Answers to Exercises for Chapter 3

1. $$\dot x = x^3-7x+6$$ 3 fixed points, $x=1,2,-3$. 1 is stable, 2 and -3 are unstable.
2. $$ \dot x = 3x^4 -2x^3 - 3 x^2 -1=f(x).$$ To find fixed points, consider $$f'(x)=12x^3-6x^2-6x=6x(2x^2-x-1)$$ so there are three stationary points, at x=0 (max), 1 (min), -1/2 (min), where $f=-1, -3, -27/16$ so there are only two fixed points, near -1 (stable) and 2 (unstable).
3. $$\dot x =e^x-x^3 =f(x). $$ For $x<0$ both terms are positive so there are no fixed points there. $ f(0)=1, \ f(1)=e-1>0, \ f(3)=e^3-3^3<0, \ f(5)>0 $. There are two fixed points. The smaller one is stable, the larger one unstable.

The fixed points of $$\dot x = 1+\cos x =f(x) $$ are at $ x=\pi +2n\pi $. Since $ f(x)>0 $ for all other x, all solutions approach one of the fixed points as $ t \to \infty$, even though all the fixed points are unstable.

For the differential equation $$\dot x = x^2-t, $$ $ \dot x = 0$ on the curve $ t=x^2$ so the direction field has horizontal line segments there. Inside this parabolic curve $ x $ decreases. For the initial condition $x(0)=0$, $x$ decreases and approaches the curve $x(t) = -\sqrt{t} $ for large $t$.

1. $$ \dot x = x^2 - 5x +7 > 0 $$ so there are no fixed points and all solutions tend to infinity.
2. For $$ \dot x = -|x| $$ there is a fixed point at 0. Positive initial conditions approach this point. Negative initial conditions approach -infinity. We can't use the usual stability rule because the function is not differentiable, but the fixed point is unstable.
3. Exact straight line solutions are $x=0$ and $x=1+t$. Near the first of these, solutions approach it for $t>0$ so it can be regarded as stable. Near the other, solutions move away from it. So all initial conditions below 1 approach 0, and all solutions with initial condition greater than 1 approach infinity.