# Suppose a spring-mass system satisfies the inital value problem

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

$$y(0)=0$$

$$y'(0)=0$$

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 : $$\mathcal {L^{-1}}\{e^{-cs}F(s)\}=u(t-c)f(t-c)$$ $$\mathcal {L^{-1}}\left \{\dfrac s {s^2+1/4}\right\}=\cos (t/2)$$
• 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 $cos(1/4(t-5/2))$? – Not2Scary Mar 19 at 1:41
• You need to shift the cosine function and note that $a^2=1/4 \implies a=1/2$ @Not2Scary What's your value of $a^2$ ? – Aryadeva Mar 19 at 1:42
• Oh it simple it means it zero for $t<3/2$ and the normal cosine graph It's just Heaviside function... @Not2Scary – Aryadeva Mar 19 at 1:47