Answer:
3...stable critical value, -5 unstable critical vale
Step-by-step explanation:
A critical value of the given differential is a value [tex]y \in \mathh{R}[/tex] such that the left hand side of the equation is equal to zero, i.e, y such that
[tex]0=15 - 2y -y^2=-(-15+2y+y^2)=-(x+5)(x-3)[/tex]
By the last equation the critical values of the given differential equation are y=-5, 3.
Now, since
[tex]\dfrac{dy}{dx}=-(y+5)(y-3)[/tex]
It holdst that
[tex]\dfrac{dy}{dx} < 0 \quad \text{for} \quad x>3[/tex]
and
[tex]\dfrac{dx}{dy}>0 \quad \text{for} \quad -5<x<3[/tex]
hence 3 is a stable critical value.
Also note that
[tex]\dfrac{dx}{dy}>0 \quad \text{for} \quad -5 < x <3[/tex]
and
[tex]\dfrac{dx}{dy} < 0 \quad \text{for } \quad x<-5[/tex]
hence -5 is an unstable critical value.
