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+r-6=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
To find the values of r that make the function y=e^(rt) satisfy the differential equation y'' + y' - 6y = 0, we can substitute y into the equation and solve for r.
First, let's find the first derivative of y=e^(rt). To differentiate e^(rt) with respect to t, we use the chain rule:
dy/dt = d/dt (e^(rt))
= re^(rt)
Now let's find the second derivative:
d^2y/dt^2 = d/dt (re^(rt))
= r * d/dt (e^(rt))
= r * re^(rt)
= r^2 * e^(rt)
Now substitute y, y', and y'' into the differential equation:
y'' + y' - 6y = r^2 * e^(rt) + r * e^(rt) - 6 * e^(rt) = 0
Next, factor out e^(rt) from the equation:
e^(rt) * (r^2 + r - 6) = 0
For this equation to hold true, either e^(rt) must be equal to zero (which is not possible), or the term inside the parentheses must be equal to zero:
r^2 + r - 6 = 0
Now we can solve this quadratic equation to find the values of r:
(r + 3)(r - 2) = 0
This equation has two solutions: r = -3 and r = 2.
Therefore, the function y=e^(rt) satisfies the given differential equation when r is equal to -3 or 2.