Differential equation with Heaviside step function

[TheFerruccio]

Homework Statement

Find the solution to the differential equation using Laplace transforms.

Homework Equations

y'' - y' - 2y = 12 \theta{(t-\pi)}\sin{t}

y(0) = 1

y'(0) = -1

The Attempt at a Solution

Using Laplace transforms...

\mathcal{L}(y'') = s^2F - s + 1
\mathcal{L}(-y') = -sF-1
\mathcal{L}(-2y) = -2F

\mathcal{L}(12\theta{t-pi}\sin{t}) = \mathcal{L}(-12\theta{(t-\pi)}\sin{(t-\pi)}) = \frac{-12e^{-\pi s}}{s^2+1}

(s^2-s-2)F = \frac{-12e^{-\pi s}}{s^2+1}

I rearrange and end up with partial fractions...

F = e^{-\pi s}[\frac{-12}{(s^2+1)(s-2)(s+1)}] = e^{-\pi s}(\frac{A}{s-2} \frac{B}{s+1} \frac{Cs+D}{s^2+1})

From this point, I can rearrange everything and end up with 4 equations, 4 unknowns. I am thinking, however, that there is an easier way, simply by using one of the common transforms found in a Laplace transform table. Is there a better way to go about this problem without having to resort to partial fractions?

 

[vela]

You seem to have omitted the terms due to the initial conditions. Also, you should check the signs of those terms. I think you flipped one.

Do you know how to find the inverse Laplace transform using complex analysis? If not, partial fractions is your best bet.

 

[gomunkul51]

where is the step function ? :)

 

[TheFerruccio]

Crap, I didn't surround the initial step function with parentheses. Also, I forgot the initial conditions. I'll fix this soon.

 

[TheFerruccio]

It's fixed. I don't see anywhere where I forgot a sign, and I do not know how to figure out the inverse using complex analysis. So, if partial fractions are the way to go, then that's what I'll do!

 

[vela]

You have a sign error in your transform for -y'. You also left out the plus signs between the terms of the partial fraction expansion.

 

[TheFerruccio]

You have a sign error in your transform for -y'. You also left out the plus signs between the terms of the partial fraction expansion.

For the second one, yikes, I wrote that on the paper, but left them out when transferring them over.F = e^{-\pi s}[\frac{-12}{(s^2+1)(s-2)(s+1)}] = e^{-\pi s}(\frac{A}{s-2}+\frac{B}{s+1}+\frac{Cs+D}{s^2+1})

As for the sign error in the transform...

\mathcal{L}(y') = s\mathcal{L}(y) - y'(0)

So

\mathcal{L}(-y') = -s\mathcal{L}(y) + y'(0)

So taking the initial condition that y'(0) = -1...

\mathcal{L}(-y') = -s\mathcal{L}(y) - 1

Three negatives = negative. Is this not right?

EDIT: wait, I see where I messed up. I got the original formula wrong, so I switched the 1 with a -1. I am fixing this now.

 

[TheFerruccio]

So, for the equations and unknowns, I am getting...

A - 2B + 0C - 2D = -12
A + B - 2C - D = 0
A - 2B - C + D = 0
A + B + C + 0D = 0

For the -12e^{\pi s} term, and

A - 2B + 0C - 2D = -2
A + B - 2C - D = 0
A - 2B - C + D = 0
A + B + C + 0D = 0

For the 2 term.