How do I compute the maximum value of the solution of this differential equation?

Solution 1:

Here is my approach to show that $y_{\infty}=\bar{y}$.

For the sake of contradiction assume that $y_{\infty}<\bar{y}$. Because $y(t)$ is a solution to your IVP we must have for any $T>0$ that $$\int_{y(0)}^{y(T)}\frac{\mathrm{d}x}{1+k\epsilon-kx-e^{-x}}=\int_0^T\mathrm{d}t$$ This implies for any $T>0$ $$\int_{\epsilon}^{y(T)}\frac{\mathrm{d}x}{1+k\epsilon-kx-e^{-x}}=T$$ Take $T\mapsto +\infty$ and get $$\int_{\epsilon}^{y_{\infty}}\frac{\mathrm{d}x}{1+k\epsilon-kx-e^{-x}}=+\infty$$ This is a contradiction since $x\mapsto \frac{1}{1+k\epsilon-kx-e^{-x}}$ is continuous on $[\epsilon,y_{\infty}]$.