here's what I'm working on:

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

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

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

Simplifying the above expression gets me to:

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

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

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

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

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

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

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

**Jhevon** · April 30th 2009, 06:57 PM

Originally Posted by BlazingFire

Here's what I'm working on:

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

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

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

Simplifying the above expression gets me to:

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

where is your 5Y??

**BlazingFire** · April 30th 2009, 07:00 PM

Originally Posted by Jhevon

where is your 5Y??

I edited my original post to put it back. Sorry about that. Still doesn't change my original question, though.

**Jhevon** · April 30th 2009, 07:04 PM

Originally Posted by BlazingFire

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

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

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

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

now, do partial fractions decomposition

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