Riccati equations: Does uniform convergence of the coefficients imply uniform convergence of the solution? For all $n \in \mathbb{N}$, I have a Riccati differential equation on $[0,T]$:
\begin{align}\tag{1}
y'(t)=f^n(t) y(t)^2+g^n(t)y(t)+h^n(t), ~~~~~y(T)=y_T \in \mathbb{R}.
\end{align}
Here, $f^n, g^n$ and $h^n$ are $\mathcal{C}^1([0,T];\mathbb{R})$-functions.
Suppose that for every $n$, equation (1) has a unique solution $y^n$; and suppose further that $f^n, g^n, h^n$ converge uniformly on $[0,T]$ to some functions $f, g, h \in \mathcal{C}^1$, and that the 'limit' Riccati equation
$$
y'(t) = f(t)y(t)^2+g(t)y(t)+h(t), ~~~~~y(T) = y_T
$$
also has a unique $\mathcal{C}^1$-solution $y^\infty$.
My question: Do the $y^n$ converge uniformly to $y^\infty$ as $n \to \infty$?
To be honest, I have no idea how to tackle this problem. 
If the differential equation were linear, I would try to use Gronwall's inequality; but I did not see how I could apply it here. 
Any help is appreciated - maybe someone knows a textbook or paper where this kind of problem has been tackled?
 A: Let me use subscripts instead of superscripts and $y$ instead of $y_\infty$. First of all, we see that $f_n$, $g_n$ and $h_n$ are uniformly bounded.  This implies also that the solutions $y_n$ and $y_\infty$, which are assumed to be defined on $[0,T]$, are also uniformly bounded. Let $M$ be a common bound of all the functions involved. Integrating the equation for $y_n$ and $y_\infty$ and subtracting we get
$$\begin{multline}
|y(t)-y_n(t)|\le\int_0^t|f(s)y(s)^2-f^n(s)y_n(s)^2|\,ds\\+\int_0^t|g(s)y(s)-g^n(s)y_n(s)|\,ds+\int_0^t|h(s)-h_n(s)|\,ds.
\end{multline}$$
Let's treat the first integral, which the most difficult.
$$\begin{align}
\int_0^t|f(s)y(s)^2-f^n(s)y_n(s)^2|\,ds&\le\int_0^t|f(s)|\,|y(s)^2-y_n(s)^2|\,ds+\int_0^t|f(s)-f^n(s)|\,|y_n(s)|\,ds\\
&\le M\int_0^t|y(s)^2-y_n(s)^2|\,ds+M\int_0^t|f(s)-f^n(s)|\,ds\\
&\le2\,M^2\int_0^t|y(s)-y_n(s)|\,ds+M\int_0^t|f(s)-f^n(s)|\,ds.
\end{align}$$
 By uniform convergence, given $\epsilon>0$ we get that for all $n$ sufficiently large
$$
|y(t)-y_n(t)|\le K\int_0^t|y(s)-y_n(s)|\,ds+\epsilon
$$
where $K$ is a constant independent of $n$. Now use Gronwall's inequality.
