Calculus
posted by Matt on .
I posted this several days ago but it hasn't been answered and I still can't figure it out:
For which values of r does the function defined by y=e^(rt) satisfy the differential equation y''+ y'6y = 0

find y', and y"
y'=r y
y"=r^2 y
now put them in the equation...
and divide both sides by y (y cann never be zero).
now solve for r.
r^2+r6=0 
how did you get ry and r^2y for y' and y''?

if y = e^rt where r is constant and t is variable
then
dy/dt = r e^rt
since e^rt = y then
dy/dt = r y
d/dy(dy/dt) = d^2y/dt^2 = r*re^rt
= r^2 e^rt
= r^2 y again since y = e^rt