$y''+0.25y=k[u_{1.5}(t)-u_{2.5}(t)]$
$y(0)=0$
$y'(0)=0$
where $k$ is a positive parameter.
a) Solve the initial value problem in terms of k.
b) Plot the solution for $k=1/2$, $k=1$, and $k=2$.
For this question I got the solution:
$y(t)=4k[1-cos(0.25t)]u_{1.5}(t-1.5)-4k[1-cos(0.25t)]u_{2.5}(t-2.5)$.
My question is, am I supposed to leave the solution in this form, or is there something I should do with the $u_{1.5}(t-1.5)$ and the $u_{2.5}(t-2.5)$?
If I do leave the solution as it is, for part b, how would I plot this function with $u_{1.5}(t-1.5)$ and the $u_{2.5}(t-2.5)$?
I got this for the first question: $$y(t)=4k u_{3/2}\left(1-\cos \left(\frac 12 (t-\frac 32)\right)\right)-4ku_{5/2}\left (1-\cos \left(\frac 12 (t-\frac 52)\right)\right)$$
Since we have : $$\mathcal {L^{-1}}\{e^{-cs}F(s)\}=u(t-c)f(t-c)$$ $$\mathcal {L^{-1}}\left \{\dfrac s {s^2+1/4}\right\}=\cos (t/2)$$