Laplace transforms of heaviside step functions

[danj303]

Homework Statement

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

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

f_k(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.

The Attempt at a Solution

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

f_k(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]

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

f_k(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

H(t-4+k)-H(t+4-k)=\left\{\begin{array}{lr}-1 &amp;, t&lt; 4-k \\ 1 &amp;, t&gt;4-k\end{array}\right.

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:

\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]

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