Find the general solution of?

Question

d^2y/dt^2 - dy/dt - 2y = 0

Answer 1

Auxiliary equation is:-
m² - m - 2 = 0
(m - 2)(m + 1) = 0
m = 2 , m = - 1
y = A e^(2t) + B e^(- t)

Answer 2

Generate an auxilary equation:
say p^2-p-2=0
(p+1)(p-2) by factorising
THerefore, p=-1 or p=2
Thus a general solution will be in the form
y=C*e^-t + D*e^2t

Answer 3

Let's try y=e^(cx)

d^2y/dt^2 - dy/dt - 2y = a^2*e^(ct) - a*e^(ct) - 2*e^(ct) = 0

Divide by e^(ct)

c^2 - c - 2 = 0

(c + 1)(c - 2) = 0

c is -1 or 2

The general solution is ae^(-t) + be^(2t)

Let's derive:

d^2y/dt^2 - dy/dt - 2y = [ae^-t + 4be^(2t)] -
- [-a*e^-t + 2be^(2t)] - 2[a^-t + be^(2t)] =
= (a + a - 2a)e^-t + (4b - 2b - 2b)e^(2t) =
= 0e^-t + 0e^(2t) = 0

Answer 4

Let the solution be y=e^mt.
m^2-m-2=0
m^2-2m+m-2=0
m(m-2)+1(m-2)=0
(m-2)(m+1)=0
m=-1,2.
so the general solution is:
y=Ae^(-t)Bê^(2t), where A&B are constants which can be determined from the initial conditions.

Answer 5

Plug in y* = e^(rt), and solve for r,
r^2-r-2 = (r-2)(r+1) = 0
r = 2, -1
y = Ae^(2t) + Be^(-t), where A and B are two constants.