Proving a solution to ODE exists without solving it I am analysing the ODE
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\ln y(x)+\frac{c}{y(x)}\right)=1,\quad y(1)=1.$$
I would like to show, without attempting to solve the ODE, that $\exists$ a solution $y(x)>c$ if $\ln c-c<-2$. I am wondering what techniques I should employ to do so? I attempted to stuff it into a form involving Grönwall's inequality, but have not had much progress with that. What should I do? Thanks!
 A: Your equation is equivalent to $\log y+\frac{c}{y}=x+\lambda$ for some $\lambda\in\mathbb{R}$. Plugging $x=1$ in the above equality gives $\lambda=c-1$. Thus your differential system is equivalent to $-\frac{c}{y}e^{-\frac{c}{y}}=-ce^{1-c-x}$. Let $\varphi(x)=xe^x$, then the above equation can be written as $\varphi\left(-\frac{c}{y(x)}\right)=-ce^{1-c-x}$.  But a quick study of $\varphi'$ shows that $\varphi$ is a bijection from $(-1,+\infty)$ to $\left(-\frac{1}{e},+\infty\right)$, therefore there exists a solution to your initial equation such that $y>c$ iff $-ce^{1-c-x}<-\frac{1}{e}$ for all $x\geqslant 0$ (is suppose that $y$ is defined on $[0,+\infty)$), which is equivalent to $\log c-c<x-2$ for all $x\geqslant 0$, and this is equivalent to $\log c-c<-2$. Therefore if $\log c-c<-2$, let $y(x)=-\frac{c}{\varphi^{-1}(-ce^{1-c-x})}$, then $y$ is well defined because of what said above and since $\varphi^{-1}>-1$ and $\varphi^{-1}(-ce^{-c})=\varphi^{-1}\circ\varphi(-c)=-c$, we have $y>c$ and $y(1)=1$ and $y$ is solution of the differentiel equation.
I don't know if this solution suits you since I found the solution of the differential equation, but we can still prove the existence without writing what $y$ is by ignoring what I wrote from "Therefore if...".
A: Your ODE is of first order:

$\frac{dy}{dx}=\frac{y^2}{y-c}, y(1)=1$

Can you take from here using Picard's existence theorem?
