Using an impulse to bring an oscillator to rest The Problem: Consider the system 2y''+y'+2y=δ(t-5), in which an oscillation is excited by a unit impulse at t=5. Suppose that it is desired to bring the system to rest again after exactly one cycle--that is, when the response first returns to equilibrium moving in the positive direction.

(a) Determine the impulse kδ(t-t0) that should be applied to the system in order to accomplish this objective. Note that k is the magnitude of the impulse and t0 is the time of its application.

(b) Solve the resulting IVP.

The book's answer for (a): -e-T/4δ(t-5-T);
The book's answer for (b): T=(8*pi)/(sqrt(15))

So far: I found part (a) on my own and it is correct. I am having trouble properly setting up the resulting IVP. Thanks for your help!
You have the differential equation for the IVP.  After $t=5$ the right side is zero.  Just after $t=5$, the oscillator can't have moved instantaneously, so $y(5)=0$.  It will have a velocity of $\frac 12$ from the impulse, so after $t=5$ you have $2y''+y'+2y=0, y(5)=0, y'(5)=\frac 12$
You have the differential equation for the IVP.  After $t=5$ the right side is zero.  Just after $t=5$, the oscillator can't have moved instantaneously, so $y(5)=0$.  It will have a velocity of $\frac 12$ from the impulse, so after $t=5$ you have $2y''+y'+2y=0, y(5)=0, y'(5)=\frac 12$