Here are some proofs I developed for differential inequalities. They may give you some more intuition on how to visualize the answer suggested above. \begin{theorem}
Let $w(t,u)$ be continuous on an open connected set D $\subset$ ${\rm I\!R^{2}}$ and be such that the initial value problem for the scalar equation,
\begin{equation}\label{thm2} u'=w(t,u) \end{equation}
has a unique solution. If $u(t)$ is a solution of (\ref{thm2}) on $a\leq t\leq b$ and $y(t)$ is a solution of __ on $a\leq t\leq b$ with $y(a)\leq u(a)$, then $y(t) \leq u(t)$ for $a\leq t\leq b$.
\end{theorem}
\begin{lemma}
Suppose w(t, u) satisfies the conditions of Theorem 2 such that, $a\leq t\leq b$, with $u\geq 0$, and let $u(t) \geq 0$ be a solution of (\ref{thm2}) on $a\leq t < b$. If $\bff: [a,b) \times \Rn \rightarrow \Rn$ is continuous and
\begin{equation}\label{lemma3} \norm{\bff (t, \bfy )}\leq w(t, \norm{\bfy}), \hspace{.8cm} a\leq t<b, \hspace{.8cm} \bfy \in \Rn , \end{equation}
then the solutions of
\begin{equation} \bfy' = \bff(t,\bfy), \hspace{.80cm} \norm{\bfy (a)|} \leq u(a) \end{equation}
exist on $[a, b)$ and $\norm{ \bfy (t)}\leq u(t)$, with t $\in$ $[a,b).$
\end{lemma}
The following lemma will be used for our differential inequalities. That is, we will use it to help us prove that we can extend our solutions to the boundary values. We will use certain parts of this lemma in our proofs.
Let S be a differentiable function on [a, b].\\
Part A: If S satisfies the differential inequality
\begin{equation}\label{219} S'(t)\leq \lambda S(t), \hspace{.80cm} a\leq t\leq b \end{equation}
where $\lambda$ is a constant and $\lambda >0$, then
\begin{equation}\label{220} S(t)\leq S(a)e^{\lambda (t-a)} \mbox{ for } a\leq t\leq b.\end{equation}
Part B: If S satisfies the differential inequality
\begin{equation}\label{221} S'(t) + \lambda S(t) \leq M_{1} , \hspace{.80cm} a\leq t\leq b.\end{equation}
where $M_{1}>0$ and $\lambda >0$ are constants, then
\begin{equation}\label{222} S(t) \leq \frac{M_{1}}{\lambda} +(S(a) -\frac{M_{1}}{\lambda})e^{\lambda (a-t)} \mbox{ for } a\leq t\leq b. \end{equation}
Part C: If S satisfies the differential inequality
\begin{equation}\label{223} S'(t)\leq (M_{1} + M_{2}e^{\lambda t})S(t), \hspace{.80cm} a\leq t\leq b \end{equation}
where $M_{1}>0, M_{2} >0,$ and $\lambda > 0$ are constants, then
\begin{equation}\label{224} S(t)\leq S(a)e^{M_{1}(t-a)+\frac{M_{2}}{\lambda}(e^{\lambda t} - e^{\lambda a})} \hspace{.1cm} \mbox{ for } a\leq t\leq b.\end{equation}
\end{lemma}
%proof of Lemma 4 part A
We prove each Part of Lemma 4 separately.
\emph{Proof of Part A.}
Let S be a differentiable function on [a,b] where $\lambda$ is a constant and $\lambda >0$.\\
Then for part A, we suppose
\begin{equation}\label{iea} S'(t) \leq \lambda S(t), \hspace{.6cm} a\geq t\geq b .\end{equation}
Rearranging (\ref{iea}) we have,
\begin{equation}\label{mul} 0\geq S'(t) - \lambda S(t) .\end{equation}
Similar to an ordinary differential equation we multiply both sides of (\ref{mul}) by the integrating factor $e^{-\lambda t}$, \cite{Vance}.
This gives us,
\begin{equation}\label{rhs5} 0 \geq S'(t)e^{-\lambda t} - \lambda S(t) e^{-\lambda t} \end{equation}
Notice that the right hand side of (\ref{rhs5}) is the derivative of the product
\begin{equation*} S(t)e^{-\lambda t}. \end{equation*}
Hence we can say,
\begin{equation}\label{hence} 0 \geq S'(t)e^{-\lambda t} - \lambda S(t)e^{-\lambda t}=\frac{d}{dt}(S(t)e^{-\lambda t})
\end{equation}
We can now form the definite integral from $a$ to $t$ on each side of (\ref{hence}) to obtain
\begin{equation} \int_{a}^{t} 0\geq \int_{a}^{t} \frac{d}{dt}(S(t)e^{-\lambda t})dt. \end{equation}
Evaluating both integrands yields,
\begin{equation} 0 \geq S(t)e^{-\lambda t} - S(a)e^{-\lambda a} .\end{equation}
%Solving for $S(t)$, we factor out $e^{-\lambda}$ which gives
%\begin{equation}\label{56} 0 \geq e^{-\lambda}(S(t)e^{t} - S(a)e^{a}) .\end{equation}
Isolating $S(t)$ offers the desired result,
\begin{equation} S(t)\leq S(a) e^{\lambda (t-a)}.\end{equation}
This completes the proof of Part A. \\
% Part B
\emph{Proof of Part B.}\\
Let S be a differentiable function on [a,b] where $\lambda >0$ and $M_{1} >0$ are constants.
Then for part B, we suppose
\begin{equation} S'(t)+\lambda S(t) \leq M_{1}, \hspace{.6cm} a\geq t\geq b, \hspace{.3cm} \mbox{is satisfied.}\end{equation}
Rearranging we have,
\begin{equation}\label{moo} 0 \geq [S'(t)+\lambda S(t) - M_{1}] .\end{equation}
Again, this is similar to an ODE, so we multiply both sides of the above by the integrating factor $e^{\lambda t}$.
This gives us,
\begin{equation}\label{pab} 0 \geq e^{\lambda t}[S'(t) + \lambda S(t) - M_{1}]. \end{equation}
Note that the right hand side of (\ref{pab}) is
\begin{equation}\label{mom2} \frac{d}{dt} [S(t)e^{\lambda t} - \frac{M_{1}}{\lambda}e^{\lambda t}]. \end{equation}
Thus we can rewrite (\ref{pab}) as
\begin{equation}\label{keira} 0 \geq \frac{d}{dt} [S(t)e^{\lambda t} - \frac{M_{1}}{\lambda}e^{\lambda t}].\end{equation}
We now form the definite integral from $a$ to $t$ on each side of (\ref{keira}),
\begin{equation}\label{99} \int_{a}^{t} 0 \geq \int_{a}^{t}\frac{d}{dt} [S(t)e^{\lambda t} - \frac{M_{1}}{\lambda}e^{\lambda t}]dt. \end{equation}
Evaluating the definite integrals in (\ref{99}), we obtain,
\begin{equation}\label{24} 0 \geq (S(t) -\frac{M_{1}}{\lambda})e^{\lambda t} - (S(a)-\frac{M_{1}}{\lambda})e^{\lambda a}.\end{equation}
%Simplifying (\ref{24}) have,
%\begin{equation} (S(t)-\frac{M_{1}}{\lambda})e^{\lambda t} \leq (S(a)-\frac{M_{1}}{\lambda})e^{\lambda a}\end{equation}
Isolating $S(t)$ gives the desired result,
\begin{equation} S(t)\leq \frac{M_{1}}{\lambda}+(S(a)-\frac{M_{1}}{\lambda})e^{\lambda (a-t)}. \end{equation}
This completes the proof of part B.
% Part C
\emph{Proof of Part C.}\\
Let S be a differentiable function on [a,b] where $\lambda >0$, $M_{1} >0$, and $M_{2}>0$ are constants.
Then for part C, we suppose,
\begin{equation}\label{333} S'(t) \leq (M_{1} + M_{2}e^{\lambda t})S(t), \hspace{.60cm} a\leq t\leq b \hspace{.3cm} \mbox{is satisfied.} \end{equation}
Rearranging (\ref{333}) we have,
\begin{equation}\label{alice} 0 \geq S'(t)-(M_{1}+M_{2}e^{\lambda t})S(t).\end{equation}
Multiplying both sides of (\ref{alice}) by the integrating factor $e^{-(M_{1}t+ \frac{M_{2}}{\lambda} e^{\lambda t})}$ we have,
\begin{equation}\label{chain} 0\geq e^{-{(M_{1}t+ \frac{M_{2}}{\lambda} e^{\lambda t)}}}[S'(t) -(M_{1} + M_{2}e^{\lambda t})S(t)] \end{equation}
Note that the right hand side of (\ref{chain}) is the derivative of the product
\begin{equation}\label{deriv} [ e^{-{(M_{1} t+\frac{M_{2}}{\lambda}e^{\lambda t})}}S(t)] .\end{equation}
Thus we can rewrite (\ref{chain}) as
\begin{equation}\label{408} 0 \geq \frac{d}{dt} [ e^{-{(M_{1} t+\frac{M_{2}}{\lambda}e^{\lambda t})}}S(t) ] .\end{equation}
We now form the definite integral from $a$ to $t$ on each side of (\ref{408}) to obtain
\begin{equation}\label{408} \int_{a}^{t} 0 \geq \int_{a}^{t}\frac{d}{dt} [ e^{-{(M_{1} t+\frac{M_{2}}{\lambda}e^{\lambda t})}}S(t)]dt\ .\end{equation}
Evaluating both integrands gives us
\begin{equation} 0 \geq [e^{-{(M_{1} t+\frac{M_{2}}{\lambda}e^{\lambda t})}}S(t)] - [e^{-{(M_{1} a+\frac{M_{2}}{\lambda}e^{\lambda a})}}S(a)] .\end{equation}
Isolating $S(t)$ offers the desired result,
\begin{equation}\label{partbb} S(t) \leq S(a)e^{M_{1}(t-a)+\frac{M_{2}}{\lambda}(e^{\lambda t}- e^{\lambda a})}. \end{equation}
This completes the proof of Lemma 4 Part C. \\
