# Differential equation with Heaviside step function

## 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?

Last edited:

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vela
Staff Emeritus
Homework Helper
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.

where is the step function ? :)

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

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
Staff Emeritus
Homework Helper
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.

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.

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.