ODE's: study of the case $x'=f(x/t)$ with $f:\Bbb{R}\to \Bbb{R}$ a $C^1$ function with $f(r)=r$ for some $r\in \Bbb{R}$... (Exercise 6, Chap. 1, from Sotomayor's ODE Lessons): Let $f:\Bbb{R}\to \Bbb{R}$ be a $C^1$ function and $r\in \Bbb{R}$ such that $f(r)=r$. Show that
a) if $f'(r)<1$, then no solution of the equation
$$x'=f\left(\frac{x}{t}\right)\,\,\,\,(*)$$
is tangent at $0$ to the solution $\varphi(t)=rt$.
b) if $f'(r)>1$, then there are infinitely many solutions of $(*)$ tangent to $\varphi(t)=rt$ at the origin.
Two functions $\varphi$ and $\psi$ defined for $t>0$ are said to be tangent at $0$ if $\lim\limits_{t\to 0}\dfrac{\psi(t)-\varphi(t)}{t}=0$.
There is a duplicate for this question here: If $f$ continuous differentiable and $f'(r) < 1,$ then $x'=f(x/t)$ has no other solution tangent at zero to $\phi(t)=rt$ but only the answer to item a) is given there and, furthermore, I could not understand it...
 A: I know this question is old, but since i'm studying the subject, i'll leave my answer here for future students like me who struggled with this problem.
First, for part a) i'll rewrite the answer given before, hopefully in a way that's easier to understand.
a) Suppose that $\phi$ is a solution tangent to $rt$ and not equal to it. Since solution curves do not cross, either:
(i) $\phi(t) > rt$ for all $t>0$, or
(ii) $\phi(t) < rt$ for all $t>0$.
I will consider (i), the other case being similar.
from our assumption that the solutions are tangent:
$$
\lim_{t \rightarrow 0^+} \frac{\phi(t) - \varphi(t)}{t} = 0 \Rightarrow \lim_{t \rightarrow 0^+} \frac{\phi(t)}{t} = r
$$
Now we look at the linearization of $f$ around $r$ and apply it to $x = \frac{\phi(t)}{t}$ to get
$$
f(\frac{\phi(t)}{t}) = r + f'(r)(\frac{\phi(t)}{t} - r) + \mathcal{O}(\frac{\phi(t)}{t} - r)
$$
Evaluating $(\frac{\phi}{t})'$
$$
(\frac{\phi(t)}{t})' = \frac{\phi'(t)t-\phi(t)}{t^2} = \frac{f(\frac{x}{t}) - \frac{\phi(t)}{t}}{t}
$$
Applying the linearization
$$
(\frac{\phi(t)}{t})' = \frac{f(\frac{\phi(t)}{t}) - \frac{\phi(t)}{t}}{t} = \frac{(f'(r) - 1)(\frac{\phi(t)}{t} -r ) + \mathcal{O}(\frac{\phi(t)}{t} - r)}{t}
$$
which is negative for small values of $t$, that means $\frac{\phi(t)}{t}$ is decreasing close to zero. A decreasing function has to increase in reverse time. That contradicts our assumpition that it converges to $r$ because we assumed $\phi(t)>rt$ for all $t>0$. The other case gives us a similar contradiction.
Now, for the second part:
b) Suposse $\phi$ is a different solution to the equation. We again, have the same two cases.(i'll again consider case (i)) Use the linearization of $f$ around $r$ applied to $x = \frac{\phi(t)}{t}$ to get:
$$
f(\frac{\phi(t)}{t}) = r + f'(r)(\frac{\phi(t)}{t} - r) + \mathcal{O}(\frac{\phi(t)}{t} - r)
$$
and hence
$$
(\frac{\phi(t)}{t})' = \frac{f(\frac{\phi(t)}{t}) - \frac{\phi(t)}{t}}{t} = \frac{(f'(r) - 1)(\frac{\phi(t)}{t} -r ) + \mathcal{O}(\frac{\phi(t)}{t} - r)}{t}
$$
Which is positive for small $t$. This means $\frac{\phi(t)}{t}$ is increasing close to zero. That means it decreases in reverse time, we then get a decreasing function bounded bellow by $r$, that means $\frac{\phi(t)}{t} \rightarrow r$ as $t \rightarrow 0^+$.
This only tells us that if another solution exists, then it has to be tangent to $rt$ at $0$. But we can create infinitely many solutions using the theorem of existence and uniqueness, by giving the equation inicial conditions.
