On the uniform convergence of a sequence of functions I have $f_n$ solution of a Cauchy problem :$y'=y^2t,y(1)=(2^{1/n})$.
I have to study the uniform convergence.
I found as solution :$f_n=\frac{2}{1+2*2^{-1/n}-t^2}$. The pointwise limit is $f(t)=\frac{2}{3-t^2}$. But for the uniform convergence? 
 A: To examine uniform convergence, we need to consider specific sets $D$ where $t \in D \subset \mathbb{R}$.  Clearly $D \neq \mathbb{R}$. While the pointwise limit can be written as $f(t)= 2/(3-t^2)$, the domain of $f$ cannot include $\pm\sqrt{3}$ where the sequence diverges to $+\infty$. 
There also are points where $1 +2/2^{1/n} - t^2 = 0$ for specific values of $n$ and $f_n(t)$ is undefined. With $t$ fixed this problem ceases as $n$ becomes large, but it should be addressed in determining uniform convergence.
Note that for $t^2 \neq 3$ and excluding points where $f_n(t)$ is not defined we have 
$$|f_n(t) - f(t)| = \left|\frac{2}{1 + 2/2^{1/n} - t^2} - \frac{2}{3 - t^2} \right| = \frac{4|1 - 1/2^{1/n}|}{|1 - 2/2^{1/n} - t^2||3 - t^2|}$$.
Consider first $D_{1} = [-\sqrt{2},\sqrt{2}]$, where $1 \leqslant 3 - t^2 \leqslant 3$ and $\sqrt{2} - 1 \leqslant 1 +2/2^{1/n} - t^2 < 3$ for $n > 1$.
For all $t \in D_1$ we have
$$|f_n(t) - f(t)| < \frac{4(1 - 1/2^{1/n})}{\sqrt{2} -1} .$$
Since the right-hand side converges to $0$ and does not depend on $t$, we have uniform convergence on $D_1 = [-\sqrt{2}, \sqrt{2}]$.
For $D_\alpha = \{t:t^2 \geqslant 3 + \alpha\}$ with $\alpha > 0$, we have $3 - t^2 \leqslant -\alpha$ and $1 +2/2^{1/n} - t^2 < -\alpha$. 
Hence, for all $t \in D_\alpha$ we have
$$|f_n(t) - f(t)| < \frac{4(1 - 1/2^{1/n})}{\alpha^2} .$$
Again, since the right-hand side converges to $0$ and does not depend on $t$, we have uniform convergence on $D_\alpha =(-\infty, -\sqrt{3 + \alpha}] \cup [\sqrt{3 + \alpha}, \infty)$ for any $\alpha > 0.$
Problems with uniform convergence arise for $D_3 = [-\sqrt{3}, -\sqrt{2}) \cup ( \sqrt{2}, \sqrt{3}]$.
See if you can finish by showing the sequence is not uniformly convergent on at least part of this interval (depending how you handle points where $f_n(t)$ is undefined.) The best way to show this is to find  a sequence $t_n \in D_3$ where 
$$\lim_{n \to \infty} |f_n(t_n) - f(t_n)| \neq 0.$$
