<s>[INST] 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 :
$$y=\frac{dp}{dy}(\sqrt{y^2+p^2}-p)$$
 [/INST] 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
$$\ln(\sec t+ \tan t)=\ln cy \implies cy=\sqrt{1+v^2}+v \implies c^2y^2-1=2cp$$
$$\implies \frac{dy}{dx}=\frac{c^2y^2-1}{2c}\implies \int\frac{2cdy}{c^2y^2-1}=\int dx+b$$
$$\implies \ln \left(\frac{cy-1}{cy+1} \right)=x+b \implies y=\frac{1}{c}\left(\frac{1+ae^x}{1-ae^x}\right)$$
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