What is the solution to this initial value problem?
y" + 2y' - 8y = 0, y(0)=1, y'(0) = 0
2007-10-19 00:07:04 · 2 answers · asked by waldeab j 1 in Science & Mathematics ➔ Mathematics
r^2+2r-8=0
r=((-2+-sqrt(36))/2 so r = 2 and r=-4
general solution
y= a e^2x+be^-4x
y(0) = a+b=1
y´= 2ae^2x-4be^-4x
y´(0) = 2a-4b=0 so a = 2b and 3b=1 so b= 1/3 and a= =2/3
y= 1/3 e^2x+2/3 e^-4x
2007-10-19 01:48:45 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
ok, figure out the general form first as i described. then since you know Ae^at + Be^bt is 1 when t = 0, so A + B = 1 b/c e^a0 is 1 regardless of a and b.
y' is Aae^at + Bbe^bt, which again at t =0 is 0.
so Aa + Bb =0. I won't solve this for you b/c you need to find a and b as i suggested in the other reply, but you have two equations and two unknowns once you get a and b.
2007-10-19 00:13:01 · answer #2 · answered by keyahnoo 2 · 0⤊ 0⤋