# Differential equation with Heaviside step function

• TheFerruccio
In summary, the solution to the given differential equation using Laplace transforms involves rearranging the equations and using partial fractions. The initial conditions and the step function are also taken into account in finding the solution. There may be other methods, such as using complex analysis, but partial fractions is a common and viable approach.
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?

Last edited:
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!

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.

vela said:
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.

## 1. What is a Heaviside step function in the context of differential equations?

A Heaviside step function is a mathematical function that represents a sudden change or discontinuity in a system. It has a value of 0 before the change and a value of 1 after the change. In the context of differential equations, it is often used to model abrupt changes in a system's behavior or to simulate the effects of external stimuli.

## 2. How is a Heaviside step function used in differential equations?

In differential equations, the Heaviside step function is typically used to define a piecewise function that represents a system's behavior before and after a specific point in time. This allows for a more accurate and realistic model of the system's dynamics, especially when sudden changes occur.

## 3. Can the Heaviside step function be used to solve differential equations?

While the Heaviside step function itself cannot be used to solve differential equations, it is often used as a part of the solution process. By incorporating the step function into the differential equation, it can help to model and analyze the behavior of a system that undergoes sudden changes.

## 4. Are there other similar functions to the Heaviside step function used in differential equations?

Yes, there are other similar functions used in differential equations, such as the Dirac delta function and the ramp function. These functions also represent discontinuities or changes in a system, but they may have different properties and uses compared to the Heaviside step function.

## 5. What are some practical applications of using the Heaviside step function in differential equations?

The Heaviside step function has various practical applications, including modeling the behavior of electrical circuits, analyzing control systems, and simulating the effects of external forces on a system. It is also commonly used in physics, engineering, and other scientific fields to model and understand sudden changes or events in a system.

Replies
1
Views
530
Replies
3
Views
911
Replies
3
Views
1K
Replies
1
Views
567
Replies
1
Views
818
Replies
7
Views
1K
Replies
6
Views
1K
Replies
1
Views
1K
Replies
18
Views
2K
Replies
2
Views
984