Calculus

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

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  1. 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

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    bobpursley
  2. how did you get ry and r^2y for y' and y''?

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  3. 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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