# Attempting to prove that the linear solution to an ODE is the unique solution

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

• In case the right hand side is lipschitz continous, maybe you can apply Piccard Lindelöfs threorem? Although I am not sure here, because you have a square in $y'$, so you might end up with a square root after rearanging, which is not lipschitz continous. Commented Oct 13, 2023 at 21:29
• My understanding is that Piccard Lindelof's theorem would prove that there is a unique solution for a given initial condition, whereas I want to show that the only solution that holds globally is the linear solution (which has no free parameters). Commented Oct 13, 2023 at 21:36
• This is a d'Alembert's equation, although if you only want linear solutions you could substitute $y=mx+c$ and solve for $m,c$. Commented Oct 13, 2023 at 21:40

$$y(x)=mx+c$$ 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}