I have to explain a student two exercises but I don't know how continue at the first and how start at the second one.
- Let $f(x)$ continously differentiable and such that $f(0)=0$. Show that the solutions of $x'=f(x)h(t)$ cannot change sign.
My attemp: As $f$ is cont. diff. we know that $f$ is continous and also $\frac{\partial{f}}{\partial{x}}=f'(x)$, then for the existence and unicity theorem we have a unique solution $x(t)$. Since $f(0)=0$ we have that $t=0$ is an extreme value for $x(t)$. On a graph, I can see that $x(t)$ cannot change sign independent of the extreme value, and I know that we have to use the product $f(x)h(t)$ and law of signs but I can see how.
- Find the limits as $t\rightarrow \pm \infty$, of the solution $\phi(t)$ of the ivp $x'=(x+2)(1-x^4)$, with $x(0)=0$.
I'm lost at this one
The first idea that came out to me are using algebra we can put: $x'=(x+2)(1+x)(1-x)(1+x^2)$ via partial fractions we can get an antiderivative and then estimate that limits, but this appears a lot of unnecessary work. There's an elegant way to get this excersice?
For the second one, simply looking at the sign of $x'$ in different intervals gives you the phase portrait, where you should be able to easily read off where you end up if you start at $x=0$: $$ \longrightarrow \quad -2 \quad \longleftarrow \quad -1 \quad \longrightarrow \quad 1 \quad \longleftarrow $$