karush said:
that was helpful
l have more to do so hope it sinks in
the next one is$\tiny{9.4. 7}$
$$y'=(y-3)(y-4)(y-5)$$
identify the equilibrium values
3,4,5
which are stable and which are unstable?
Okay, 3, 4, and 5
are equilibrium values because:
when y= 3, y'= (3- 3)(3- 4)(3- 5)= 0
when y= 4, y'= (4- 3)(4- 4)(4- 5)= 0
when y= 5, y'= (5- 3)(5- 4)(5- 5)= 0.
Now, if y< 3, all of y- 3, y- 4, and y- 5 are
negative and the product of three negative numbers is negative. For y< 3, y'< 0.
If 3< y< 4, y- 3 is positive while y- 4 and y- 5 are still negative. The product of two negative and one positive number is positive. For 3< y< 4, y'> 0.
Think about what that means. If y is close to 3 but a little
less than 3, y' is negative so y is decreasing, moving away from 3. If y is close to 3 but a little
more than 3, y' is positive so y is increasing, moving away from 3. In either case, y is moving
away from 3. y= 3 is an
unstable equilibrium.
If 4< y< 5, both y- 3 and y- 4 are positive while y- 5 is still negative. The product of two positive and one negative number is negative. For 4< y< 5, y'< 0.
Now, if y is close to 4 but less than 4, y' is positive so y is increasing, moving
toward 4. If y is close to 4 but a little more than 4, y' is negative so y is decreasing, moving toward 4. 4 is a
stable equilibrium.
Finally, for y> 5, all three of y- 3, y- 4, and y- 5 are positive. The product of three positive numbers is positive. For y> 5, y'> 0.
So if y is close to 5 but less than 5, y'< 0 so y is decreasing, moving
away from 5. If y' is close to 5 but more than 5, y'> 0 so y is increasing, moving away from 5. y= 5 is an
unstable equilibrium.