Solve $yy' = \sqrt{y^2+y'^2}y''-y'y''$ Solve $$yy' = \sqrt{y^2+y'^2}y''-y'y''$$
First I set $p = y'$ and $p' = \frac{dp}{dy}p$ to form:
$$yp=\sqrt{y^2+p^2}\frac{dp}{dy}p-p\frac{dp}{dy}p \rightarrow y=\sqrt{y^2+p^2}\frac{dp}{dy}-\frac{dp}{dy}p$$
I am trying to come up with a clever substitution to deal with the square root and $p$ of :
$$y=\frac{dp}{dy}(\sqrt{y^2+p^2}-p)$$
 A: After OP's work
$$y=\frac{dp}{dy}(\sqrt{y^2+p^2}-p)$$
This is a homogeneous equation for $p(y)$
$$\frac{dp}{dy}=\frac{y}{\sqrt{y^2+p^2}-p}=\frac{\sqrt{y^2+p^2}+p}{y}$$
Let $p=vy$, then get
$$v+y\frac{dv}{dy}=\sqrt{1+v^2}+v \implies \int \frac{dv}{\sqrt{1+v^2}}= \int \frac{dy}{y}$$
Let $v=\tan t$, thenwe get
$$\ln(\sec t+ \tan t)=\ln cy \implies cy=\sqrt{1+v^2}+v \implies c^2y^2-1=2cp$$
$$\implies \frac{dy}{dx}=\frac{c^2y^2-1}{2c}\implies \int\frac{2cdy}{c^2y^2-1}=\int dx+b$$
$$\implies \ln \left(\frac{cy-1}{cy+1} \right)=x+b \implies y=\frac{1}{c}\left(\frac{1+ae^x}{1-ae^x}\right)$$
A: Let $y=e^z$ which makes the equation to be
$$z'+(z'-\sqrt{z'^2+1}) \left(z''+z'^2\right)
   =0\implies z''=z'\sqrt{1+z'^2}$$ Reduction of order $(p=z')$ leads to
$$p'=p\sqrt{1+p^2}$$ Switching variables gives
$$x'=\frac 1 {p\sqrt{1+p^2}}\implies x+c_1=\log (p)-\log \left(1+\sqrt{1+p^2}\right)$$ Solving for $p$
$$p=\frac{2 e^{(x+c_1)}}{1-e^{2(x+c_1)}}=-\text{csch}(x+c_1)$$
Half-angle substitution leads to
$$z=\log \left(\coth \left(\frac{x+c_1}{2}\right)\right)+c_2$$ Back to $y$
$$y=c_3 \coth \left(\frac{x+c_1}{2}\right)$$
Hoping no mistake !
A: Note that
$$
(y^2+y'^2)'=2y'(y''+y)=2\sqrt{y^2+y'^2}y''
$$
so that separation and integrating once gives
$$
y'+C=\sqrt{y^2+y'^2}
$$
If $C=0$ this only has the zero solution. For $C\ne 0$ square it and simplify
$$
2Cy'+C^2=y^2\implies \frac{y'}{C+y}+\frac{y'}{C-y}=-1,
$$
if $y$ is not a constant function with value $\pm C$. Otherwise this integrates to
$$
\ln|y-C|-\ln|y+C|=x+d\implies \frac{y-C}{y+C}=De^x\implies y(x)=C\frac{1+De^x}{1-De^x}.
$$
Depending on the sign of $D$, this can be expressed via the hyperbolic tangent or cotangent.
