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

b

step-by-step explanation:

this is because the profit is decreasing and turning into debt

third choice down

step-by-step explanation:

the first inequality is based on the number of adults and campers, and the second inequality is based on the cost of feeding both the adult and the camper.

we know that there is room for 525 people max.   it's ok to have less than 525, but it's not ok to have more than 525, so the inequality is less than or equal to.   the number combination of adults (x) and campers (y) cannot exceed this 525 max.   therefore:

x + y ≤ 525

now we talk about cost.   it costs $9 to feed one adult, so depending upon the number of adults you have (x) depends upon how much it will cost to feed that number of adults.   same with the campers, who cost $5 to feed.   the adults can be represented by 9x and the campers can be represented by 5y.   since the budget cannot exceed 1,200, the second inequality is:

9x + 5y ≤ 1200