> #OK to post #HW 16, Timothy Nasralla, 10/31/21
> #Question 1, confirm the calculated fixed point to the given equations. #Let z
> = [x,y], so z[1] = x and z[2] = y.
> evalf(Orbk(2,z,(5/3)*z[1]-[z1]*z[2],[1,2],1000,1010))
> 
> evalf(Orbk(2,z,(2)*z[1]-[z1]*z[2],[1,2],1000,1010))
> #Question 4 #Equation1, x'(t) = x(t)*(5-x(t))*(3-x(t)) #Based on the
> calculations on paper the equilibrium points are x=0, x=3, and x=5 but the
> only stable fixed point is x = 3. plot(Dis1(x*(3 - x)*(5 - x), x, 0.01, 0.01,
> 20))

> plot(Dis1(x*(3 - x)*(5 - x), x, 1.0, 3.01, 20))

> plot(Dis1(x*(3 - x)*(5 - x), x, 0.01, 5.01, 20))

> #Equation 2, x'(t) = x(t)^2 * (3-x(t)) * (5-x(t)) * (7-x(t)), after solving
> for equilbriium points I found x = 0, x = 3, x = 5, and x= 7 as equilibrium
> points, as well as that x = 3 and x = 7 are stable.
> plot(Dis1(x^2*(7-x)*(3 - x)*(5 - x), x, 0.01, 0.01, 20))

> plot(Dis1(x^2*(7-x)*(3 - x)*(5 - x), x, 0.01, 3.01, 20))

> plot(Dis1(x^2*(7-x)*(3 - x)*(5 - x), x, 0.01, 5.01, 20))

> plot(Dis1(x^2*(7-x)*(3 - x)*(5 - x), x, 0.01, 7.01, 20))