<s>[INST] Solving the nonlinear ODE: $ \frac{1}{y^{\prime}}+\left(x-\frac{y}{y^{\prime}}\right)^{2}+1=0$ $$\frac{1}{y^{\prime}}+\left(x-\frac{y}{y^{\prime}}\right)^{2}+1=0$$
I am trying to solve this non-linear ODE and have tried all sorts of substitutions; any hints on how I should progress?
Thanks.
 [/INST] Let $y = a x + b$, then $y' = a$, and you have the equation:
$$\begin{align} \dfrac{1}{a} + \left(x - x - \dfrac{b}{a}\right)^2 + 1 &  =  0\\ \dfrac{1}{a} + \frac{b^2}{a^2} + 1 &  =  0 \\ a + b^2 + a^2 &  =  0 \\ \left(a + \frac{1}{2}\right)^2 + b^2 & =  \left(\frac{1}{2}\right)^2 \end{align} $$
So the solutions are lines with coefficients lying on a circle
To quickly check if there are any more solutions, we may wish to differentiate the equation and see if $y'' = 0$.  Multiplying first by $(y')^2$, we have
$$(1 - 2xy)y' + (x^2 + 1)(y')^2 + y^2 = 0 $$ 
Differentiating,
$$(-2y)y' - 2x (y')^2 + (1 - 2xy) y'' + 2x(y')^2 + 2(x^2 + 1)y' y'' + 2y y' = 0 $$
Simplifying,
$$(1 - 2xy)y'' + 2(x^2 + 1)y' y'' = 0 $$
Leaving us with
$$y'' = 0$$ or $$1 - 2xy + 2(x^2 + 1)y' = 0 $$
In the second case, we have an ODE which we can solve with an integrating factor
$$y' - \frac{x}{x^2 + 1} y = \frac{-1}{2(x^2 + 1)}$$
This gives us equations of the form $$y = -\frac{x}{2} + C \sqrt{1 + x^2}$$
Substituting this back into the original equation, we see this has a solution when $4 C^2 = 1$, or when $C = \pm \dfrac{1}{2}$.
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