Uniqueness of Solution to a Boundary Value Problem Question

Let $f:\mathbb R_+ \to \mathbb R_+$ be a function twice continuously differentiable (with derivative $f'$ and second derivative $f''$), and $a$, and $b$ be parameters in $\mathbb R_+$. Consider the system 

\begin{align}
\dfrac{dx(t)}{dt}=&amp;\; f\left(x(t)\right)-y(t), \\[2ex]
\dfrac{dy(t)}{dt}=&amp;\; ay(t)\left(f'(x(t))-b\right),
\end{align}

with $x(t)\ge 0$ and $y(t)\ge 0$ for all $t$, and  boundary conditions

\begin{equation}
x(0)= x_0, \qquad\text{and}\qquad\lim_{t\to\infty}e^{-bt}x(t)y(t)^{-a}=0.
\end{equation}

By choosing $f$ appropriately it is possible to show that this system, without the initial condition, can have multiple stationary points. I would like to show, that even if that is the case, the following conjecture it true: 


  Conjecture $\;$ Given an $x_0$, there is a unique solution to the system that converges to one of the stationary points.




Why I think the conjecture is true

If $f'(0)&gt;0$ and $f''(x)&lt;0$ for all $x\in\mathbb R_+$, then there exists a unique stationary point and there is a simple proof of uniqueness of the solution given an arbitrary $x_0$ that involves drawing a phase diagram in the space $(x,y)$ and showing that there is a unique saddle path that the solution must be at all times (otherwise the second boundary condition would be violated) and that converges to the stationary point. 

The reason why I think the conjecture is true is because, given an $f$ in $\mathcal C^2$, a similar phase diagram can be drawn. Here is a sketch of an example of a phase diagram with multiple stationary points:



Notice that the arrows flip between regions in a way that, given an $x_0$, there is only one saddle path (the lines in blue) that the solution could follow. My guess is that a proof for the conjecture would involve showing this is always the case.



Background

If $x$ denotes the capital stock, $y$ the consumption level, $f$ is the production function, $a$ the inverse of the intertemporal elasticity of substitution, and $b$ is the discount rate, then this system is describes the equilibrium allocation of a simple economic growth model.
Multiplying of the system equations leads to the equation in the form of
$$\left(f(x(t))-y(t)\right)\dfrac{dy}{dt} = ay(t)\left(f'(x(t))-b\right)\dfrac{dx(t)}{dt}.\tag1$$
Presentation of the function $y(t)$ as the superposition $y(x(t)),$
$$\dfrac{dy}{dx} = \dfrac{y'_t}{x'_t}$$
 allows to eliminate variable $t$ from $(1):$
$$(f(x)-y)\dfrac{dy}{dx} = ay\dfrac{d}{dx}(f(x)-bx).\tag2$$
Then
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+(y-f(x))f'(x)+ay(f'(x)-b)=0,$$
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)y -f(x)f'(x)=0,$$
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)(y-f(x))+a(f'(x)-b)f(x)=0.\tag3$$
Equation $(3)$ allows the substitution
$$z(x)= y-f(x)\tag4,$$
$$\dfrac{dx}{dt}= -z(x(t)),$$
which changes the issue system to the form of
\begin{cases}
\dfrac{dt}{dx}= -\dfrac1z\\[4pt]
z\dfrac{dz}{dx}+((a+1)f'(x)-ab)z+a(f'(x)-b)f(x)=0,
\end{cases}
\begin{cases}
t= -\int\limits_{x_0}^x\dfrac{\,\mathrm d\xi}{z(\xi)}\\[4pt]
\left(\dfrac{dz}{dx}+((a+1)f'(x)-ab)\right)z+a(f'(x)-b)f(x)=0.\tag5
\end{cases}
Easy to see that the variable $t$ has the gaps in the points $x,$ where $z(x)= 0,$ or $y(x)=f(x).$ 

On the other hand, such points presents as the stationary points by the coordinate $t.$

Let us consider behavior of the equation $(5.2)$ at the simple case 
$$f(x)= cx+d,\quad s=h(cx+d).\tag6$$
Then
$$zz'+(a(c-b)+c)z+a(c-b)(cx+d)=0,$$
$$\dfrac{dz}{dx}=\dfrac{dz}{ds}\dfrac{ds}{dx}= ch\dfrac{dz}{ds},$$
or
$$z\dfrac{dz}{ds}+gz+s=0,\tag7$$
where
$$\quad h = \sqrt{a\left(1-\dfrac bc\right)},\quad a(c-b)=ch^2,\quad  g=h+\dfrac1h.\tag8$$
Equation $(7)$ has the common solution 
$$\begin{cases}
\ln|z^2+gsz+s^2|-\dfrac{2g}{\sqrt{4-g^2}}\arctan
\dfrac{2z+gs}{s\sqrt{4-g^2}}=c,\text{ if }|g|&lt;2,\\
(s\pm z)\exp\dfrac{s}{s\pm z}=C,\text{ if }g=\pm2,\\
C_1|z-gs|^\alpha+C_2|z-gs|^\beta,\text{ if }|g|&gt;2,\tag9
\end{cases}$$
wherein
$$|C_1|+|C_2|&gt;0,\quad \alpha,\ \beta = -\dfrac12\pm\dfrac12\sqrt{g^2-4}.\tag{10}$$
Exact solution $(9)-(10)$ has two branches. The reason is that the equation $(5.2)$ can be written in the form of
$$\dfrac12\dfrac{dz^2}{dx}+((a+1)f'(x)-ab)\sqrt{z^2}+a(f'(x)-b)f(x)=0,$$
which allows existance of multiple solutions.
Multiplying of the system equations leads to the equation in the form of
$$\left(f(x(t))-y(t)\right)\dfrac{dy}{dt} = ay(t)\left(f'(x(t))-b\right)\dfrac{dx(t)}{dt}.\tag1$$
Presentation of the function $y(t)$ as the superposition $y(x(t)),$
$$\dfrac{dy}{dx} = \dfrac{y'_t}{x'_t}$$
 allows to eliminate variable $t$ from $(1):$
$$(f(x)-y)\dfrac{dy}{dx} = ay\dfrac{d}{dx}(f(x)-bx).\tag2$$
Then
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+(y-f(x))f'(x)+ay(f'(x)-b)=0,$$
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)y -f(x)f'(x)=0,$$
$$(y-f(x))\dfrac{d}{dx}(y-f(x))+((a+1)f'(x)-ab)(y-f(x))+a(f'(x)-b)f(x)=0.\tag3$$
Equation $(3)$ allows the substitution
$$z(x)= y-f(x)\tag4,$$
$$\dfrac{dx}{dt}= -z(x(t)),$$
which changes the issue system to the form of
\begin{cases}
\dfrac{dt}{dx}= -\dfrac1z\\[4pt]
z\dfrac{dz}{dx}+((a+1)f'(x)-ab)z+a(f'(x)-b)f(x)=0,
\end{cases}
\begin{cases}
t= -\int\limits_{x_0}^x\dfrac{\,\mathrm d\xi}{z(\xi)}\\[4pt]
\left(\dfrac{dz}{dx}+((a+1)f'(x)-ab)\right)z+a(f'(x)-b)f(x)=0.\tag5
\end{cases}
Easy to see that the variable $t$ has the gaps in the points $x,$ where $z(x)= 0,$ or $y(x)=f(x).$ 

On the other hand, such points presents as the stationary points by the coordinate $t.$

Let us consider behavior of the equation $(5.2)$ at the simple case 
$$f(x)= cx+d,\quad s=h(cx+d).\tag6$$
Then
$$zz'+(a(c-b)+c)z+a(c-b)(cx+d)=0,$$
$$\dfrac{dz}{dx}=\dfrac{dz}{ds}\dfrac{ds}{dx}= ch\dfrac{dz}{ds},$$
or
$$z\dfrac{dz}{ds}+gz+s=0,\tag7$$
where
$$\quad h = \sqrt{a\left(1-\dfrac bc\right)},\quad a(c-b)=ch^2,\quad  g=h+\dfrac1h.\tag8$$
Equation $(7)$ has the common solution 
$$\begin{cases}
\ln|z^2+gsz+s^2|-\dfrac{2g}{\sqrt{4-g^2}}\arctan
\dfrac{2z+gs}{s\sqrt{4-g^2}}=c,\text{ if }|g|&lt;2,\\
(s\pm z)\exp\dfrac{s}{s\pm z}=C,\text{ if }g=\pm2,\\
C_1|z-gs|^\alpha+C_2|z-gs|^\beta,\text{ if }|g|&gt;2,\tag9
\end{cases}$$
wherein
$$|C_1|+|C_2|&gt;0,\quad \alpha,\ \beta = -\dfrac12\pm\dfrac12\sqrt{g^2-4}.\tag{10}$$
Exact solution $(9)-(10)$ has two branches. The reason is that the equation $(5.2)$ can be written in the form of
$$\dfrac12\dfrac{dz^2}{dx}+((a+1)f'(x)-ab)\sqrt{z^2}+a(f'(x)-b)f(x)=0,$$
which allows existance of multiple solutions.