Answer:
[tex]Y_{n}[/tex]=π/2 + nπ =(π/2)*(1+2n) where n∈Z (integers)
if n=even, Y is stable
if n=odd, Y is unstable
Step-by-step explanation:
y' = x^2 cos^3 y
Equilibrium solutions occur when y'(y,x) =f(y,x)=0, we need to find y=Y=constant and satisfy this.
[tex]x^{2} cos^{3}(y)=0[/tex] ⇒ [tex]cos^{3}(y)=0[/tex] ⇒ [tex]cos(y)=0[/tex]
[tex]Y_{n}[/tex]=π/2 + nπ =(π/2)*(1+2n) where n∈Z (integers)
Stability analisys, f(y,x)= [tex]x^{2} cos^{3}(y)=0[/tex]
(i.) If f(y,x)) < 0 on the left of Y, and f(y) > 0 on the right of Y, then the equilibrium solution is unstable.
(ii.) If f(y,x) > 0 on the left of Y, and f(y) < 0 on the right of Y, then the equilibrium solution y = c is stable.
(iii.) If f(x) > 0 on both sides of X, or f(x) < 0 on both sides of c, then the equilibrium solution y = X is semi-stable.
So, In the graph annexed we see f(y,x)= [tex]x^{2} cos^{3}(y)=0[/tex] , we can verify that:
if n=even, Y is stable
if n=odd, Y is unstable
