Finite limit of function in ODE Given
$$y'(x)=a(x)y(x)+b(x);$$
$y(x_0)=y_0$, $a(x)=k>0$, $b(x)\leq M e^{\epsilon x}$, $0<\epsilon<k$, $M>0$, then $\frac{y(x)}{e^{k x}}$ approaches a finite limit as $x \rightarrow \infty$ with $x_0=0$, $x_2=\infty$.
 A: This is not a full solution (see the comments by @LutzLehmann on the post) but here is what I could prove. Multiply the whole equation by $e^{-\epsilon x}$ to get $$e^{-\epsilon x}y'(x)=e^{-\epsilon x}ky(x)+c(x)$$ where $c(x)=b(x)e^{-\epsilon x}\le M$ as given. The above equation gives $$e^{-\epsilon x}y'(x)-\epsilon e^{-\epsilon x}y(x)=e^{-\epsilon x}(k-\epsilon)y(x)+c(x)$$ so, taking $h(x)=e^{-\epsilon x} y(x)$, the LHS above is $h'(x)$ and the equation can be rewritten as $$h'(x)=(k-\epsilon)h(x)+c(x)$$ Take note that you want $\lim\limits_{x\to\infty}h(x)$ to be finite. $\qquad \qquad \qquad (*)$  Now, from the condition $0<\epsilon < k$, you have $p=k-\epsilon>0$, so you can rewrite the last equation as $$\begin{aligned} h'(x)-ph(x)&=c(x) \\ \implies e^{-px}h'(x)-pe^{-px}h(x)&=e^{-px}c(x) \\ \implies \dfrac{d}{dx}(e^{-px}h(x))&=e^{-px}c(x) \qquad \qquad (1)\end{aligned}$$ Now, the RHS above is bounded above as follows: $$\begin{aligned}e^{-px}c(x)&\le Me^{-px}\ (\because c(x)\le M \text{ from above})\\ 
\implies \int_{x_0}^{\infty} e^{-px}c(x)dx &\le \int_{x_0}^{\infty} Me^{-px}dx \\
\implies \int_{x_0}^{\infty} \dfrac{d}{dx}(e^{-px}h(x))dx & \le M\left[-\dfrac{e^{-px}}p\right]_{x_0}^{\infty} \ (\text{from }(1))\\
\implies \left[e^{-px}h(x)\right]_{x_0}^{\infty} & \le M\left(0+\dfrac{e^{-px_0}}p\right) \\ 
\implies \lim_{x\to\infty}(e^{-px}h(x))-h(0) & \le \dfrac{M}p \ (\because x_0=0) \\
\implies \lim_{x\to\infty}(e^{-px}h(x)) & \le \dfrac{M}p+y \ \left(\because h(0)=e^{-\epsilon \cdot 0}y(0)=y\right) \qquad (2)\\
\implies \lim_{x\to\infty}(e^{-kx}y(x))&\le \dfrac{M}p+y \qquad \qquad \qquad (3)\end{aligned}$$
We want the last step to have $\epsilon$ instead of $k$, and this is the closest I could get.
A: The solution for $y'(x)$ is
\begin{eqnarray*}
y(x)&=&y_0 e^{\int a(x) dx} + \int_0^{\infty} e^{\int_s^x a(u)du} b(s) ds \\
&=& y_0 e^{\int k dx} + \int_0^{\infty} e^{\int_s^x k du} b(s) ds, \textrm{ since } a(x)=k \\
&=& y_0 e^{kx} + \int_0^x e^{kx-ks} b(s) ds \\
&=& e^{kx} \left( y_0 +  \int_0^{\infty} e^{-ks} b(s) ds   \right) \\
&\leq& e^{kx} \left( y_0 +  \int_0^{\infty} e^{-ks} Me^{\epsilon s} ds   \right), \textrm{ since } b(x)\leq Me^{\epsilon x} \\
&=& e^{kx} \left( y_0 +  M \int_0^{\infty} e^{(\epsilon-k)s} ds   \right) \\
&=& e^{kx} \left( y_0 +  M \left[ \frac{e^{(\epsilon-k)s}}{\epsilon - k} \right]_0^{\infty}   \right) \\
&=& e^{kx} \left( y_0 - \frac{M}{\epsilon-k} \right) \\
&=& e^{kx} \left( y_0 + \frac{M}{k-\epsilon} \right).
\end{eqnarray*}
Multiplying both sides by $\frac{1}{e^{kx}}$, we have
\begin{eqnarray*}
& & \frac{y(x)}{e^{kx}} \leq \frac{e^{kx}}{e^{kx}} \left( y_0 + \frac{M}{k-\epsilon} \right) \\
& \Rightarrow & \frac{y(x)}{e^{kx}} \leq \left( y_0 + \frac{M}{k-\epsilon} \right).
\end{eqnarray*}
Applying limit as $x \rightarrow \infty$,
\begin{eqnarray*}
\lim_{x \to \infty} \frac{y(x)}{e^{kx}} &\leq& \lim_{x \to \infty} \left( y_0 + \frac{M}{k-\epsilon} \right)\\
&=& y_0 + \frac{M}{k-\epsilon}.
\end{eqnarray*}
Hence, as $ x \to \infty$, $\frac{y(x)}{e^{kx}}$ approaches a finite limit.
