Asymptotic Behaviour of Solution x(t) for dx/dt = x^4 +4(x^3) - 60(x^2)

[dopey9]

i have the ode

dx/dt = x^4 +4(x^3) - 60(x^2)

generally the solution s x(t) satisfy x(0) = x[0]

and i found out that the
attactor is -10
repellor is 6
and 0 is niether

however i want to describe the asymptotic behaviour of the solution satisfying x(0) = 1/2 , which is were i got stuck??

 

[HallsofIvy]

dx/dt= x^4 +4(x^3) - 60(x^2) = x^2(x^2+ 4x- 60)= x^2(x-6)(x+10).

If x(0)= 1/2, between 0 and 6, then three of the factors, x, x, and (x+ 10) are positive while the fourth, x- 6, is negative. That means that the solution is decreasing- and, as long as it stays between 0 and 6, is always decreasing. Since the solution curve cannot cross x= 0 (because of uniqueness) this solution will go to 0 as x-> infinity.

 

[dopey9]

Thanks

dx/dt= x^4 +4(x^3) - 60(x^2) = x^2(x^2+ 4x- 60)= x^2(x-6)(x+10).

If x(0)= 1/2, between 0 and 6, then three of the factors, x, x, and (x+ 10) are positive while the fourth, x- 6, is negative. That means that the solution is decreasing- and, as long as it stays between 0 and 6, is always decreasing. Since the solution curve cannot cross x= 0 (because of uniqueness) this solution will go to 0 as x-> infinity.

...Thankz