I am working with an ODE of the following form:
$$y\left(x\right)=\frac{x-ay'\left(x\right)}{1+b y'\left(x\right)^{2}},$$
where $a,b>0$.
This equation clearly has a linear solution. I would like to show that this is the only solution. My intuition says this is the case, and I have been able to prove it in the special case in which $a=0$. Though, my proof in the special case in non-trivial and I am not sure it is the ideal approach, so I won't include it here.
Thanks
$y(x)=mx+c$ is not the most general solution, this can be seen by plugging it into the equation and you'll find it only works for certain $m$ and $c$.
As @user10354138 points out, this is a d'Alembert's equation: \begin{align} y=xf(p)+g(p)&;\qquad [y'_x=p]\\\\ f(p)=\frac{1}{1+bp^2},\quad g(p)&=\frac{-ap}{1+bp^2}. \end{align} To solve we will take the Legendre Transformation, $x=Y'_x=P$, $y=XP-Y$, $p=X$ to get \begin{align} XP-Y=f(X)P+g(X), \end{align} which is linear in $X$. Under the integrating factor $E(X)=-\smallint \mathrm dX/(X-f(X))$ we have \begin{align} Y=\frac{1}{E(X)}\left(C+\int\frac{g(X)E(X)}{X-f(X)}\mathrm dX\right). \end{align} Now, take $X$ to be a parameter for our parametric solution, it follows that \begin{align} X(t)=t,\quad Y(t)=\frac{1}{E(t)}\left(C+\int\frac{g(t)E(t)}{t-f(t)}\mathrm dt\right),\quad E(t)=-\int\frac{\mathrm dt}{t-f(t)}. \end{align} In terms of $x$ and $y$ we then have our general solution: \begin{align} x(t)=\frac{Y(t)+g(t)}{t-f(t)},\quad y(t)=\frac{tg(t)-f(t)Y(t)}{t-f(t)}. \end{align}