Posted by Matt on Saturday, January 12, 2008 at 1:22pm.
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

Calculus  bobpursley, Saturday, January 12, 2008 at 1:43pm
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

Calculus  Matt, Saturday, January 12, 2008 at 2:28pm
how did you get ry and r^2y for y' and y''?

Calculus  Damon, Saturday, January 12, 2008 at 4:01pm
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
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