I have 2 Integral equation: $$y(t)= e + \int_1^t y(s) + e^s ds$$ $$z(t)= e + \int_1^t sin z(s) + e^s ds$$
The solution for the first one ist $y(t)= te^t$ Now I want to calculate $$|y(t) -z(t)|= |\int_1^t y(s) + e^s -sin z(s) - e^s ds |\leq \int_1^t |(y(s) -sin z(s) ds| $$ Now I can use the following $| (y(t) -sin z(t) | \leq \frac{|y|^3}{6}$, so I get: $$|y(t) -z(t)|\leq \int_1^t \frac{|y|^3}{6} ds = \int_1^t \frac{|e^tt|^3}{6} ds$$
How can I apply Grönwall's inequality now?
Using $|y-\sin y|\le \frac{|y|^3}6$, you get that $$ |y-\sin z|\le|\sin y-\sin z|+|y-\sin y|\le \min(2,|y-z|)+\frac{|y|^3}6. $$ By the Grönwall inequality, the exact solution of $$ u(t)=\int_1^t\left(u(s)+\frac{s^3e^{3s}}6\right)\,ds\iff u'(t)=u(t)+\frac{t^3e^{3t}}6,~~u(1)=0, $$ will be an upper bound for $|z(t)-y(t)|\le u(t)$. This gives $$ (e^{-t}u(t))'=\frac{t^3e^{2t}}6\implies u(t)=\frac{e^{3t} (t-1)(4t^2-2t+4)}{48}+\frac{e^{3t}-e^{2+t}}{48} $$