Convergence of parameterized differential equations Consider an ordinary differential equation whose coefficients are parameterized by $\epsilon$:
$y'(x) = f_\epsilon(x) y(x) + g_\epsilon(x) $
and use $y_\epsilon(x)$ to denote its solution.
Suppose $f_\epsilon(x)$ and $g_\epsilon(x)$ converges to $f_0(x)$ and $g_0(x)$ pointwise as $\epsilon\to 0$. Is there a theorem that asserts the corresponding solution of the ODE also has such a convergence, i.e. $y_\epsilon(x)\to y_0(x)$ pointwise? 
 A: First of all, you should fix some initial value. But even then, pointwise convergence isn't enough. Consider  the simplest ODE $y'=g_\epsilon(x)$ (i.e., with $f_\epsilon\equiv 0$), where
$$g_\epsilon(x)= \begin{cases} \epsilon^{-3}x, \quad & 0<x<\epsilon \\
2\epsilon^{-2}-\epsilon^{-3}x, \quad & \epsilon \le x<2\epsilon \\
0 & \text{otherwise} \end{cases}$$
Then the solution $y_\epsilon$ with initial value $y_\epsilon(0)=0$ has $y_\epsilon(x)=\epsilon^{-1}$ for $x>\epsilon$. The solutions blow up as $\epsilon\to 0$, even though $y_\epsilon\to 0$ pointwise. 
Uniform convergence would be enough: just write down $y_\epsilon$ using the integral formula for solution of first order linear inhomogeneous equation and pass to the limit in each integral.  
A: Here is a theorem on Page 58 of Theory of Ordinary Differential Equations by Earl Coddington and Norman Levinson (1955). Note: the notations are different than those in the question.
Let $D$ be a domain of $(t, x)$ space, $I_\mu$ the domain $|\mu-\mu_0|<c$, $c>0$, and $D_\mu$ the set of all $(t, x, \mu)$ satisfying $(t,x)\in D$, $\mu\in I_\mu$. Suppose $f$ is a continuous function on $D_\mu$ bounded by a constant $M$ there. For $\mu=\mu_0$ let 
$$x'=f(t,x,\mu)\quad x(\tau)=\xi$$
have a unique solution $\varphi_0$ on the interval $[a,b]$, where $\tau\in[a,b]$. Then there exists a $\delta>0$ such that, for any fixed $\mu$ satisfying $|\mu-\mu_0|<\delta$, every solution $\varphi_\mu$ of the displayed ODE exists over $[a,b]$ and as $\mu\to\mu_0$
$$ \varphi_\mu\to\varphi_0$$
uniformly over $[a,b]$.
