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.
 A: 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 $C = \pm \dfrac{1}{2}$.
A: Try solving for $x$ in terms of $y$ instead, to avoid fractions. If $x'$ means $dx/dy$, then we have 
$$x'+\left(x-x'y\right)^{2}+1=0$$
Differentiating, 
$$x''-2\left(x-x'y\right)(x''y)=0$$
$$x''(1-2xy+2x'y^2)=0$$
Now at any point, either the second derivative $x''$ is $0$ or $1-2xy+2x'y^2=0$. Away from $y=0$, the latter equation is equivalent to $x'=\frac{x}{y}-\frac{1}{2y^2}$ which is first order linear.
If $x''=0$ in a neighborhood, then in that neighborhood the solution is a straight line, and @muzzlator's answer puts constraints on the coefficients.
If $x'=\frac{x}{y}-\frac{1}{2y^2}$ in some neighborhood, then we can combine this with the original equation to get 
$$\frac{x}{y}-\frac{1}{2y^2}+\left(x-\left(\frac{x}{y}-\frac{1}{2y^2}\right)y\right)^{2}+1=0$$
$$\frac{x}{y}-\frac{1}{2y^2}+\frac{1}{4y^2}+1=0$$
$$x-\frac{1}{4y}+y=0$$
$$x=\frac{1}{4y}-y$$
which also checks against the original equation. The full solution set consists of all the lines from @muzzlator's answer, spliced in all possible differentiable ways with the curve $x=\frac{1}{4y}-y$.
A: By inspection, we can see that $y=-x$ is one of the solutions because, $y'=-1$ which implies
$$\frac{1}{y^{\prime}}+\left(x-\frac{y}{y^{\prime}}\right)^{2}+1
=-1+(x-x)^2+1=0.$$
