Find the equilibrium solutions of the ordinary differential equation
y' = x^2 cos^3 y
which of these solutions are stable?

=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.

⇒ ⇒

=π/2 + nπ =(π/2)*(1+2n)    where n∈Z (integers)

Stability analisys, f(y,x)=

(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)= , we can verify that:

if n=even, Y is stable

if n=odd, Y is unstable

all you have to do is multiply them both to see how many out comes you can put:

3 x 2 = 6

you can make 6 diferent combinations of crusts and toppings.

~batmans wife dun dun