Second Order differential equations?

Question

I'm having problem with some basic 2nd order equations. The problem states, find the solution of the given initial value problem: 4y'' - y = 0; y(-2) = 1 and y'(-2) = -1. The ANSWER is y = -1/2e^((t+2)/2) + 3/2e^-((t+2)/2) Thanks for the help.

Answer 1

This type of DE usually has an exponential function as a solution so make a guess that

y=Ae^st+Be^(-ts)

y' =Ase^st -Bse^(-st)

y'' = s^2A^(st)+s^2Be^(-st)

So now 4y" - y = 0

(4s^2-1)Ae^(st) + (4s^2-1)Be^(-st) = 0

Solve the neg and pos exp separately: s^2 = 1/4 --> s =+/-1/2

Now apply IC.

y(-2) = Ae^-1+Be^1=1

y'(-2) = A/2e^(-1) -B/2e^(1) =-1 mult by 2 and add to above

2Ae^(-1) = -1 ---> A = -1/2*e^(1)

Then -1/2+Be^(1) = 1 ---> B =3/2e^(-1)

y(t) = -1/2e(t/2+1)+3/2e^(-t/2-1)