Perturbed differential equation Given a differential equation $y'(t)=f(t,y(t))$, where f satisfies the condition $(u-v)(f(t,u)-f(t,v))\le0$ for all $u$ and $v$. Suppose $W$ satisfies a perturbed differential equation $W'=f(t,W(t))+r(t)$ for $t\ge0$. I need to show that $|U(t)-W(t)|\le|U(0)-W(0)|+\int_0^t|r(s)|ds$. I think that to show $|U(t)-W(t)|\le|U(0)-W(0)|+\int_0^t|r(s)|ds$, I can first show that $|U(t)-W(t)|\le|U(0)-W(0)|$ and then show that $\int_0^t|r(s)|ds\ge0$. Am I on the right track?

$$\begin{align} (U-W)(U'-W')&=(U-W)\bigl((f(y,U)-f(t,W)-r(t)\bigr)\\\ &=(U-W)(f(y,U)-f(t,W))-(U-W)\,r(t)\\\ &\le-(U-W)\,r(t) \end{align}$$ From here $$ |U'-W'|\le r(t) $$ and $$\begin{align} |U(t)-W(t)|&\le|U(0)-W(0)|+\int_0^t|U'(s)-W'(s)|\,ds\\\ &\le|U(0)-W(0)|+\int_0^t|r(s)|\,ds. \end{align}$$