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Laplace transforms of heaviside step functions

  • Thread starter danj303
  • Start date
15
0
1. Homework Statement

Consider the initial value problem
y'' + 1/3y' + 4y = fk(t)

with y(0) = y'(0) = 0,

fk(t) = 1/2k for 4 - k < t < 4 + k
0 otherwise

and 0 < k < 4.



(a) Write fk(t) in terms of Heaviside step functions and then solve the initial value problem.





3. The Attempt at a Solution

I can convert it to a heaviside function and do the laplace transform and get

fk(t)= 1/2k H(t-(4-k)) - 1/2k H(t+(4-k))

and then taking the laplace transform of the entire equation get

Y(s) = 1/2k(s2+1/3 s + 4) * (e-(4-k)s/s - e-(4+k)s/s)

Im assuming this is corrent. But then how do I take the inverse laplace transform of this???
 

gabbagabbahey

Homework Helper
Gold Member
5,001
6
I can convert it to a heaviside function and do the laplace transform and get

fk(t)= 1/2k H(t-(4-k)) - 1/2k H(t+(4-k))
I think if you graph this out, you'll find it is not the function you want

[tex]H(t-4+k)-H(t+4-k)=\left\{\begin{array}{lr}-1 &, t< 4-k \\ 1 &, t>4-k\end{array}\right.[/tex]

and then taking the laplace transform of the entire equation get

Y(s) = 1/2k(s2+1/3 s + 4) * (e-(4-k)s/s - e-(4+k)s/s)

Im assuming this is corrent. But then how do I take the inverse laplace transform of this???
Much like the other problem you posted, you'll want to use the linearity of the inverse LT:

[tex]\mathcal{L}^{-1}\left[\frac{e^{-(4-k)s}-e^{-(4+k)s}}{2ks\left(s^2+\frac{1}{3}s+4\right)}\right]=\mathcal{L}^{-1}\left[\frac{e^{-(4-k)s}}{2ks\left(s^2+\frac{1}{3}s+4\right)}\right]-\mathcal{L}^{-1}\left[\frac{e^{-(4+k)s}}{2ks\left(s^2+\frac{1}{3}s+4\right)}\right][/tex]

For each of the two individual inverse LTs, you might want to use the method of partial fraction decomposition
 

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