Given $y'(t)=\frac{t}{4}-\frac{1}{y^2+1},y(0)=1$ and let $x(t)=1-\frac{t}{2}$ approx $|x(t)-y(t)|$ Given $y'(t)=\frac{t}{4}-\frac{1}{y^2+1},y(0)=1$ and let $x(t)=1-\frac{t}{2}$
Approximate $|x(t)-y(t)|,|t|<1/2$
I need to do this with $\epsilon$ solution which is corollary of Gronwall thm but I don't know how to use it.
 A: If I understand you right, you apply the decomposition
$$
(y-x)'(t)=[f(t,y(t))-f(t,x(t))]+[f(t,x(t))-x'(t)]
$$
and use it to get the inequality
$$
|(y-x)'(t)|\le L|y(t)-x(t)|+|f(t,x(t))-x'(t)|
$$
which can then be treated via the Grönwall lemma. One can easily conclude $L=1$, using $2|y|\le 1+y^2$, and with some effort $L=0.65$. For the second term,
$$
f(t,x(t))-x'(t)=\frac{t}4-\frac1{1+1-t+\frac{t^2}4}+\frac12=\frac{(1+\frac t2)(1-\frac t2+\frac{t^2}8)-1}{2-t+\frac{t^2}4}
\\
=\frac{-\frac{t^2}8+\frac{t^3}{16}}{2-t+\frac{t^2}4}=-\frac{t^2}{16}\frac{2-t}{2-t+\frac{t^2}4}
$$
so that
$$
|f(t,x(t))-x'(t)|\le \frac{t^2}{16}
$$
for $2-t>0$. Thus by applying Grönwall you get
\begin{align}
e^{-Lt}|y(t)-x(t)|&\le\frac1{16}\int_0^ts^2e^{-Ls}\,ds
\\
&=\frac{6}{16L^4}-e^{-L t} \frac{L^3 t^3 + 3 L^2 t^2 + 6 L t + 6}{16L^4}
\\
|y(t)-x(t)|&\le\frac{3}{8L^4}\left(e^{Lt}-1-Lt-\frac12(Lt)^2- \frac16(Lt)^3\right)
=\frac3{32}t^4+\frac3{160}Lt^5+...
\end{align}
A: First let us find lipschitz constant    $|f(t,y)-f(t,z)|=|1/4-\frac{1}{y^2+1}-1/4+\frac{1}{z^2+1}|\leq|\frac{1}{(z^2+1)(y^2+1)}|\leq 1$ therefore we can choose $L=1$
by the $\epsilon$ solutions $|y(t)-x(t)|\leq|x_0-y_0|e^{L|t-t_0}+2\epsilon\frac{e^{L|t-t_0|}-1}{L}=2\epsilon\frac{e^{L|t-t_0|}-1}{L}\leq2\epsilon(e-1)$.
Let us find $\epsilon$ $|x'(t)-f(t,x(t))|=|1/2-t/4+\frac{1}{x^2}|\leq1/2+\frac{|t|}{4}+|\frac{1}{x^2+1}|\leq1/2+1/8+1=1.625$ therefore we overall got
$\forall t ,|t|<1/2,|y(t)-x(t)|\leq2*1.625(e-1)$.
if anyone can varify this?
