Differential Equation Involving Heaviside function

In summary: That makes it clear that the argument of H is t - 3##\pi##.In summary, the given problem involves a differential equation and initial conditions, and the solution involves taking the Laplace transform and using partial fractions to find the inverse Laplace transform. The final solution may differ from that given by Wolfram due to different approaches and notations used.
  • #1
1MileCrash
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Homework Statement


y''+y= H(3pi)(t)

y(0)=1

y'(0)=0

H(3pi)(t) means the heaviside function such that it is 1 above 3pi and 0 otherwise.


Homework Equations





The Attempt at a Solution



Taking the laplace of both sides, I have that:

L(y) = (L(H(3pi)(t))+s) / (s^2 + 1)

The laplace transform of H(3pi)(t) is e^(-3pi(s))/s, so plugging that in and doing algebra I get:

e^(-3pis)(1/(s(s^2+1)) + s/(s^2+1)

For the first, I split into partial fractions. The second is ready to be inverted already.

e^(-3pis)((1/s) - s/(s^2+1)) + s/(s^2 + 1)

I find the inverse laplace of this to be:


H(3pi)(t)(1-cos(t-3pi)) + cos(t)

or

H(3pi)(t)(1+cos(t)) + cos(t)

Wolfram has a coefficient of 2 on the conditional cosine. How do I get that?


Thanks again!
 
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  • #2
1MileCrash said:

Homework Statement


y''+y= H(3pi)(t)

y(0)=1

y'(0)=0

H(3pi)(t) means the heaviside function such that it is 1 above 3pi and 0 otherwise.


Homework Equations





The Attempt at a Solution



Taking the laplace of both sides, I have that:

L(y) = (L(H(3pi)(t))+s) / (s^2 + 1)

The laplace transform of H(3pi)(t) is e^(-3pi(s))/s, so plugging that in and doing algebra I get:

e^(-3pis)(1/(s(s^2+1)) + s/(s^2+1)

For the first, I split into partial fractions. The second is ready to be inverted already.

e^(-3pis)((1/s) - s/(s^2+1)) + s/(s^2 + 1)

I find the inverse laplace of this to be:


H(3pi)(t)(1-cos(t-3pi)) + cos(t)

or

H(3pi)(t)(1+cos(t)) + cos(t)

Wolfram has a coefficient of 2 on the conditional cosine. How do I get that?


Thanks again!

Your work looks OK to me. Without seeing what you entered into Wolfram it's hard to say where they got that 2 you mention.

Your notation for the Heaviside function is a bit cumbersome. Instead of H(3pi)(t), which suggests the product of H(3pi) and t, I would write it as H(t - 3##\pi##).
 

Question 1: What is a Heaviside function?

The Heaviside function, also known as the unit step function, is a mathematical function that is defined as 0 for negative input values and 1 for positive input values. It is used to model sudden changes in a system, such as when a switch is turned on or off.

Question 2: How is the Heaviside function used in differential equations?

The Heaviside function is often used to define boundary conditions or initial conditions in differential equations. It can also be used to represent discontinuous terms in the equations, such as when a system experiences a sudden change in behavior.

Question 3: Can the Heaviside function be differentiated?

Yes, the Heaviside function can be differentiated using the derivative of a piecewise function. The derivative of the Heaviside function is 0 for all input values except for 0, where it is undefined.

Question 4: How does the Heaviside function impact the solution of a differential equation?

The Heaviside function can change the form of the solution to a differential equation. It can introduce discontinuities or jumps in the solution, depending on how it is used in the equation.

Question 5: Are there any real-world applications of differential equations involving the Heaviside function?

Yes, the Heaviside function is commonly used in engineering and physics to model systems that experience sudden changes, such as in electric circuits or control systems. It is also used in economics and finance to model unexpected events or changes in market behavior.

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