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$y''+0.25y=k[u_{1.5}(t)-u_{2.5}(t)]$

$y(0)=0$

$y'(0)=0$

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 $u_{2.5}(t-2.5)$?

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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 : $$\mathcal {L^{-1}}\{e^{-cs}F(s)\}=u(t-c)f(t-c)$$ $$\mathcal {L^{-1}}\left \{\dfrac s {s^2+1/4}\right\}=\cos (t/2)$$

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  • $\begingroup$ Oh, okay, I know what I did wrong. But how did you get $cos(1/2(t-3/2))$ and $cos(1/2(t-5/2))$? Shouldn't it be $cos(1/4(t-3/2))$ and $cos(1/4(t-5/2))$? $\endgroup$ – Not2Scary Mar 19 at 1:41
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    $\begingroup$ You need to shift the cosine function and note that $a^2=1/4 \implies a=1/2$ @Not2Scary What's your value of $a^2$ ? $\endgroup$ – Aryadeva Mar 19 at 1:42
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    $\begingroup$ Right! Thank you! Do you know how I would go about plotting this with the different values of k? $\endgroup$ – Not2Scary Mar 19 at 1:43
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    $\begingroup$ You can plot it on DESMOS for different values of k....@Not2Scary Take a look here desmos.com/calculator $\endgroup$ – Aryadeva Mar 19 at 1:45
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    $\begingroup$ Oh it simple it means it zero for $t<3/2$ and the normal cosine graph It's just Heaviside function... @Not2Scary $\endgroup$ – Aryadeva Mar 19 at 1:47

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