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Differential Equation Involving Heaviside function

  • Thread starter 1MileCrash
  • Start date
  • #1
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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!!
 

Answers and Replies

  • #2
33,632
5,289

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##).
 

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