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+r-6=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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