Differential Equation with unit step function and Laplace transforms

In summary, the student made a small error in their final answer when using the transform of the unit step function. They forgot to include the 1/s term in the transform.
  • #1
Mangoes
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1
EDIT:

Nevermind I see what I did wrong near the end.

Homework Statement



[tex] x'' + 4x = f(t) [/tex]

Where f(t) is 1 if t is between 0 and π, 0 if t > π. Initial conditions are x(0) = x'(0) = 0.

Homework Equations



Transform of a derivative:
[tex] L(f^{(n)}(t)) = s^nF(s) - s^{n-1}f(0) -...-f^{n-1}(0) [/tex]

Transform of unit step function:
[tex] L(u(t-a)) = \frac{e^{-as}}{s} [/tex]

The Attempt at a Solution



Letting u(t) be the step function, I wrote:

[tex] f(t) = 1 - u(t - π) [/tex]

So I rewrote the equation as

[tex] x'' + 4x = 1 - u(t - π) [/tex]

To take the Laplace transform of the left side, I used the transform of a derivative. Since the initial conditions are 0, the transform simplifies quite a bit. For the right side of the equation, I had to use the transform of the unit step function and the transform of 1.

[tex] s^2X(s) + 4X(s) = \frac{1}{s} - \frac{e^{-πs}}{s} [/tex]

Factored and moved around a bit to get:

[tex] X(s)(s^2 + 4) = \frac{1}{s}(1 - e^{-πs}) [/tex]

[tex] X(s) = \frac{1 - e^{-πs}}{(s)(s^2 + 4)} [/tex]

[tex] X(s) = \frac{1}{s(s^2 + 4)} - \frac{e^{-πs}}{s(s^2 + 4)} [/tex]

At this point I had a choice to do either a convolution or use partial fractions. Since I hate partial fractions, I went for a convolution, but it didn't work out and I had to end up using partial fractions.

Making up a new name:

[tex] G(s) = \frac{1}{s(s^2 + 4)} [/tex]

I used partial fractions and got this:

[tex] G(s) = \frac{1}{4s} - \frac{s}{4(s^2 + 4)} [/tex]

With the above I can easily do the inverse Laplace transform of the first term in X(s). But for the second term, I wanted to use the fact that

[tex] u(t-a)f(t-a) = e^{-as}F(s) [/tex]

where a = π and F(s) would be the same as G(s). The inverse Laplace transform of G(s) is

[tex] 4 - \frac{1}{4}cos(2t) [/tex]

And by just following the fact I just mentioned above, I would think that the Laplace transform of the second term in X(s) would be:

[tex] u(t - π)(4 - \frac{1}{4}cos(2(t - π)) [/tex]

cos(2t - 2pi) simplifies to cos(2t) due to periodicity of cosine.

So finally my final answer would be

[tex] f(t) = 4 - \frac{1}{4}cos(2t) - u(t - π)(4 - \frac{1}{4}cos(2t)) [/tex]

When I checked the answers, my answer was a little off. Normally, this really wouldn't bother me too much because there's a very good chance I made a small error somewhere, but I'm not too used to doing questions like these and I was wondering if anyone could find a really obvious glaring error the procedure I did. I have an exam tomorrow and it would be nice to make sure.
 
Last edited:
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  • #2


Hi there,

Thank you for posting your solution and explaining your thought process. It seems like you have a good understanding of the concepts and equations involved in solving this problem. However, I did notice a small error in your final answer.

When you used the fact that u(t-a)f(t-a) = e^{-as}F(s), you forgot to account for the 1/s term in the transform of the unit step function. So your final answer should be:

f(t) = 4 - \frac{1}{4}cos(2t) - u(t - π)(4 - \frac{1}{4}cos(2(t - π)))\frac{1}{s}

I hope this helps and good luck on your exam!
 

1. What is the unit step function in differential equations?

The unit step function, also known as the Heaviside function, is a mathematical function that returns 0 for negative inputs and 1 for positive inputs. It is often used in differential equations to model abrupt changes or discontinuities in a system.

2. How is the unit step function used in differential equations?

The unit step function is used in differential equations to represent a sudden change or event that affects the behavior of a system. It is commonly used in modeling physical systems such as electrical circuits, mechanical systems, and chemical reactions.

3. What is the Laplace transform in differential equations?

The Laplace transform is a mathematical operation that converts a function of time into a function of complex frequency. It is often used in solving differential equations, as it can simplify the equations and make them easier to solve.

4. How is the Laplace transform applied to differential equations with unit step functions?

The Laplace transform can be applied to differential equations with unit step functions by using the properties of the Laplace transform, such as the shifting property and the unit step function property. These properties allow us to transform the differential equation into an algebraic equation, which can then be solved using standard algebraic techniques.

5. What are some real-world applications of differential equations with unit step functions and Laplace transforms?

Differential equations with unit step functions and Laplace transforms have many real-world applications, including in systems engineering, control systems, signal processing, and circuit analysis. They are also used in physics, chemistry, and biology to model and predict the behavior of complex systems.

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