Question

The number 0 is a critical point of the autonomous differential equation$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{x}}^{\mathbf{n}}$, where n is a positive integer. For what values of n is 0 asymptotically stable? Semi-stable? Unstable? Repeat for the differential equation$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{-}{\mathit{x}}^{\mathbf{n}}$ .

Short answer

Hence, for xⁿ

0 is unstable when nⁿ is odd

0 is semi-stable when n is even

Hence, for -xⁿ

0 is stable when n is odd

0 is semi-stable when n is even

Step-by-step solution

Determine the value of n

Consider the differential equation $\frac{\mathrm{dx}}{\mathrm{dt}}=x^{\mathrm{n}}$where n is a positive integer.

Consider the number 0 is a critical point of the autonomous given differential equation.

The objective is to find for what values of nis 0 asymptotically stable? Semi-stable?

Unstable? Repeat for the differential equation$\frac{\mathrm{dx}}{\mathrm{dt}}=-x^{\mathrm{n}}$

Determine the value of n for xn

Consider

For n is odd xⁿ < 0 for x < 0 and xⁿ > 0 for x > 0

The phase portrait is

0 is unstable whennis odd

Consider

For n is even xⁿ > 0 for all values of x

The phase portrait is

0 is semi-stable when n is even

Hence, for xⁿ

0 is unstable when nⁿ is odd

0 is semi-stable when n is even

Determine the value of n for -xn

Let xⁿ is replaced by -xⁿ

Consider n is odd then -xⁿ < 0 for x < 0 and -xⁿ > 0 for x > 0

The phase portrait is

0 is stable when n is odd

Let n is even then -xⁿ < 0 for all values of x

The phase portrait is

0 is semi-stable when n is even

Hence, for -xⁿ

0 is stable when n is odd

0 is semi-stable when n is even