Solve the initial value problem y"-y'-2y=0, y(0)=α, y'(0)=2. Then find α so that the solution approaches zero as t→∞.

Here we have 2 distinct roots to the characteristic equations r = -1, 2 so the general solution becomes:

y(t) = C_{1}e^{-t} + C_{2}e^{2t}

plugging in our initial conditions we get:

y(0) = C_{1} + C_{2} = a

y'(0) = -C_{1} + 2C_{2} = 2

After solving this 2 X 2 linear system (preferably using the elimination method) we should get C_{1} = 2(a-1)/3 and C_{2} = (a+2)/3.

since e^{2t} → ∞ as t → ∞ we need to force that C_{2
}= 0, doing so gets you a = -2