Solve y'' - y' - 2y = 0?

Question

Solve this second order differential such that y -> 0 as x -> ∞ and y(0)=1

Answer 1

given y''-y'-2y=0

y'=dy/dx

and y''=d/dx(dy/dx)

let the operator d/dx be m then y'=my

and y"=(m^2)y

hence m^2-m-2=0

=>(m-2)*(m+1)=0

=>m=2 or m=-1

the solution is  y= Ae^2x+Be^-x

y(0)=1 =>A+B=1

y->0 as x->infinity =>A=0

=>B=1

=>y=e^-x

Answer 2

m^2 - m - 2 = 0

=> (m - 2)(m + 1) = 0

=> m = 2, -1

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

y->0 as x-> inf => A=0

y(0)=1 => B = 1

y = e^(-x)