Fix $a>0$ and $b>0$. Does the following ODE
\begin{equation}
 G(x)^2+2axG(x)G'(x)+2aG'(x)(x-b)=0 \tag{\*}
\end{equation}
have a solution, say, $F(x)$, that satisfies $F(x)>0$ and $F'(x)<0$ on $(b,\infty)$?


I tried to solve it by Mathematica, and it gives $G$ as solution to
\begin{equation} x=e^{-2a\left(\text{log}\,G(x)-\frac{1}{G(x)}\right)}\left(C\_0+b(2a)^{2a}\text{Gamma}\left(1-2a,\frac{2a}{G(x)}\right)\right) \tag{\*\*}
\end{equation}


for some free parameter $C\_0$, where Gamma$(\cdot,\cdot)$ is the upper incomplete gamma function. However I'm not sure if this is a proper inverse function or if $G$ given by this equation has domain $(b,\infty)$.


Thanks!!



A first order ODE problem
The answer is yes: indeed, for any $a>0$ and $b>0$, your equation $(\*\*)$ with any $C\_0>0$ determines  

a solution $G$ to your ODE $(\*)$ such that $G>0$ and $G'<0$ on $(b,\infty)$ -- and even on $(0,\infty)$. 


Indeed, rewrite $(\*\*)$ as 
\begin{equation}
 x=X(G(x)),\quad X(g):=
 e^{-2a\left(\text{log}\,g-\frac{1}{g}\right)}\left(C\_0+b(2a)^{2a}\text{Gamma}\left(1-2a,\frac{2a}{g}\right)\right). 
\end{equation}
Take any $C\_0>0$. 
Then $X(0+)=\infty$ and $X(\infty-)=0$. Let 
\begin{equation}
 x\_1(g):=X'(g)\,\frac{g^2 e^{2 a \left(\log g-\frac{1}{g}\right)}}{g+1}. 
\end{equation}
Then 
\begin{equation}
 x'\_1(g)=-\frac{2 a b e^{-\frac{2 a}{g}} g^{2 a}}{(g+1)^2}, 
\end{equation}
which is manifestly negative for positive $a,b,g$, so that $X'<0$ on the interval $(0,\infty)$ and  

hence the function $X\colon(0,\infty)\to(0,\infty)$ continuously decreases from $\infty$ to $0$ on $(0,\infty)$. So, the inverse function $G:=X^{-1}$ exists and solves the equation $x=X(G(x))$ for $x>0$. Moreover, $G>0$ and $G'=1/(X'\circ G)<0$, as desired. 


It remains to verify that this function $G$ is a solution to your ODE $(\*)$. To do this, just replace, in $(\*)$, $G(x)$ by $g$ and $G'(x)$ by $1/X'(g)$, and then $x$ by $X(g)$, so that $(\*)$ is rewritten as $g^2 + 2 a X(g) \frac g{X'(g)} + 2 \frac a{X'(g)}\, (X(g) - b)=0$, which is straightforward to check. 


Details of calculations can be seen in [the Mathematica notebook](https://my.pcloud.com/publink/show?code=XZbSXR7ZEjFeVXkzBKpK1YnEb5en2BDhcle7)
and/or [its pdf image](https://my.pcloud.com/publink/show?code=XZjSXR7ZDm1GQhtTgUJrkH75RGVkA4tRfa47). 


